{"id":6042,"date":"2023-08-12T04:55:15","date_gmt":"2023-08-12T04:55:15","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6042"},"modified":"2023-08-12T04:55:15","modified_gmt":"2023-08-12T04:55:15","slug":"ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Chapter &#8211; 5 (Arithmetic Progressions)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 5 Arithmetic Progressions <\/strong>Exercise 5.2 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-5-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 5.3<\/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;\"><span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">Exercise &#8211; 5.2<\/span><\/strong><\/span><\/h2>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Fill in the blanks in the following table, given that\u00a0<em>a<\/em>\u00a0is the first term,\u00a0<em>d<\/em>\u00a0the common difference and\u00a0<em>a<sub>n<\/sub><\/em>\u00a0the\u00a0<em>n<\/em><sup>th<\/sup>\u00a0term of the A.P.<br \/>\n<\/strong><\/span><\/p>\n<table style=\"width: 50%;\" border=\"1\">\n<tbody>\n<tr style=\"height: 27px;\">\n<td style=\"height: 27px; text-align: center;\"><\/td>\n<td style=\"height: 27px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">a<\/span><\/td>\n<td style=\"height: 27px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">d<\/span><\/td>\n<td style=\"height: 27px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">n<\/span><\/td>\n<td style=\"height: 27px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">a<sub>n<\/sub><\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">(i)<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">7<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">3<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">8<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">\u2013<\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">(ii)<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">-18<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">\u2013<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">10<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">0<\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">(iii)<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">\u2013<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">-3<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">18<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">-5<\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">(iv)<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">-18.9<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">2.5<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">\u2013<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">3.6<\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">(v)<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">3.5<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">0<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">105<\/span><\/td>\n<td style=\"height: 24px; text-align: center;\"><span style=\"font-family: Georgia, Palatino;\">\u2013<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><strong>(i)\u00a0<\/strong><\/span><span style=\"color: #000000;\">First term,\u00a0<em>a<\/em>\u00a0= 7<br \/>\n<\/span><span style=\"color: #000000;\">Common difference,\u00a0<em>d<\/em>\u00a0= 3<br \/>\n<\/span><span style=\"color: #000000;\">Number of terms,\u00a0<em>n<\/em>\u00a0= 8,<br \/>\n<\/span><span style=\"color: #000000;\">We have to find the n<sup>th<\/sup> term,\u00a0<em>a<sub>n<\/sub><\/em>\u00a0= ?