{"id":6043,"date":"2023-08-12T05:04:33","date_gmt":"2023-08-12T05:04:33","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6043"},"modified":"2023-08-12T05:04:33","modified_gmt":"2023-08-12T05:04:33","slug":"ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 5 (Arithmetic Progressions)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 5 Arithmetic Progressions <\/strong>Exercise 5.3 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 5 Arithmetic Progressions<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 5.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 5.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 5.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 5.3<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Find the sum of the following APs.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i)<\/strong> 2, 7, 12 ,\u2026., to 10 terms.<br \/>\n<strong>(ii)<\/strong> \u2212 37, \u2212 33, \u2212 29 ,\u2026, to 12 terms<br \/>\n<strong>(iii)<\/strong> 0.6, 1.7, 2.8 ,\u2026\u2026.., to 100 terms<br \/>\n<strong>(iv)<\/strong> 1\/15, 1\/12, 1\/10, \u2026\u2026 , to 11 terms<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) 2, 7, 12 ,\u2026., to 10 terms.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">First term, <em>a<\/em> = 2<br \/>\n<\/span><span style=\"color: #000000;\">Common difference, <em>d<\/em> = <em>a<sub>2<\/sub>\u00a0\u2212\u00a0a<sub>1<\/sub>\u00a0<\/em>= 7 \u2212 2 = 5<br \/>\n<\/span><span style=\"color: #000000;\">n\u00a0= 10<br \/>\n<\/span><span style=\"color: #000000;\">Sum of <em>n<\/em><sup>th<\/sup> term in AP series is,<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\">S<sub>10<\/sub> = 10\/2 [2(2) + (10 -1)\u00d75]<br \/>\n<\/span><span style=\"color: #000000;\">= 5[4 + (9)\u00d7(5)]<br \/>\n<\/span><span style=\"color: #000000;\">= 5 \u00d7 49<br \/>\n= 245<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) \u2212 37, \u2212 33, \u2212 29 ,\u2026, to 12 terms<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">First term, <em>a<\/em> = \u221237<br \/>\n<\/span><span style=\"color: #000000;\">Common difference, <em>d<\/em> = <em>a<sub>2 <\/sub>\u2212\u00a0 a<sub>1<br \/>\n<\/sub><\/em><\/span><span style=\"color: #000000;\">d = (\u221233) \u2212 (\u221237)<br \/>\n<\/span><span style=\"color: #000000;\">d = \u2212 33 + 37<br \/>\nd = 4<br \/>\n<\/span><span style=\"color: #000000;\">n\u00a0= 12<br \/>\nSum of <em>n<\/em><sup>th<\/sup> term in AP series is,<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\">S<sub>12<\/sub> = 12\/2 [2(-37) + (12 &#8211; 1)\u00d74]<br \/>\n<\/span><span style=\"color: #000000;\">= 6[-74 + 11\u00d74]<br \/>\n<\/span><span style=\"color: #000000;\">= 6[-74 + 44]<br \/>\n<\/span><span style=\"color: #000000;\">= 6(-30)<br \/>\n= -180<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 0.6, 1.7, 2.8 ,\u2026\u2026.., to 100 terms<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">First term, <em>a<\/em> = 0.6<br \/>\n<\/span><span style=\"color: #000000;\">Common difference, <em>d\u00a0<\/em>=\u00a0<em>a<sub>2<\/sub>\u00a0\u2212\u00a0a<sub>1<\/sub>\u00a0<\/em>= 1.7 \u2212 0.6 = 1.1<br \/>\n<\/span><span style=\"color: #000000;\">n\u00a0= 100<br \/>\nSum of <em>n<\/em><sup>th<\/sup> term in AP series is,<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>12<\/sub><\/em> = 50\/2 [1.2 + (99)\u00d71.1]<br \/>\n<\/span><span style=\"color: #000000;\">= 50[1.2 + 108.9]<br \/>\n<\/span><span style=\"color: #000000;\">= 50[110.1]<br \/>\n<\/span><span style=\"color: #000000;\">= 5505<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 1\/15, 1\/12, 1\/10, \u2026\u2026 , to 11 terms<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">First term, <em>a<\/em> = 1\/5<br \/>\n<\/span><span style=\"color: #000000;\">Common difference, <em>d<\/em> = <em>a<sub>2\u00a0<\/sub>\u2013 a<sub>1<\/sub>\u00a0<\/em>= (1\/12) &#8211; (1\/5) = 1\/60<br \/>\n<\/span><span style=\"color: #000000;\">n = 11<br \/>\nSum of <em>n<\/em><sup>th<\/sup> term in AP series is,<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><em>S<sub>11<\/sub><\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{11}{2}\\left&amp;space;[&amp;space;2\\times&amp;space;\\frac{1}{15}&amp;space;+&amp;space;\\frac{(11-1)1}{60}\\right&amp;space;]\" alt=\"\\frac{11}{2}\\left [ 2\\times \\frac{1}{15} + \\frac{(11-1)1}{60}\\right ]\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= 11\/2(2\/15 + 10\/60)<br \/>\n<\/span><span style=\"color: #000000;\">= 11\/2 (9\/30)<br \/>\n<\/span><span style=\"color: #000000;\">= 33\/20<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the sums given below<\/strong>:<br \/>\n<strong>(i) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?7&amp;space;+&amp;space;10\\frac{1}{2}&amp;space;+&amp;space;14+&amp;space;......&amp;space;+&amp;space;84\" alt=\"7 + 10\\frac{1}{2} + 14+ ...... + 84\" align=\"absmiddle\" \/><\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>(ii)<\/strong> 34 + 32 + 30 + \u2026\u2026\u2026.. + 10<br \/>\n<strong>(iii)<\/strong> \u2212 5 + (\u2212 8) + (\u2212 11) + \u2026\u2026\u2026\u2026 + (\u2212 230)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?7&amp;space;+&amp;space;10\\frac{1}{2}&amp;space;+&amp;space;14+&amp;space;......