<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<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;\">Putting the values,<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 7 + (8 \u22121) 3<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 7 + (7) 3<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 7 + 21<br \/>\n\u21d2 28<br \/>\n<\/span><span style=\"color: #000000;\">Hence,<em>\u00a0a<sub>n<\/sub><\/em>\u00a0= 28<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong>\u00a0<\/span><span style=\"color: #000000;\">First term,\u00a0<em>a<\/em>\u00a0= -18<br \/>\n<\/span><span style=\"color: #000000;\">Common difference,\u00a0<em>d<\/em>\u00a0= ?<br \/>\n<\/span><span style=\"color: #000000;\">Number of terms,\u00a0<em>n<\/em>\u00a0= 10<br \/>\n<\/span><span style=\"color: #000000;\">n<sup>th<\/sup> term, <em>a<sub>n<\/sub><\/em>\u00a0= 0<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<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;\">\u21d2 0 = \u2212 18 +(10\u22121)<em>d<br \/>\n\u21d2 <\/em><\/span><span style=\"color: #000000;\">18 = 9<em>d<br \/>\n\u21d2 <\/em><\/span><span style=\"color: #000000;\"><em>d<\/em>\u00a0= 18\/9<br \/>\n\u21d2 <em>d <\/em>= 2<br \/>\n<\/span><span style=\"color: #000000;\">Hence, common difference,\u00a0<em>d\u00a0<\/em>= 2<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><strong>(iii)<\/strong>\u00a0<\/span><span style=\"color: #000000;\">First term,\u00a0<em>a<\/em>\u00a0= ?<br \/>\n<\/span><span style=\"color: #000000;\">Common difference,\u00a0<em>d<\/em>\u00a0= -3<br \/>\n<\/span><span style=\"color: #000000;\">Number of terms,\u00a0<em>n<\/em>\u00a0= 18<br \/>\n<\/span><span style=\"color: #000000;\">n<sup>th<\/sup> term, <em>a<sub>n<\/sub><\/em>\u00a0= -5<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<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;\">\u22125 =\u00a0<em>a<\/em>+(18\u22121) (\u22123)<br \/>\n<\/span><span style=\"color: #000000;\">\u22125 =\u00a0<em>a<\/em>+(17) (\u22123)<br \/>\n<\/span><span style=\"color: #000000;\">\u22125 =\u00a0<em>a<\/em>\u221251<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em>\u00a0= 51\u22125<br \/>\n<em>a <\/em>= 46<br \/>\n<\/span><span style=\"color: #000000;\">Hence,\u00a0<em>a<\/em>\u00a0= 46<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><strong>(iv)<\/strong> <\/span><span style=\"color: #000000;\">First term,\u00a0<\/span><em style=\"color: #000000;\">a<\/em><span style=\"color: #000000;\">\u00a0= -18.9<br \/>\n<\/span><span style=\"color: #000000;\">Common difference,\u00a0<em>d<\/em>\u00a0= 2.5<br \/>\n<\/span><span style=\"color: #000000;\">Number of terms,\u00a0<em>n<\/em>\u00a0= ?<br \/>\n<\/span><span style=\"color: #000000;\">n<sup>th<\/sup> term, <em>a<sub>n<\/sub><\/em>\u00a0= 3.6<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a<\/em> + (<em>n<\/em>\u00a0\u22121)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">3.6 = \u2212 18.9 + (<em>n<\/em> \u2212 1)2.5<br \/>\n<\/span><span style=\"color: #000000;\">3.6 + 18.9 = (<em>n <\/em>\u2212 1)2.5<br \/>\n<\/span><span style=\"color: #000000;\">22.5 = (<em>n<\/em>\u22121)2.5<br \/>\n<\/span><span style=\"color: #000000;\">(<em>n<\/em>\u00a0\u2013 1) = 22.5\/2.5<br \/>\n<\/span><span style=\"color: #000000;\"><em>n<\/em>\u00a0\u2013 1 = 9<br \/>\n<\/span><span style=\"color: #000000;\"><em>n<\/em>\u00a0= 10<br \/>\n<\/span><span style=\"color: #000000;\">Hence,\u00a0<em>n<\/em>\u00a0= 10<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><strong>(v)<\/strong>\u00a0<\/span><span style=\"color: #000000;\">First term,\u00a0<em>a<\/em>\u00a0= 3.5<br \/>\n<\/span><span style=\"color: #000000;\">Common difference,\u00a0<em>d<\/em>\u00a0= 0<br \/>\n<\/span><span style=\"color: #000000;\">Number of terms,\u00a0<em>n<\/em>\u00a0= 105<br \/>\n<\/span><span style=\"color: #000000;\">n<sup>th<\/sup> term, <em>a<sub>n<\/sub><\/em>\u00a0= ?