&amp;space;+&amp;space;84\" alt=\"7 + 10\\frac{1}{2} + 14+ ...... + 84\" align=\"absmiddle\" \/><br \/>\n<\/strong><\/span><span style=\"color: #000000;\">First term, <em>a\u00a0<\/em>= 7<br \/>\nCommon difference, <em>d = a<sub>2<\/sub> &#8211; a<sub>1<\/sub> = <\/em>10<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> &#8211; 7<br \/>\n<em>d<\/em> = 21\/2 &#8211; 7<br \/>\n<em>d<\/em> = 7\/2<br \/>\n<\/span><span style=\"color: #000000;\">n<sup>th<\/sup>\u00a0term, a<sub>n\u00a0<\/sub>= 84<br \/>\n<\/span><span style=\"color: #000000;\">a<sub>n\u00a0<\/sub>= a(n &#8211; 1)d<br \/>\n<\/span><span style=\"color: #000000;\">84 = 7 + (n \u2013 1) \u00d7 7\/2<br \/>\n<\/span><span style=\"color: #000000;\">77 = (n &#8211; 1) \u00d7 7\/2<br \/>\n<\/span><span style=\"color: #000000;\">22 = n \u2212 1<br \/>\n<\/span><span style=\"color: #000000;\">n\u00a0= 23<br \/>\n<\/span><span style=\"color: #000000;\">We know that the sum of n term is;<br \/>\n<em>S<sub>n<\/sub><\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;l)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + l) \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\"><em>l<\/em> = 84<br \/>\n<em>S<sub>n<\/sub><\/em> = 23\/2 (7 + 84)<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em>\u00a0 = (23 \u00d7 91\/2)<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0 =\u00a0 2093\/2<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0 = 1046 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 34 + 32 + 30 + \u2026\u2026\u2026.. + 10<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">First term, <em>a<\/em> = 34<br \/>\n<\/span><span style=\"color: #000000;\">Common difference, <em>d<\/em> =<em> a<sub>2 <\/sub>\u2212 a<sub>1<\/sub>\u00a0<\/em>= 32 \u2212 34 = \u22122<br \/>\n<\/span><span style=\"color: #000000;\">a<sub>n <\/sub>= <em>l <\/em>=\u00a0 10<br \/>\n<\/span><span style=\"color: #000000;\">a<sub>n <\/sub>= a + (n \u2212 1)d<br \/>\n<\/span><span style=\"color: #000000;\">10 = 34 + (n \u2212 1)(\u22122)<br \/>\n<\/span><span style=\"color: #000000;\">\u221224 = (n\u00a0\u22121)(\u22122)<br \/>\n<\/span><span style=\"color: #000000;\">12 =\u00a0n\u00a0\u22121<br \/>\n<\/span><span style=\"color: #000000;\">n\u00a0= 13<br \/>\n<\/span><span style=\"color: #000000;\">We know that the sum of n terms is;<br \/>\n<em>S<sub>n<\/sub><\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;l)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + l) \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>n<\/sub><\/em> = <\/span><span style=\"color: #000000;\">13\/2 (34\u00a0+ 10)<br \/>\n<em>S<sub>n<\/sub><\/em> = <\/span><span style=\"color: #000000;\">(13 \u00d7 44\/2)<br \/>\n<em>S<sub>n<\/sub><\/em> = 13 \u00d7 22<br \/>\n<\/span><span style=\"color: #000000;\">= 286<\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">(iii) \u2212 5 + (\u2212 8) + (\u2212 11) + \u2026\u2026\u2026\u2026 + (\u2212 230)<br \/>\n<\/span><\/strong><span style=\"color: #000000;\"><em>a\u00a0<\/em>= \u22125<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub><em>n<\/em> <\/sub>= <em>l =<\/em> \u2212230<br \/>\n<\/span><span style=\"color: #000000;\">Common difference, <em>d\u00a0<\/em>=\u00a0<em>a<sub>2 <\/sub>\u2212 a<sub>1<\/sub>\u00a0<\/em>= (\u22128)\u2212(\u22125)<br \/>\n<\/span><span style=\"color: #000000;\"><em>d<\/em> = \u2212 8 + 5 = \u22123<br \/>\n<\/span><span style=\"color: #000000;\">Let us assume \u2212230 be the nth term of this A.P.<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n <\/sub><\/em>=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">\u2212230 = \u2212 5 + (<em>n <\/em>\u2212 1)(\u22123)<br \/>\n<\/span><span style=\"color: #000000;\">\u2212225 = (<em>n <\/em>\u2212 1)(\u22123)<br \/>\n<\/span><span style=\"color: #000000;\">(<em>n <\/em>\u2212 1) = 75<br \/>\n<\/span><span style=\"color: #000000;\"><em>n<\/em>\u00a0= 76<br \/>\n<\/span><span style=\"color: #000000;\">And the sum of n term,<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;l)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + l) \\right ]\" align=\"absmiddle\" data-wp-editing=\"1\" \/><br \/>\n<em>S<sub>n<\/sub><\/em> =<\/span><span style=\"color: #000000;\">\u00a076\/2\u00a0[(-5) +\u00a0(-230)]<br \/>\n<em>S<sub>n<\/sub><\/em> = <\/span><span style=\"color: #000000;\">38(-235)<br \/>\n<em>S<sub>n<\/sub><\/em> = <\/span><span style=\"color: #000000;\">-8930<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. In an AP<br \/>\n(i)<\/strong> Given\u00a0<em>a<\/em>\u00a0= 5,\u00a0<em>d<\/em>\u00a0= 3,\u00a0<em>a<sub>n<\/sub><\/em>\u00a0= 50, find\u00a0<em>n<\/em>\u00a0and\u00a0<em>S<sub>n<\/sub><\/em>.<br \/>\n<strong>(ii)<\/strong> Given\u00a0<em>a<\/em>\u00a0= 7,\u00a0<em>a<\/em><sub>13<\/sub>\u00a0= 35, find\u00a0<em>d<\/em>\u00a0and\u00a0<em>S<\/em><sub>13<\/sub>.<br \/>\n<strong>(iii)<\/strong> Given\u00a0<em>a<\/em><sub>12<\/sub>\u00a0= 37,\u00a0<em>d<\/em>\u00a0= 3, find\u00a0<em>a<\/em>\u00a0and\u00a0<em>S<\/em><sub>12<\/sub>.<br \/>\n<strong>(iv)<\/strong> Given\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= 15,\u00a0<em>S<\/em><sub>10<\/sub>\u00a0= 125, find\u00a0<em>d<\/em>\u00a0and\u00a0<em>a<\/em><sub>10<\/sub>.<br \/>\n<strong>(v)<\/strong> Given\u00a0<em>d<\/em>\u00a0= 5,\u00a0<em>S<\/em><sub>9<\/sub>\u00a0= 75, find\u00a0<em>a<\/em>\u00a0and\u00a0<em>a<\/em><sub>9<\/sub>.<br \/>\n<strong>(vi)<\/strong> Given\u00a0<em>a<\/em>\u00a0= 2,\u00a0<em>d<\/em>\u00a0= 8,\u00a0<em>S<sub>n<\/sub><\/em>\u00a0= 90, find\u00a0<em>n<\/em>\u00a0and\u00a0<em>a<sub>n<\/sub><\/em>.<br \/>\n<strong>(vii)<\/strong> Given\u00a0<em>a<\/em>\u00a0= 8,\u00a0<em>a<sub>n<\/sub><\/em>\u00a0= 62,\u00a0<em>S<sub>n<\/sub><\/em>\u00a0= 210, find\u00a0<em>n<\/em>\u00a0and\u00a0<em>d<\/em>.<br \/>\n<strong>(viii)<\/strong> Given\u00a0<em>a<sub>n<\/sub><\/em>\u00a0= 4,\u00a0<em>d<\/em>\u00a0= 2,\u00a0<em>S<sub>n<\/sub><\/em>\u00a0= \u2212 14, find\u00a0<em>n<\/em>\u00a0and\u00a0<em>a<\/em>.<br \/>\n<strong>(ix)<\/strong> Given\u00a0<em>a<\/em>\u00a0= 3,\u00a0<em>n<\/em>\u00a0= 8,\u00a0<em>S<\/em>\u00a0= 192, find\u00a0<em>d<\/em>.<br \/>\n<strong>(x)<\/strong> Given\u00a0<em>l<\/em>\u00a0= 28,\u00a0<em>S<\/em>\u00a0= 144 and there are total 9 terms. Find\u00a0<em>a<\/em>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) Given\u00a0<em>a<\/em>\u00a0= 5,\u00a0<em>d<\/em>\u00a0= 3,\u00a0<em>a<sub>n<\/sub><\/em>\u00a0= 50, find\u00a0<em>n<\/em>\u00a0and\u00a0<em>S<sub>n<\/sub><\/em>.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><em>a<\/em> = 5<br \/>\n<em>d<\/em> = 3<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0= 50<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a<\/em> + (<em>n<\/em> \u2212 1)<em>d<\/em>,<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 50 = 5 + (<em>n<\/em> &#8211; 1)\u00d73<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 3(<em>n<\/em> &#8211; 1) = 45<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2\u00a0<em>n<\/em> &#8211; 1 = 15<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2\u00a0<em>n<\/em>\u00a0= 16<br \/>\n<\/span><span style=\"color: #000000;\">Now, sum of n terms,<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0=\u00a0<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;a_n)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + a_n) \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em>\u00a0= 16\/2 (5\u00a0+ 50)<br \/>\n<em>S<sub>n<\/sub><\/em> = 8 \u00d7 55<br \/>\n<em>S<sub>n<\/sub><\/em> = 440<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) Given\u00a0<em>a<\/em>\u00a0= 7,\u00a0<em>a<\/em><sub>13<\/sub>\u00a0= 35, find\u00a0<em>d<\/em>\u00a0and\u00a0<em>S<\/em><sub>13<\/sub>.<br \/>\n<\/strong><em>a<\/em> = 7<br \/>\n<em>a<\/em><sub>13<\/sub> = 35<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<\/em>,<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 35 = 7 + (13 &#8211; 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">\u21d2 12<em>d<\/em>\u00a0= 28<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2\u00a0<em>d<\/em>\u00a0= 28\/12 = 2.33<br \/>\nNow, sum of n terms,<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0=\u00a0<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;a_n)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + a_n) \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>13<\/sub><\/em> = 13\/2 (7 + 35)<br \/>\n<em>S<sub>13<\/sub><\/em> = 6.5 \u00d7 42<br \/>\n<em>S<sub>13<\/sub><\/em> = 273<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) Given\u00a0<em>a<\/em><sub>12<\/sub>\u00a0= 37,\u00a0<em>d<\/em>\u00a0= 3, find\u00a0<em>a<\/em>\u00a0and\u00a0<em>S<\/em><sub>12<\/sub>.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>12<\/sub> = 37<br \/>\n<em>d<\/em>\u00a0= 3<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n<\/em> \u2212 1)<em>d<\/em>,<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2\u00a0<em>a<\/em><sub>12<\/sub>\u00a0=\u00a0<em>a <\/em>+ (12 \u2212 1)3<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 37 =\u00a0<em>a <\/em>+ 33<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2\u00a0<em>a<\/em>\u00a0= 4<br \/>\n<\/span><span style=\"color: #000000;\">Now, sum of n<sup>th<\/sup> term,<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub>\u00a0=\u00a0<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;a_n)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + a_n) \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em>\u00a0=\u00a0<em>12<\/em>\/2 (4 + 37)<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em> = 6 \u00d7 42<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0=\u00a0246<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) Given\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= 15,\u00a0<em>S<\/em><sub>10<\/sub>\u00a0= 125, find\u00a0<em>d<\/em>\u00a0and\u00a0<em>a<\/em><sub>10<\/sub>.<br \/>\n<\/strong><em>a<\/em><sub>3<\/sub> = 15<br \/>\n<em>S<\/em><sub>10<\/sub>\u00a0= 125<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a<\/em> + (<em>n <\/em>\u2212 1)<em>d<\/em>,<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub>\u00a0=\u00a0<em>a <\/em>+ (3\u22121)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">15 =\u00a0<em>a <\/em>+ 2<em>d\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;- <\/em>(i)<br \/>\n<\/span><span style=\"color: #000000;\">Sum of the n<sup>th<\/sup> term,<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>S<sub>10<\/sub><\/em>\u00a0=\u00a010\/2\u00a0[2<em>a <\/em>+ (10 &#8211; 1)<em>d<\/em>]<br \/>\n<\/span><span style=\"color: #000000;\">125 = 5(2<em>a <\/em>+ 9<em>d<\/em>)<br \/>\n<\/span><span style=\"color: #000000;\">25 = 2<em>a <\/em>+ 9<em>d\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- <\/em>(ii)<br \/>\n<\/span><span style=\"color: #000000;\">On multiplying equation\u00a0(i)\u00a0by\u00a0(ii), we will get;<br \/>\n<\/span><span style=\"color: #000000;\">30 = 2<em>a <\/em>+ 4<em>d<\/em> <em>&#8212;&#8212;&#8212;&#8212;&#8212;- <\/em>(iii)<br \/>\n<\/span><span style=\"color: #000000;\">By subtracting equation\u00a0(iii)\u00a0from\u00a0(ii), we get,<br \/>\n<\/span><span style=\"color: #000000;\">\u22125 = 5<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>d<\/em>\u00a0= \u22121<br \/>\n<\/span><span style=\"color: #000000;\">From equation\u00a0(i),<br \/>\n<\/span><span style=\"color: #000000;\">15 =\u00a0<em>a <\/em>+ 2(\u22121)<br \/>\n<\/span><span style=\"color: #000000;\">15 =\u00a0<em>a<\/em>\u22122<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em>\u00a0= 17 = First term<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>10<\/sub>\u00a0=\u00a0<em>a <\/em>+ (10 \u2212 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>10<\/sub> = 17 + (9)(\u22121)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>10<\/sub> = 17 \u2212 9 = 8<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) Given\u00a0<em>d<\/em>\u00a0= 5,\u00a0<em>S<\/em><sub>9<\/sub>\u00a0= 75, find\u00a0<em>a<\/em>\u00a0and\u00a0<em>a<\/em><sub>9<\/sub>.<br \/>\n<\/strong><em>d<\/em> = 5<br \/>\n<em>S<\/em><sub>9<\/sub>\u00a0= 75<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>S<sub>9<\/sub><\/em>\u00a0= 9\/2\u00a0[2<em>a<\/em> +(9 &#8211; 1)<em>5<\/em>]<br \/>\n<\/span><span style=\"color: #000000;\">25 = 3(<em>a <\/em>+ 20)<br \/>\n<\/span><span style=\"color: #000000;\">25 = 3<em>a <\/em>+ 60<br \/>\n<\/span><span style=\"color: #000000;\">3<em>a<\/em> = 25 \u2212 60<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em>\u00a0= -35\/3<br \/>\n<\/span><span style=\"color: #000000;\">As we know, the n<sup>th<\/sup>\u00a0term can be written as;<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n<\/em>\u22121)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>9<\/sub>\u00a0=\u00a0<em>a <\/em>+ (9 \u2212 1)(5)<br \/>\n<\/span><span style=\"color: #000000;\">= -35\/3 + 8(5)<br \/>\n<\/span><span style=\"color: #000000;\">= -35\/3 + 40<br \/>\n<\/span><span style=\"color: #000000;\">= (-35 + 120)\/3<br \/>\n= 85\/3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) Given\u00a0<em>a<\/em>\u00a0= 2,\u00a0<em>d<\/em>\u00a0= 8,\u00a0<em>S<sub>n<\/sub><\/em>\u00a0= 90, find\u00a0<em>n<\/em>\u00a0and\u00a0<em>a<sub>n<\/sub><\/em>.