<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n<\/em>\u00a0\u22121)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em> = 3.5 + (105 \u2212 1)0<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em> = 3.5 + 104 \u00d7 0<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<sub>n<\/sub><\/em>\u00a0= 3.5<br \/>\n<\/span><span style=\"color: #000000;\">Hence,\u00a0<em>a<sub>n<\/sub><\/em>\u00a0= 3.5<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. Choose the correct choice in the following and justify:<br \/>\n(i) 30<sup>th<\/sup>\u00a0term of the A.P: 10,7, 4, \u2026, is<br \/>\n<\/strong>(A)\u00a097<br \/>\n(B)\u00a077<br \/>\n(C)\u00a0\u221277<br \/>\n(D) \u221287<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><strong>Solutions &#8211;\u00a0 <\/strong><\/span><span style=\"color: #000000;\">Given<br \/>\n<\/span><span style=\"color: #000000;\">A.P. = 10, 7, 4, \u2026<br \/>\n<\/span><span style=\"color: #000000;\">First term,\u00a0<em>a<\/em>\u00a0= 10<br \/>\n<\/span><span style=\"color: #000000;\">Common difference,\u00a0<em>d<\/em>\u00a0=\u00a0<em>a<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>1\u00a0<\/sub>= 7 \u2212 10 = \u22123<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<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>30<\/sub> = 10 + (30 \u2212 1)(\u22123)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>30<\/sub> = 10 + (29)(\u22123)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>30<\/sub> = 10 \u2212 87<br \/>\n<em>a<\/em><sub>30<\/sub> = \u221277<br \/>\n<\/span><span style=\"color: #000000;\">Hence, the correct answer is option\u00a0C.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">(ii) 11<sup>th\u00a0<\/sup>term of the A.P. -3, -1\/2, ,2 \u2026. is<\/span><\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(A) 28<br \/>\n(B) 22<br \/>\n(C) \u2013 38<br \/>\n(D) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?-48\\frac{1}{2}\" alt=\"-48\\frac{1}{2}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><strong>Solutions &#8211; <\/strong><\/span><span style=\"color: #000000;\">Given<br \/>\n<\/span><span style=\"color: #000000;\">A.P. = -3, -1\/2 ,2 \u2026<br \/>\n<\/span><span style=\"color: #000000;\">First term\u00a0<em>a<\/em>\u00a0= \u2013 3<br \/>\n<\/span><span style=\"color: #000000;\">Common difference,\u00a0<em>d<\/em>\u00a0=\u00a0<em>a<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>1<\/sub><br \/>\n= (-1\/2) &#8211; (-3)<br \/>\n<\/span><span style=\"color: #000000;\">= (-1\/2) + 3<br \/>\n= 5\/2<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<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>11<\/sub> = -3 + (11 &#8211; 1)(5\/2)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>11<\/sub> = -3 + (10)(5\/2)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>11<\/sub> = -3 + 25<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>11<\/sub>\u00a0= 22<br \/>\n<\/span><span style=\"color: #000000;\">Hence, the answer is option B.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><strong>3. In the following APs find the missing term in the boxes.<br \/>\n<\/strong><\/span><strong>(i)<\/strong> 2, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 26<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 13, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 3<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino;\"><strong>(iii)<\/strong> 5, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?9\\frac{1}{2}\" alt=\"9\\frac{1}{2}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"font-family: Georgia, Palatino;\"><strong>(iv)<\/strong> -4, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 6<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino;\"><strong>(v)<\/strong> <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 38, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, -22<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><strong>Solutions &#8211;<br \/>\n(i) 2, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 26<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">Given<br \/>\n<em>a<\/em>\u00a0= 2<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub>\u00a0= 26<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<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<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub> = 2 + (3 &#8211; 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">26 = 2 + 2<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">24 = 2<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>d<\/em>\u00a0= 12<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub> = 2 + (2 &#8211; 1)12<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub> = 14<br \/>\n<\/span><span style=\"color: #000000;\">Therefore, The Series is &#8211; <strong>2, 14, 26<\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 13, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 3<br \/>\n<\/strong><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub> = 13<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>4<\/sub>\u00a0= 3<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<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<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub>\u00a0=\u00a0<em>a<\/em> + (2 &#8211; 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">13 =\u00a0<em>a <\/em>+ <em>d<\/em>\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>4<\/sub>\u00a0=\u00a0<em>a <\/em>+ (4 &#8211; 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">3 =\u00a0<em>a <\/em>+ 3<em>d<\/em>\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<br \/>\n<\/span><span style=\"color: #000000;\">On subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get,<br \/>\n<\/span><span style=\"color: #000000;\">\u2013 10 = 2<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>d<\/em>\u00a0= \u2013 5<br \/>\n<\/span><span style=\"color: #000000;\">From equation\u00a0(i), putting the value of d,we get<br \/>\n<\/span><span style=\"color: #000000;\">13 =\u00a0<em>a <\/em>+ (-5)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em>\u00a0= 18<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub> = 18 + (3 &#8211; 1)(-5)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub> = 18 + 2(-5)<br \/>\n<em>a<\/em><sub>3<\/sub> = 18 &#8211; 10<br \/>\n<em>a<\/em><sub>3<\/sub> = 8<br \/>\nTherefore, The Series is <strong>18, 13, 8, 3.<\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(iii) 5, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?9\\frac{1}{2}\" alt=\"9\\frac{1}{2}\" align=\"absmiddle\" \/><br \/>\n<\/strong><span style=\"color: #000000;\"><em>a\u00a0<\/em>= 5<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>4<\/sub>\u00a0= <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?9\\frac{1}{2}\" alt=\"9\\frac{1}{2}\" align=\"absmiddle\" \/> = <\/strong>19\/2<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<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<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>4<\/sub>\u00a0=\u00a0<em>a <\/em>+ (4 &#8211; 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">19\/2 =\u00a0<em>5 <\/em>+ 3d<br \/>\n<\/span><span style=\"color: #000000;\">(19\/2) \u2013 5 = 3d<br \/>\n<\/span><span style=\"color: #000000;\">3d =\u00a09\/2<br \/>\n<\/span><span style=\"color: #000000;\">d = 3\/2<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub>\u00a0=\u00a0<em>a <\/em>+ (2 &#8211; 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub> = 5 + 3\/2<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub> =\u00a013\/2<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub>\u00a0=\u00a0<em>a <\/em>+ (3 &#8211; 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub> = 5 + 2 \u00d7 3\/2<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub>\u00a0=\u00a0<em>8<br \/>\n<\/em><\/span><span style=\"color: #000000;\">Therefore, The Series is <strong>5, 13\/2, 8, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?9\\frac{1}{2}\" alt=\"9\\frac{1}{2}\" align=\"absmiddle\" \/>.