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><em>a<\/em> = 2<br \/>\n<em>d<\/em> = 8<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= 90<br \/>\n<\/span><span style=\"color: #000000;\">As, sum of n terms in an AP is,<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><\/span><span style=\"color: #000000;\">90 = <em>n<\/em>\/2\u00a0[2<em>a<\/em> + (<em>n<\/em> &#8211; 1)<em>d<\/em>]<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 180 =\u00a0<em>n<\/em>(4 + 8<em>n<\/em>\u00a0-8)<br \/>\n\u21d2 180 = <em>n<\/em>(8<em>n <\/em>&#8211; 4)<br \/>\n\u21d2 180 = 8<em>n<\/em><sup>2 <\/sup>&#8211; 4<em>n<br \/>\n<\/em><\/span><span style=\"color: #000000;\">\u21d2 8<em>n<\/em><sup>2 <\/sup>&#8211; 4<em>n \u2013<\/em>180 = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 2<em>n<\/em><sup>2 <\/sup>\u2013 <em>n <\/em>&#8211; 45 = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 2<em>n<\/em><sup>2 <\/sup>&#8211; 10<em>n <\/em>+ 9<em>n <\/em>&#8211; 45 = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 2<em>n<\/em>(<em>n<\/em> &#8211; 5) + 9(<em>n<\/em> &#8211; 5) = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 (<em>n <\/em>&#8211; 5)(2<em>n <\/em>+ 9) = 0<br \/>\n<\/span><span style=\"color: #000000;\">So,\u00a0<em>n<\/em>\u00a0= 5 (as n only is a positive integer)<br \/>\n<\/span><span style=\"color: #000000;\">\u2234\u00a0<em>a<sub>5<\/sub><\/em><sub>\u00a0<\/sub>= 8 + 5 \u00d7 4<br \/>\n<em>a<sub>5<\/sub><\/em><sub>\u00a0<\/sub>= 34<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vii) Given\u00a0<em>a<\/em>\u00a0= 8,\u00a0<em>a<sub>n<\/sub><\/em>\u00a0= 62,\u00a0<em>S<sub>n<\/sub><\/em>\u00a0= 210, find\u00a0<em>n<\/em>\u00a0and\u00a0<em>d<\/em>.<br \/>\n<\/strong><em>a<\/em> = 8<br \/>\n<em>a<sub>n<\/sub><\/em> = 62<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= 210<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub>\u00a0=\u00a0<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;a_n)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + a_n) \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><\/span><span style=\"color: #000000;\">210 = <em>n<\/em>\/2 (8 + 62)<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 35<em>n<\/em>\u00a0= 210<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2\u00a0<em>n<\/em>\u00a0= 210\/35<br \/>\n\u21d2\u00a0<em>n<\/em>\u00a0= 6<br \/>\n<\/span><span style=\"color: #000000;\">Now, 62 = 8 + 5<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">\u21d2 5<em>d<\/em> = 62 &#8211; 8<br \/>\n\u21d2 5<em>d<\/em> = 54<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2\u00a0<em>d<\/em>\u00a0= 54\/5<br \/>\n\u21d2 <em>d<\/em> = 10.8<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(viii) Given\u00a0<em>a<sub>n<\/sub><\/em>\u00a0= 4,\u00a0<em>d<\/em>\u00a0= 2,\u00a0<em>S<sub>n<\/sub><\/em>\u00a0= \u2212 14, find\u00a0<em>n<\/em>\u00a0and\u00a0<em>a<\/em>.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em> = 4<br \/>\n<em>d<\/em> = 2<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= \u221214.<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub>\u00a0=\u00a0<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;a_n)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + a_n) \\right ]\" align=\"absmiddle\" \/><br \/>\na<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n<\/em>\u00a0\u22121)<em>d<\/em>,<br \/>\n\u21d2 <\/span><span style=\"color: #000000;\">4 =\u00a0<em>a <\/em>+ (<em>n\u00a0<\/em>\u2212 1)2<br \/>\n\u21d2 <\/span><span style=\"color: #000000;\">4 =\u00a0<em>a <\/em>+ 2<em>n <\/em>\u2212 2<br \/>\n\u21d2 <\/span><span style=\"color: #000000;\"><em>a <\/em>+ 2<em>n<\/em>\u00a0= 6<br \/>\n\u21d2 <\/span><span style=\"color: #000000;\"><em>a\u00a0<\/em>= 6 \u2212 2<em>n<\/em> &#8212;&#8212;&#8212;&#8212;&#8212;- (i)<br \/>\n<\/span><span style=\"color: #000000;\">As we know, the sum of n terms is;<br \/>\n<em>S<sub>n<\/sub>\u00a0=\u00a0<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;a_n)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + a_n) \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><\/span><span style=\"color: #000000;\">-14 =\u00a0<em>n<\/em>\/2 (<em>a <\/em>+ <em>4<\/em>)<br \/>\n<\/span><span style=\"color: #000000;\">\u221228 =\u00a0<em>n\u00a0<\/em>(<em>a <\/em>+ 4)<br \/>\n<\/span><span style=\"color: #000000;\">\u221228 =\u00a0<em>n\u00a0<\/em>(6 \u22122<em>n<\/em> + 4)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0{From equation (i)}<br \/>\n<\/span><span style=\"color: #000000;\">\u221228 =\u00a0<em>n\u00a0<\/em>(\u2212 2<em>n<\/em> + 10)<br \/>\n<\/span><span style=\"color: #000000;\">\u221228 = \u2212 2<em>n<\/em><sup>2 <\/sup>+ 10<em>n<br \/>\n<\/em><\/span><span style=\"color: #000000;\">2<em>n<\/em><sup>2<\/sup>\u00a0\u221210<em>n<\/em>\u00a0\u2212 28 = 0<br \/>\n<\/span><span style=\"color: #000000;\"><em>n<\/em><sup>2<\/sup> \u2212 5<em>n\u00a0<\/em>\u221214 = 0<br \/>\n<\/span><span style=\"color: #000000;\"><em>n<\/em><sup>2<\/sup> \u2212 7<em>n + <\/em>2<em>n<\/em>\u00a0\u221214 = 0<br \/>\n<\/span><span style=\"color: #000000;\"><em>n\u00a0<\/em>(<em>n <\/em>\u2212 7) + 2(<em>n<\/em> \u2212 7) = 0<br \/>\n<\/span><span style=\"color: #000000;\">(<em>n<\/em> \u2212 7)(<em>n<\/em> + 2) = 0<br \/>\n<\/span><span style=\"color: #000000;\">Either\u00a0<em>n<\/em>\u00a0\u2212 7 = 0 or\u00a0<em>n<\/em>\u00a0+ 2 = 0<br \/>\n<\/span><span style=\"color: #000000;\"><em>n<\/em>\u00a0= 7 or\u00a0<em>n<\/em>\u00a0= \u22122<br \/>\n<\/span><span style=\"color: #000000;\">However,\u00a0<em>n<\/em>\u00a0can neither be negative nor fractional.