\u00a0<\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">(iv) -4, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 6<br \/>\n<\/span><\/strong><span style=\"color: #000000;\"><em>a<\/em> = \u22124<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>6<\/sub>\u00a0= 6<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<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<br \/>\n<\/em><\/span><span style=\"color: #000000;\">a<sub>6<\/sub> = a + (6 \u2212 1)d<br \/>\n<\/span><span style=\"color: #000000;\">6 = \u2212 4 + 5<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">10 = 5<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>d<\/em>\u00a0= 2<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub>\u00a0=\u00a0<em>a <\/em>+ <em>d<\/em> = \u2212 4 + 2 = \u22122<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub>\u00a0=\u00a0<em>a <\/em>+ 2<em>d<\/em> = \u2212 4 + 2(2) = 0<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>4<\/sub>\u00a0=\u00a0<em>a <\/em>+ 3<em>d<\/em> = \u2212 4 + 3(2) = 2<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>5<\/sub>\u00a0=<em> a\u00a0<\/em>+ 4<em>d<\/em> = \u2212 4 + 4(2) = 4<br \/>\n<\/span><span style=\"color: #000000;\">Therefore, The Series is <strong>-4, -2, 0, 2, 4, 6.<\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino;\"><strong>(v) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, 38, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\square\" alt=\"\\square\" align=\"absmiddle\" \/>, -22<br \/>\n<\/strong><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub>\u00a0= 38<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>6<\/sub>\u00a0= \u221222<br \/>\n<\/span><span style=\"color: #000000;\">As we know, for an A.P.,<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<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>a<\/em><sub>2<\/sub>\u00a0=\u00a0<em>a <\/em>+ (2 \u2212 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">38 =\u00a0<em>a <\/em>+ <em>d<\/em>\u00a0 &#8212;&#8212;&#8212;&#8212;- (i)<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>6<\/sub>\u00a0=\u00a0<em>a <\/em>+ (6 \u2212 1)<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">\u221222 =\u00a0<em>a <\/em>+ 5<em>d<\/em>\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;- (ii)<br \/>\n<\/span><span style=\"color: #000000;\">On subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get<br \/>\n<\/span><span style=\"color: #000000;\">\u2212 22 \u2212 38 = 4<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\">\u221260 = 4<em>d<br \/>\n<\/em><\/span><span style=\"color: #000000;\"><em>d<\/em>\u00a0= \u221215<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em>\u00a0=\u00a0<em>a<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>d<\/em>\u00a0= 38 \u2212 (\u221215) = 53<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>3<\/sub>\u00a0=\u00a0<em>a\u00a0<\/em>+ 2<em>d\u00a0<\/em>= 53 + 2 (\u221215) = 23<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>4<\/sub>\u00a0=\u00a0<em>a<\/em>\u00a0+ 3<em>d<\/em>\u00a0= 53 + 3 (\u221215) = 8<br \/>\n<\/span><span style=\"color: #000000;\"><em>a<\/em><sub>5<\/sub>\u00a0=\u00a0<em>a<\/em>\u00a0+ 4<em>d<\/em>\u00a0= 53 + 4 (\u221215) = \u22127<br \/>\n<\/span><span style=\"color: #000000;\">Therefore, The Series is\u00a0<strong>53, 38, 23, 8, -7, -22.<\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Which term of the A.P. 3, 8, 13, 18, \u2026 is 78?