<br \/>\n<\/span><span style=\"color: #000000;\">Therefore,\u00a0<em>n<\/em>\u00a0= 7<br \/>\n<\/span><span style=\"color: #000000;\">From equation\u00a0(i), we get<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em> = 6 \u2212 2<em>n<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em> = 6 \u2212 2(7)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em> = 6\u221214<br \/>\n<em>a<\/em> <\/span><span style=\"color: #000000;\">= \u22128<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ix) Given\u00a0<em>a<\/em>\u00a0= 3,\u00a0<em>n<\/em>\u00a0= 8,\u00a0<em>S<\/em>\u00a0= 192, find\u00a0<em>d<\/em>.<br \/>\n<\/strong><em>a<\/em> = 3<br \/>\n<\/span><span style=\"color: #000000;\"><em>n<\/em>\u00a0= 8<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em>\u00a0= 192<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\">192 = 8\/2 [2 \u00d7 3 + (8 -1)<em>d<\/em>]<br \/>\n<\/span><span style=\"color: #000000;\">192 = 4[6 + 7<em>d<\/em>]<br \/>\n<\/span><span style=\"color: #000000;\">48 = 6 + 7<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">42 = 7<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>d =\u00a0<\/em>6<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(x) Given\u00a0<em>l<\/em>\u00a0= 28,\u00a0<em>S<\/em>\u00a0= 144 and there are total 9 terms. Find\u00a0<em>a<\/em>.<br \/>\n<\/strong><em>l<\/em> = <em>a<sub>n<\/sub><\/em>= 28<br \/>\n<em>S<sub>n<\/sub><\/em> = 144<br \/>\n<em>n<\/em> = 9<br \/>\n<\/span><span style=\"color: #000000;\"><em>S<sub>n<\/sub>\u00a0=\u00a0<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;a_n)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + a_n) \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><\/span><span style=\"color: #000000;\">144 = 9\/2 (<em>a <\/em>+ 28)<br \/>\n<\/span><span style=\"color: #000000;\">16 \u00d7 2 = <em>a <\/em>+ 28<br \/>\n<\/span><span style=\"color: #000000;\">32 =\u00a0<em>a <\/em>+ 28<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em>\u00a0= 4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. How many terms of the AP. 9, 17, 25 \u2026 must be taken to give a sum of 636?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Let there be\u00a0<em>n<\/em>\u00a0terms of the AP. 9, 17, 25 \u2026<br \/>\n<em>a<\/em>\u00a0= 9<br \/>\n<em>d<\/em>\u00a0=\u00a0<em>a<\/em><sub>2 <\/sub>\u2212 <em>a<\/em><sub>1<\/sub> = 17 \u2212 9 = 8<br \/>\n<em>S<sub>n<\/sub><\/em> =<br \/>\n636 =\u00a0<em>n<\/em>\/2\u00a0[2\u00d79 + (n &#8211; 1)\u00d78]<br \/>\n636 =\u00a0<em>n<\/em>\/2 [18 + (<em>n <\/em>&#8211; 1)\u00d78]<br \/>\n636 =\u00a0<em>n\u00a0<\/em>[9 + 4<em>n<\/em>\u00a0\u22124]<br \/>\n636 =\u00a0<em>n\u00a0<\/em>(4<em>n<\/em>\u00a0+5)<br \/>\n4<em>n<\/em><sup>2<\/sup> + 5<em>n<\/em>\u00a0\u2212636 = 0<br \/>\n4<em>n<\/em><sup>2<\/sup> + 53<em>n<\/em>\u00a0\u221248<em>n<\/em>\u00a0\u2212636 = 0<br \/>\n<em>n\u00a0<\/em>(4<em>n<\/em> + 53) \u2212 12 (4<em>n<\/em>\u00a0+ 53) = 0<br \/>\n(4<em>n<\/em> + 53)(<em>n<\/em> \u2212 12) = 0<br \/>\n4<em>n <\/em>+ 53 = 0 or <em>n <\/em>\u2212 12 = 0<br \/>\n<em>n<\/em>\u00a0= (-53\/4) or\u00a0<em>n<\/em>\u00a0= 12<br \/>\n<em>n\u00a0<\/em>cannot be negative or fraction, therefore,\u00a0<strong><em>n<\/em>\u00a0= 12<\/strong> only.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\n<em>a<\/em>\u00a0= 5<br \/>\n<em>l<\/em>\u00a0= 45<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= 400<br \/>\n<em>S<sub>n<\/sub><\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;l)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + l) \\right ]\" align=\"absmiddle\" \/><br \/>\n400 =\u00a0<em>n<\/em>\/2 (5 + 45)<br \/>\n400 =\u00a0<em>n<\/em>\/2 (50)<br \/>\n<em>n<\/em> = 16<br \/>\n<em>a<sub>n<\/sub> = l = a + <\/em>(<em>n<\/em>\u00a0\u22121)<em>d<br \/>\n<\/em>45 = 5 +(16 \u22121)<em>d<br \/>\n<\/em>40 = 15<em>d<br \/>\n<\/em><em>d<\/em>\u00a0= 40\/15 = 8\/3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. The first and the last term of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there and what is their sum?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\n<em>a<\/em>\u00a0= 17<br \/>\n<em>d<\/em>\u00a0= 9<br \/>\n<em>l<\/em> = 350<br \/>\n<em>a<sub>n<\/sub> = l = a + <\/em>(<em>n<\/em>\u00a0\u22121)<em>d<br \/>\n<\/em>350 = 17 + (<em>n<\/em>\u00a0\u22121)9<br \/>\n333 = (<em>n <\/em>\u2212 1)9<br \/>\n(<em>n<\/em>\u22121) = 37<br \/>\n<em>n<\/em>\u00a0= 38<br \/>\n<em>S<sub>n<\/sub><\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;l)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + l) \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>38<\/sub><\/em> = 38\/2 (17 + 350)<br \/>\n<em>S<sub>38<\/sub><\/em> = 19\u00d7367<br \/>\n<em>S<sub>38<\/sub><\/em> = 6973<br \/>\nThus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. Find the sum of first 22 terms of an AP in which\u00a0<em>d<\/em>\u00a0= 7 and 22<sup>nd<\/sup> term is 149.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Given:<br \/>\n<em>d<\/em> = 7<br \/>\n<em>a<sub>22<\/sub>\u00a0<\/em>= 149<br \/>\n<em>S<sub>22<\/sub>\u00a0<\/em>= ?<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<br \/>\n<\/em><em>a<\/em><sub>22<\/sub>\u00a0=\u00a0<em>a <\/em>+ (22 \u2212 1)<em>d<br \/>\n<\/em>149 =\u00a0<em>a <\/em>+ 21\u00d77<br \/>\n149 =\u00a0<em>a <\/em>+ 147<br \/>\n<em>a<\/em> = 2<br \/>\n<em>S<sub>n<\/sub><\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;l)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + l) \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>22<\/sub><\/em> = 22\/2 (2 + 149)<br \/>\n<em>S<sub>22<\/sub><\/em> = 11\u00d7151<br \/>\n<em>S<sub>22<\/sub><\/em> = 1661<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18, respectively.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0= 14<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0= 18<br \/>\n<em>d<\/em>\u00a0=\u00a0<em>a<\/em><sub>3 <\/sub>\u2212 <em>a<\/em><sub>2<\/sub> = 18 \u2212 14 = 4<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0=\u00a0<em>a <\/em>+ <em>d<br \/>\n<\/em>14 =\u00a0<em>a <\/em>+ 4<br \/>\n<em>a<\/em> = 10<br \/>\nSum of n terms<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>51<\/sub><\/em> = 51\/2 [2\u00d710 + (51 &#8211; 1) 4]<br \/>\n<em>S<sub>51<\/sub><\/em> = 51\/2 [20 + (50)\u00d74]<br \/>\n<em>S<sub>51<\/sub><\/em> = 51 \u00d7 220\/2<br \/>\n<em>S<sub>51<\/sub><\/em> = 51 \u00d7 110<br \/>\n<em>S<sub>51<\/sub><\/em> = 5610<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first\u00a0<em>n<\/em>\u00a0terms.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\n<em>S<\/em><sub>7<\/sub>\u00a0= 49<br \/>\nS<sub>17<\/sub>\u00a0= 289<br \/>\nWe know, sum of n terms;<br \/>\n<em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><em>S<sub>7 <\/sub><\/em>=\u00a0<em>7<\/em>\/2\u00a0[2<em>a<\/em> + (<em>n<\/em> &#8211; 1)<em>d<\/em>]<br \/>\n49 = 7\/2\u00a0[2<em>a<\/em> + (7 &#8211; 1)<em>d<\/em>]<br \/>\n49 = 7\/2\u00a0[2<em>a\u00a0<\/em>+ 6<em>d<\/em>]<br \/>\n7 = (<em>a <\/em>+ 3<em>d<\/em>)<br \/>\n<em>a<\/em>\u00a0+ 3<em>d<\/em> = 7\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;(i)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><em>S<sub>17<\/sub><\/em>\u00a0= 17\/2\u00a0[2<em>a <\/em>+ (17 &#8211; 1)<em>d<\/em>]<br \/>\n289 = 17\/2 (2<em>a<\/em> + 16<em>d<\/em>)<br \/>\n17 = (<em>a <\/em>+ 8<em>d<\/em>)<br \/>\n<em>a<\/em> + 8<em>d<\/em> = 17\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;(ii)<br \/>\nSubtracting equation\u00a0(i)\u00a0from equation\u00a0(ii),<br \/>\n(<em>a<\/em> + 8<em>d<\/em>) &#8211; (<em>a<\/em>\u00a0+ 3<em>d<\/em>) = 17 &#8211; 7<br \/>\n5<em>d<\/em> = 10<br \/>\n<em>d<\/em>\u00a0= 2<br \/>\nFrom equation\u00a0(i), we can write it as;<br \/>\n<em>a <\/em>+ 3(2) = 7<br \/>\n<em>a + <\/em>6 = 7<br \/>\n<em>a =\u00a0<\/em>1<br \/>\nHence,<br \/>\n<em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><em>S<sub>n<\/sub>\u00a0<\/em>=\u00a0<em>n<\/em>\/2[2(1) + (<em>n<\/em>\u00a0\u2013 1)\u00d72]<br \/>\n=\u00a0<em>n<\/em>\/2(2 + 2<em>n <\/em>&#8211; 2)<br \/>\n=\u00a0<em>n<\/em>\/2(2<em>n<\/em>)<br \/>\n=\u00a0<em>n<\/em><sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. Show that <em>a<\/em><sub>1<\/sub>,\u00a0<em>a<\/em><sub>2\u00a0<\/sub>\u2026 ,\u00a0<em>a<sub>n<\/sub><\/em>\u00a0, \u2026 form an AP where\u00a0<em>a<sub>n<\/sub><\/em>\u00a0is defined as below<br \/>\n<\/strong><strong>(i)\u00a0<\/strong><em>a<sub>n<\/sub><\/em>\u00a0= 3+4<em>n<\/em><br \/>\n<strong>(ii)\u00a0<\/strong><em>a<sub>n<\/sub><\/em> = 9 \u2212 5<em>n<\/em><br \/>\n<strong>Also, find the sum of the first 15 terms in each case.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)\u00a0<em>a<sub>n<\/sub><\/em>\u00a0= 3+4<em>n<br \/>\n<\/em><\/strong><em>a<\/em><sub>1<\/sub> = 3 + 4(1) = 7<br \/>\n<em>a<\/em><sub>2<\/sub> = 3 + 4(2) = 3 + 8 = 11<br \/>\n<em>a<\/em><sub>3<\/sub> = 3 + 4(3) = 3 + 12 = 15<br \/>\n<em>a<\/em><sub>4<\/sub> = 3 + 4(4) = 3 + 16 = 19<br \/>\nWe can see here, the common difference between the terms are;<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>1<\/sub> = 11 \u2212 7 = 4<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>2<\/sub> = 15 \u2212 11 = 4<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>3<\/sub> = 19 \u2212 15 = 4<br \/>\nHence,\u00a0<em>a<sub>k<\/sub><\/em><sub>\u00a0+ 1<\/sub>\u00a0\u2212\u00a0<em>a<sub>k<\/sub><\/em> is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.<br \/>\nNow, we know, the sum of nth term is;<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 15\/2 [2(7) + (15 &#8211; 1)\u00d74]<br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 15\/2[(14) + 56]<br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 15\/2 (70)<br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 15 \u00d7 35<br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 525<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)\u00a0<em>a<sub>n<\/sub><\/em> = 9 \u2212 5<em>n<br \/>\n<\/em><\/strong><em>a<\/em><sub>1<\/sub> = 9 \u2212 5\u00d71 = 9\u22125 = 4<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0= 9\u22125\u00d72 = 9\u221210 = \u22121<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0= 9\u22125\u00d73 = 9\u221215 = \u22126<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0= 9\u22125\u00d74 = 9\u221220 = \u221211<br \/>\nWe can see here, the common difference between the terms are;<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= \u22121\u22124 = \u22125<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>2<\/sub>\u00a0= \u22126\u2212(\u22121) = \u22125<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= \u221211\u2212(\u22126) = \u22125<br \/>\nHence,\u00a0<em>a<sub>k<\/sub><\/em><sub>\u00a0+ 1<\/sub>\u00a0\u2212\u00a0<em>a<sub>k<\/sub><\/em> is same every time. Therefore, this is an A.P. with common difference as \u22125 and first term as 4.<br \/>\nNow, we know, the sum of nth term is;<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 15\/2[2(4) + (15 &#8211; 1)(-5)]<br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 15\/2[8 + 14(-5)]<br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 15\/2(8 &#8211; 70)<br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 15\/2(-62)<br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= 15(-31)<br \/>\n<em>S<sub>15\u00a0<\/sub><\/em>= -465<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>11. If the sum of the first\u00a0<em>n<\/em>\u00a0terms of an AP is 4<em>n<\/em>\u00a0\u2212\u00a0<em>n<\/em><sup>2<\/sup>, what is the first term (that is\u00a0<em>S<\/em><sub>1<\/sub>)? What is the sum of first two terms? What is the second term? Similarly find the 3<sup>rd<\/sup>, the 10<sup>th<\/sup>\u00a0and the\u00a0<em>n<\/em><sup>th<\/sup>\u00a0terms.