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Given the A.P. series as 3, 8, 13, 18, \u2026<br \/>\nFirst term, a\u00a0= 3<br \/>\nCommon difference, d\u00a0=\u00a0a<sub>2<\/sub>\u00a0\u2212\u00a0a<sub>1<\/sub>\u00a0= 8 \u2212 3 = 5<br \/>\n<em>a<sub>n<\/sub> <\/em>= 78<br \/>\n<em>n<\/em> = ?<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<br \/>\n<\/em>78 = 3 + (<em>n<\/em>\u00a0\u22121)5<br \/>\n75 = (<em>n <\/em>\u2212 1)5<br \/>\n(<em>n <\/em>\u2212 1) = 15<br \/>\n<em>n<\/em>\u00a0= 16<br \/>\nHence, 16<sup>th<\/sup>\u00a0term of this A.P. is 78.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Find the number of terms in each of the following A.P.<br \/>\n<\/strong><strong>(i) <\/strong>7, 13, 19, \u2026, 205<br \/>\n<strong>(ii) <\/strong>18, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{15\\frac{1}{2}}\" alt=\"\\mathbf{15\\frac{1}{2}}\" align=\"absmiddle\" \/>, 13, \u2026\u2026., -47<\/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) 7, 13, 19, \u2026, 205<br \/>\n<\/strong>First term, <em>a\u00a0<\/em>= 7<br \/>\nCommon difference, <em>d<\/em>\u00a0=\u00a0<em>a<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= 13 \u2212 7 = 6<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0= 205<br \/>\nn = ?<br \/>\nAs we know, for an A.P.,<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a<\/em>\u00a0+ (<em>n<\/em>\u00a0\u2212 1)\u00a0<em>d<br \/>\n<\/em>205 = 7 + (<em>n\u00a0<\/em>\u2212 1) 6<br \/>\n198 = (<em>n<\/em>\u00a0\u2212 1) 6<br \/>\n33 = (<em>n<\/em>\u00a0\u2212 1)<br \/>\n<em>n<\/em>\u00a0= 34<br \/>\nSo, there are 34 terms in the given A.P.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 18, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{15\\frac{1}{2}}\" alt=\"\\mathbf{15\\frac{1}{2}}\" align=\"absmiddle\" \/>, 13, \u2026\u2026., -47<br \/>\n<\/strong>First term, <em>a\u00a0<\/em>= 18<br \/>\nCommon difference, <em>d<\/em> = <em>a<sub>2 <\/sub>&#8211; a<sub>1\u00a0<\/sub><\/em>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?15\\frac{1}{2}\" alt=\"15\\frac{1}{2}\" align=\"absmiddle\" \/> &#8211; 18<br \/>\nd = (31 &#8211; 36)\/2 = -5\/2<br \/>\na<sub>n<\/sub>\u00a0= -47<br \/>\nAs we know, for an A.P.,<br \/>\na<sub>n<\/sub> = a + (n \u2212 1)d<br \/>\n-47 = 18 + (n &#8211; 1)(-5\/2)<br \/>\n-47 &#8211; 18 = (n &#8211; 1)(-5\/2)<br \/>\n-65 = (n &#8211; 1)(-5\/2)<br \/>\n(n &#8211; 1) = -130\/-5<br \/>\n(n &#8211; 1) = 26<br \/>\nn\u00a0= 27<br \/>\nHence, there are 27 terms in the given A.P.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6.\u00a0Check whether -150 is a term of the A.P. 11, 8, 5, 2, \u2026<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Given :<br \/>\nFirst term<em>, a<\/em>\u00a0= 11<br \/>\nCommon difference<em>, d<\/em>\u00a0=\u00a0<em>a<\/em><sub>2 <\/sub>\u2212 <em>a<\/em><sub>1<\/sub> = 8 \u2212 11 = \u22123<br \/>\nSuppose -150 is the nth term of the given A.P. then it can be written as<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<br \/>\n<\/em>-150 = 11 + (<em>n<\/em> &#8211; 1)(-3)<br \/>\n-150 = 11 &#8211; 3<em>n<\/em> + 3<br \/>\n-164 = -3<em>n<br \/>\n<\/em><em>n<\/em>\u00a0= 164\/3<br \/>\nClearly,\u00a0<em>n<\/em>\u00a0is not an integer but a fraction.<br \/>\nTherefore, \u2013 150 is <strong>not a term of this A.P.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. Find the 31<sup>st<\/sup>\u00a0term of an A.P. whose 11<sup>th<\/sup>\u00a0term is 38 and the 16<sup>th<\/sup>\u00a0term is 73.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Given :<br \/>\na<sub>11<\/sub>\u00a0= 38<br \/>\na<sub>16<\/sub>\u00a0= 73<br \/>\nWe know that,<br \/>\na<sub>n<\/sub>\u00a0=\u00a0a+(n\u22121)d<br \/>\na<sub>11<\/sub> = a + (11 \u2212 1)d<br \/>\n38 = a + 10d\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (i)<br \/>\n<em>a<\/em><sub>16<\/sub>\u00a0=\u00a0<em>a<\/em> + (16 \u2212 1)<em>d<br \/>\n<\/em>73 =\u00a0<em>a <\/em>+ 15<em>d<\/em>\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get<br \/>\n15d \u2013 10d = 73 \u2013 38<br \/>\n5<em>d = <\/em>35<br \/>\n<em>d<\/em>\u00a0= 7<br \/>\nFrom equation\u00a0(i), we can write,<br \/>\n38 =\u00a0<em>a <\/em>+ 10 \u00d7 (7)<br \/>\n38 \u2212 70 =\u00a0a<br \/>\na\u00a0= \u221232<br \/>\na<sub>31<\/sub> = a + (31 \u2212 1) d<br \/>\na<sub>31<\/sub> = \u2212 32 + 30 (7)<br \/>\na<sub>31<\/sub> = \u2212 32 + 210<br \/>\na<sub>31<\/sub> = 178<br \/>\nHence, 31<sup>st<\/sup> term of the given A.P. is 178.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. An A.P. consists of 50 terms of which 3<sup>rd<\/sup>\u00a0term is 12 and the last term is 106. Find the 29<sup>th<\/sup>\u00a0term.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; \u00a0<\/strong>Given :<br \/>\na<sub>3<\/sub>\u00a0= 12<br \/>\na<sub>50<\/sub>\u00a0= 106<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>3<\/sub>\u00a0=\u00a0<em>a <\/em>+ (3 \u2212 1)<em>d<br \/>\n<\/em>12 =\u00a0<em>a <\/em>+ 2<em>d<\/em>\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<br \/>\n<em>a<\/em><sub>50\u00a0<\/sub>=\u00a0<em>a <\/em>+ (50 \u2212 1)<em>d<br \/>\n<\/em>106 =\u00a0<em>a <\/em>+ 49<em>d<\/em>\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<br \/>\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get<br \/>\n<em>(a + 49d) &#8211; (a <\/em>+ 2<em>d)<\/em> = 106 &#8211; 12<br \/>\n47<em>d = 94<br \/>\n<\/em><em>d<\/em> = 2<br \/>\nFrom equation\u00a0(i), we can write now,<br \/>\n12 =\u00a0<em>a <\/em>+ 2(2)<br \/>\n<em>a<\/em> = 12 \u2212 4 = 8<br \/>\n<em>a<\/em><sub>29<\/sub>\u00a0=\u00a0<em>a <\/em>+ (29 \u2212 1)<em>d<br \/>\n<\/em><em>a<\/em><sub>29<\/sub> = 8 + (28)2<br \/>\n<em>a<\/em><sub>29<\/sub> = 8 + 56<br \/>\n<em>a<\/em><sub>29<\/sub> = 64<br \/>\nSo the 29<sup>th<\/sup> term of the given A.P. will be 64.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. If the 3<sup>rd<\/sup>\u00a0and the 9<sup>th<\/sup> terms of an A.P. are 4 and \u2212 8, respectively. Which term of this A.P. is zero?<\/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>3<\/sub>\u00a0= 4<br \/>\n<em>a<\/em><sub>9<\/sub>\u00a0= \u22128<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n<\/em>\u22121)<em>d<br \/>\n<\/em><em>a<\/em><sub>3<\/sub>\u00a0=\u00a0<em>a <\/em>+ (3 \u2212 1)<em>d<br \/>\n<\/em>4 =\u00a0<em>a <\/em>+ 2<em>d<\/em>\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;(i)<br \/>\n<em>a<\/em><sub>9<\/sub>\u00a0=\u00a0<em>a <\/em>+ (9 \u2212 1)<em>d<br \/>\n<\/em>\u22128 =\u00a0<em>a <\/em>+ 8<em>d<\/em> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;(ii)<br \/>\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we will get here,<br \/>\n\u22128 \u22124 = (<em>a <\/em>+ 8<em>d<\/em>) &#8211; (<em>a <\/em>+ 2<em>d<\/em>)<br \/>\n\u221212 = 6<em>d<br \/>\n<\/em><em>d<\/em>\u00a0= \u22122<br \/>\nFrom equation\u00a0(i), we can write,<br \/>\n4 =\u00a0<em>a <\/em>+ 2(\u22122)<br \/>\n4 =\u00a0<em>a <\/em>\u2212 4<br \/>\n<em>a<\/em>\u00a0= 8<br \/>\nLet\u00a0<em>n<\/em><sup>th<\/sup>\u00a0term of this A.P. be zero.<br \/>\n<em>a<sub>n\u00a0<\/sub><\/em>=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<br \/>\n<\/em>0 = 8 + (<em>n <\/em>\u2212 1)(\u22122)<br \/>\n0 = 8 \u2212 2<em>n <\/em>+ 2<br \/>\n2<em>n\u00a0<\/em>= 10<br \/>\n<em>n<\/em>\u00a0= 5<br \/>\nHence, the 5<sup>th<\/sup> term of the given A.P. will be zero.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. If 17<sup>th<\/sup>\u00a0term of an A.P. exceeds its 10<sup>th<\/sup>\u00a0term by 7. Find the common difference.