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= 4<em>n <\/em>\u2212 <em>n<\/em><sup>2<br \/>\n<\/sup><em>a<\/em>\u00a0=\u00a0<em>S<\/em><sub>1<\/sub>\u00a0= 4(1) \u2212 (1)<sup>2<\/sup> = 4 \u2212 1 = 3<br \/>\n<em>S<\/em><sub>2<\/sub>= 4(2) \u2212 (2)<sup>2<\/sup> = 8 \u2212 4 = 4<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0=\u00a0<em>S<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>S<\/em><sub>1<\/sub> = 4 \u2212 3 = 1<br \/>\n<em>d<\/em>\u00a0=\u00a0<em>a<\/em><sub>2 <\/sub>\u2212 <em>a<\/em> = 1 \u2212 3 = \u22122<br \/>\nN<sup>th<\/sup>\u00a0term,\u00a0<em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<br \/>\n<\/em>= 3 + (<em>n<\/em> \u2212 1)(\u22122)<br \/>\n= 3 \u2212 2<em>n<\/em> + 2<br \/>\n= 5 \u2212 2<em>n<br \/>\n<\/em>Therefore,\u00a0<em>a<\/em><sub>3<\/sub> = 5 \u2212 2(3) = 5 &#8211; 6 = \u22121<br \/>\n<em>a<\/em><sub>10<\/sub> = 5 \u2212 2(10) = 5 \u2212 20 = \u221215<br \/>\nHence, the sum of first two terms is 4. The second term is 1.<br \/>\nThe 3<sup>rd<\/sup>, the 10<sup>th<\/sup>, and\u00a0the\u00a0<em>n<\/em><sup>th<\/sup>\u00a0terms are \u22121, \u221215, and 5 \u2212 2<em>n<\/em>\u00a0respectively.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>12. Find the sum of the first 40 positive integers divisible by 6.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>The first positive integers that are divisible by 6 are 6, 12, 18, 24 \u2026.<br \/>\n<em>a<\/em>\u00a0= 6<br \/>\n<em>d<\/em>\u00a0= 6<br \/>\n<em>S<\/em><sub>40<\/sub><em>\u00a0=\u00a0<\/em>?<br \/>\n<em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em>Therefore, putting n = 40, we get,<br \/>\n<em>S<sub>40\u00a0<\/sub><\/em>= 40\/2 [2(6) + (40 &#8211; 1)6]<br \/>\n<em>S<sub>40\u00a0<\/sub><\/em>= 20[12 + (39)(6)]<br \/>\n<em>S<sub>40\u00a0<\/sub><\/em>= 20(12 + 234)<br \/>\n<em>S<sub>40\u00a0<\/sub><\/em>= 20 \u00d7 246<br \/>\n<em>S<sub>40\u00a0<\/sub><\/em>= 4920<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>13. Find the sum of first 15 multiples of 8.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>The first few multiples of 8 are 8, 16, 24, 32\u2026<br \/>\n<em>a<\/em>\u00a0= 8<br \/>\n<em>d<\/em>\u00a0= 8<br \/>\n<em>S<\/em><sub>15<\/sub>\u00a0= ?<br \/>\n<em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><em>S<sub>15<\/sub>\u00a0<\/em>= 15\/2 [2(8) + (15 &#8211; 1)8]<br \/>\n<em>S<sub>15<\/sub>\u00a0<\/em>= 15\/2[16 + (14)(8)]<br \/>\n<em>S<sub>15<\/sub>\u00a0<\/em>= 15\/2[16 + 112]<br \/>\n<em>S<sub>15<\/sub>\u00a0<\/em>= 15 (128)\/2<br \/>\n<em>S<sub>15<\/sub>\u00a0<\/em>= 15 \u00d7 64<br \/>\n<em>S<sub>15<\/sub>\u00a0<\/em>= 960<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>14. Find the sum of the odd numbers between 0 and 50.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 \u2026 49.<br \/>\n<em>a\u00a0<\/em>= 1<br \/>\n<em>d<\/em> = 2<br \/>\n<em>l<\/em>\u00a0= 49<br \/>\n<em>l<\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1) <em>d<br \/>\n<\/em>49 = 1 + (<em>n <\/em>\u2212 1)2<br \/>\n48 = 2(<em>n<\/em> \u2212 1)<br \/>\n<em>n<\/em>\u00a0\u2212 1 = 24<br \/>\n<em>n<\/em>\u00a0= 25<br \/>\nBy the formula of sum of nth term, we know,<br \/>\n<em>S<sub>n<\/sub><\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;l)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + l) \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>25<\/sub><\/em> = 25\/2 (1 + 49)<br \/>\n<em>S<sub>25<\/sub><\/em> = 25 (50)\/2<br \/>\n<em>S<sub>25<\/sub><\/em> = (25)(25)<br \/>\n<em>S<sub>25<\/sub><\/em> = 625<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money does the contractor have to pay as penalty, if he has delayed the work by 30 days.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given:<br \/>\n<em>a<\/em> = 200<br \/>\n<em>d<\/em> = 50<br \/>\nThe penalty that has to be paid if the contractor has delayed the work by 30 days =\u00a0<em>S<\/em><sub>30<br \/>\n<\/sub><em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<\/em><em>S<\/em><sub>30 <\/sub>= 30\/2[2(200) + (30 \u2013 1)50]<br \/>\n<em>S<\/em><sub>30 <\/sub>= 15[400+1450]<br \/>\n<em>S<\/em><sub>30 <\/sub>= 15(1850)<br \/>\n<em>S<\/em><sub>30 <\/sub>= 27750<br \/>\nTherefore, the contractor has to pay Rs 27750 as penalty.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Let us assume the cost of 1st prize be Rs. P.<br \/>\nCost of 2<sup>nd<\/sup>\u00a0prize =\u00a0Rs.\u00a0<em>P<\/em>\u00a0\u2212 20<br \/>\ncost of 3<sup>rd<\/sup>\u00a0prize =\u00a0Rs.\u00a0<em>P<\/em>\u00a0\u2212 40<br \/>\nThese prizes form an A.P., with common difference, d = \u221220<br \/>\nFirst term, <em>a<\/em> = P<br \/>\n<em>S<\/em><sub>7<\/sub>\u00a0= 700<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n700 = 7\/2 [2<em>a<\/em>\u00a0+ (7 \u2013 1)<em>d<\/em>]\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">100 = <em>a<\/em> + 3(-20)<br \/>\n<em>a<\/em> \u2212 60 = 100<br \/>\n<em>a<\/em>\u00a0= 160<br \/>\nTherefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant, will be the same as the class, in which they are studying. E.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>The number of trees planted by the students form an AP, 1, 2, 3, 4, 5\u2026\u2026\u2026\u2026\u2026\u2026..12<br \/>\n<em>a<\/em>\u00a0= 1<br \/>\n<em>d<\/em> = 2 \u2212 1 = 1<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>12<\/sub>\u00a0<\/em>= 12\/2 [2(1) + (12 &#8211; 1)(1)]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><em>S<sub>12<\/sub>\u00a0<\/em>= 6(2 + 11)<br \/>\n<em>S<sub>12<\/sub>\u00a0<\/em>= 6(13)<br \/>\n<em>S<sub>12<\/sub>\u00a0<\/em>= 78<br \/>\nTherefore, the number of trees planted by 1 section of the classes = 78<br \/>\nThe number of trees planted by 3 sections of the classes = 3 \u00d7 78 = 234<br \/>\nTherefore, 234 trees will be planted by the students.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, \u2026\u2026\u2026 as shown in the figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \u03c0 = 22\/7)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6294\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.3-Que18.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"334\" height=\"157\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.3-Que18.png 334w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.3-Que18-300x141.png 300w\" sizes=\"auto, (max-width: 334px) 100vw, 334px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>We know<br \/>\nPerimeter of a semi-circle = <em>\u03c0r<br \/>\n<\/em><i>P<\/i><sub>1<\/sub>\u00a0= \u03c0(0.5) = \u03c0\/2 cm<br \/>\n<i>P<\/i><sub>2<\/sub>\u00a0= \u03c0(1) = \u03c0 cm<br \/>\n<i>P<\/i><sub>3<\/sub>\u00a0= = \u03c0(1.5) = 3\u03c0\/2 cm<br \/>\nWhere, P<sub>1,<\/sub>\u00a0<em>P<\/em><sub>2<\/sub>,\u00a0<em>P<\/em><sub>3<\/sub>\u00a0are the lengths of the semi-circles.<br \/>\nHence we got a series here, as,<br \/>\n\u03c0\/2, \u03c0, 3\u03c0\/2, 2\u03c0, \u2026.<br \/>\nP<sub>1<\/sub><em>\u00a0<\/em>= \u03c0\/2 cm<br \/>\n<em>P<\/em><sub>2<\/sub> = \u03c0 cm<br \/>\n<em>d<\/em>\u00a0=\u00a0<em>P<sub>2\u00a0<\/sub>\u2013 P<\/em><sub>1<\/sub>\u00a0=\u00a0\u03c0 \u2013 \u03c0\/2 = \u03c0\/2<br \/>\nP<sub>1<\/sub>=\u00a0<em>a<\/em>\u00a0=\u00a0\u03c0\/2 cm<br \/>\nBy the sum of n term formula, we know,<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\nTherefor, the sum of the length of 13 consecutive circles is;<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><em>S<sub>13<\/sub><\/em>\u00a0=\u00a0<em>13<\/em>\/2\u00a0[2(\u03c0\/2)\u00a0+ (13\u00a0\u2013 1)\u03c0\/2]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><em>S<sub>13<\/sub><\/em>\u00a0= \u00a0<em>13<\/em>\/2\u00a0[\u03c0\u00a0+ 6\u03c0]<br \/>\n<em>S<sub>13<\/sub> = 13<\/em>\/2\u00a0(7\u03c0)<br \/>\n<em>S<sub>13<\/sub><\/em>\u00a0=\u00a0<em>13<\/em>\/2 \u00d7 7 \u00d7\u00a0<em>22<\/em>\/7<br \/>\n<em>S<sub>13<\/sub><\/em>\u00a0= 143 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6295\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.3-Que19.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"533\" height=\"107\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.3-Que19.png 533w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.3-Que19-300x60.png 300w\" sizes=\"auto, (max-width: 533px) 100vw, 533px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 \u00a0<\/strong>The numbers of logs in rows are in the form of an A.P. 20, 19, 18\u2026<br \/>\n<em>a<\/em> = 20<br \/>\n<em>d<\/em>\u00a0=\u00a0<em>a<\/em><sub>2 <\/sub>\u2212 <em>a<\/em><sub>1<\/sub> = 19 \u2212 20 = \u22121<br \/>\nLet a total of 200 logs be placed in\u00a0<em>n<\/em>\u00a0rows.<br \/>\nThus,<em>\u00a0S<sub>n<\/sub><\/em>\u00a0= 200<br \/>\n<em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>12<\/sub>\u00a0<\/em>= 12\/2 [2(20) + (<em>n<\/em>\u00a0-1)(-1)]<br \/>\n400 =\u00a0<em>n<\/em> (40 \u2212 <em>n <\/em>+ 1)<br \/>\n400 =\u00a0<em>n\u00a0<\/em>(41 &#8211; <em>n<\/em>)<br \/>\n400 = 41<em>n <\/em>\u2212 <em>n<\/em><sup>2<br \/>\n<\/sup><em>n<\/em><sup>2 <\/sup>\u2212 41<em>n\u00a0<\/em>+ 400 = 0<br \/>\n<em>n<\/em><sup>2 <\/sup>\u2212 16<em>n <\/em>\u2212 25<em>n <\/em>+ 400 = 0<br \/>\n<em>n<\/em>(<em>n<\/em> \u221216) \u2212 25(<em>n<\/em>\u00a0\u221216) = 0<br \/>\n(<em>n\u00a0<\/em>\u221216)(<em>n<\/em>\u00a0\u221225) = 0<br \/>\n(<em>n<\/em>\u00a0\u221216) = 0 or\u00a0<em>n <\/em>\u2212 25 = 0<br \/>\n<em>n<\/em>\u00a0= 16 or\u00a0<em>n<\/em>\u00a0= 25<br \/>\nBy the nth term formula,<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<br \/>\n<\/em><em>a<\/em><sub>16<\/sub> = 20 + (16 \u2212 1)(\u22121)<br \/>\n<em>a<\/em><sub>16<\/sub> = 20 \u2212 15<br \/>\n<em>a<\/em><sub>16<\/sub>\u00a0= 5<br \/>\nSimilarly, the 25<sup>th<\/sup>\u00a0term could be written as;<br \/>\n<em>a<\/em><sub>25<\/sub> = 20 + (25 \u2212 1)(\u22121)<br \/>\n<em>a<\/em><sub>25<\/sub> = 20 \u2212 24<br \/>\n<em>= \u2212<\/em>4<br \/>\nIt can be seen, the number of logs in the 16<sup>th<\/sup>\u00a0row is 5 as the numbers cannot be negative.<br \/>\nTherefore, 200 logs can be placed in 16 rows and the number of logs in the 16<sup>th<\/sup>\u00a0row is 5.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6296\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.3-Que20.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"621\" height=\"119\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.3-Que20.png 621w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.3-Que20-300x57.png 300w\" sizes=\"auto, (max-width: 621px) 100vw, 621px\" \/><br \/>\n<\/strong><strong>A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?<br \/>\n[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 \u00d7 5 + 2 \u00d7 (5 + 3)]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>The distance of the potatoes from the bucket are 5, 8, 11, 14\u2026, which is in the form of AP.<br \/>\nNow, we know that \u00a0the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, distances to be run w.r.t distances of potatoes is equivalent to,\u00a0 <\/span><br \/>\n<span style=\"color: #000000;\">10, 16, 22, 28, 34, \u2026\u2026\u2026.<\/span><br \/>\n<span style=\"color: #000000;\"><em>a<\/em> = 10 <\/span><br \/>\n<span style=\"color: #000000;\"><em>d<\/em> = 16 \u2212 10 = 6<\/span><br \/>\n<span style=\"color: #000000;\">Sum of n<sup>th<\/sup> term, we get,<\/span><br \/>\n<span style=\"color: #000000;\"><em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\nS<sub>10<\/sub> = 12\/2 [2(20) + (n &#8211; 1)(-1)]<\/span><br \/>\n<span style=\"color: #000000;\">S<sub>10<\/sub> = 5[20 + 54]<\/span><br \/>\n<span style=\"color: #000000;\">S<sub>10<\/sub> = 5(74)<\/span><br \/>\n<span style=\"color: #000000;\">S<sub>10<\/sub> = 370<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the competitor will run a total distance of 370 m.<\/span><\/p>\n<p><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 5 (Arithmetic Progressions)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 5 Arithmetic Progressions Exercise 5.3 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 5 Arithmetic Progressions NCERT Class 10 Maths Solution Ex &#8211; 5.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1074,1077,1044,1049,1048],"class_list":["post-6043","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-5-arithmetic-progressions-solutions","tag-ncert-class-10-mathematics-exercise-5-3-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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