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>According to the question<\/span><br \/>\n<span style=\"color: #000000;\"><em>a<sub>17<\/sub>\u00a0<\/em>= <em>a<sub>10<\/sub>\u00a0<\/em>+ 7\u00a0<\/span><br \/>\n<span style=\"color: #000000;\"><em>a<\/em> + (17 \u2013 1)<em>d<\/em> = <em>a<\/em> + (10 \u2013 1)<em>d<\/em> + 7<\/span><br \/>\n<span style=\"color: #000000;\">(<em>a<\/em> + 16<em>d<\/em>) \u2212 (<em>a <\/em>+ 9<em>d<\/em>) = 7<br \/>\n7<em>d<\/em>\u00a0= 7<br \/>\n<em>d<\/em>\u00a0= 1<br \/>\nHence, the common difference of the given A.P. will be 1.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>11.\u00a0Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54<sup>th<\/sup>\u00a0term?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\n<em>a <\/em>= 3<br \/>\n<em>d<\/em>\u00a0=\u00a0<em>a<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= 15 \u2212 3 = 12<br \/>\nLet the nth term of the A.P. will be 132 more than its 54th term<\/span><br \/>\n<span style=\"color: #000000;\"><em>a<sub>n<\/sub>\u00a0<\/em>= <em>a<sub>54<\/sub>\u00a0<\/em>+ 132<\/span><br \/>\n<span style=\"color: #000000;\"><em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d <\/em>=\u00a0<em>a <\/em>+ (54 \u2212 1)<em>d +\u00a0<\/em>132<br \/>\n(<em>n <\/em>\u2212 1)12 = (53)(12) + 132<br \/>\n(<em>n <\/em>\u2212 1)12 = 636 + 132<br \/>\n(<em>n <\/em>\u2212 1)12 = 768<br \/>\n(<em>n <\/em>\u2212 1) = 64<br \/>\n<em>n<\/em> = 64 + 1<br \/>\n<em>n<\/em>\u00a0= 65<br \/>\nHence, 65<sup>th<\/sup> term will be 132 more than its 54<sup>th<\/sup> term.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>12. Two APs have the same common difference. The difference between their 100<sup>th<\/sup>\u00a0term is 100, what is the difference between their 1000<sup>th<\/sup>\u00a0terms?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Let, the first term of two APs be\u00a0<em>a<\/em><sub>1<\/sub>\u00a0and\u00a0<em>a<\/em><sub>2<\/sub>\u00a0respectively<br \/>\nAnd the common difference of these APs be\u00a0<em>d<\/em>.<br \/>\nFor First <em>A.P., we know,<br \/>\na<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<br \/>\n<\/em><em>a<\/em><sub>100<\/sub>\u00a0=\u00a0<em>a<\/em><sub>1 <\/sub>+ (100 \u2212 1)<em>d<br \/>\n<\/em><em>a<sub>100<\/sub>\u00a0=\u00a0a<\/em><sub>1<\/sub>\u00a0+ 99d<br \/>\n<em>a<\/em><sub>1000<\/sub>\u00a0=\u00a0<em>a<\/em><sub>1 <\/sub>+ (1000 \u2212 1)<em>d<br \/>\n<\/em><em>a<\/em><sub>1000<\/sub>\u00a0=\u00a0<em>a<\/em><sub>1 <\/sub>+ 999<em>d<\/em><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">For second A.P., we know,<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>100<\/sub>\u00a0=\u00a0<em>a<\/em><sub>2 <\/sub>+ (100 \u2212 1)<em>d<br \/>\na<sub>100<\/sub> =\u00a0<\/em><em>a<\/em><sub>2<\/sub>+99<em>d<br \/>\n<\/em><em>a<\/em><sub>1000<\/sub>\u00a0=\u00a0<em>a<\/em><sub>2 <\/sub>+ (1000\u22121)<em>d<br \/>\n<\/em><em>a<\/em><sub>1000<\/sub> =\u00a0<em>a<\/em><sub>2 <\/sub>+ 999<em>d<br \/>\n<\/em>According to the question, the difference between their 100th term is 100<br \/>\n(<em>a<sub>1<\/sub>\u00a0<\/em>+ 99<em>d<\/em>) \u2013 (<em>a<sub>2<\/sub>\u00a0<\/em>+ 99<em>d<\/em>) = 100<br \/>\n<em>a<\/em><sub>1 <\/sub>\u2212 <em>a<\/em><sub>2<\/sub> = 100\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<br \/>\nDifference between 1000<sup>th<\/sup>\u00a0terms of the two APs<br \/>\n(<em>a<\/em><sub>1 <\/sub>+ 999<em>d<\/em>) \u2212 (<em>a<\/em><sub>2<\/sub>+ 999<em>d<\/em>) = 100<br \/>\nFrom equation\u00a0(i),<br \/>\n(<em>a<\/em><sub>1 <\/sub>+ 999<em>d<\/em>) \u2212 (<em>a<\/em><sub>2<\/sub>+ 999<em>d<\/em>) = <em>a<\/em><sub>1 <\/sub>\u2212 <em>a<\/em><sub>2<br \/>\n<\/sub>Hence, the difference between their 1000<sup>th<\/sup> term will be 100.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>13. How many three digit numbers are divisible by 7?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>First three-digit number that is divisible by 7 are;<br \/>\nFirst number = 105<br \/>\nSecond number = 105 + 7 = 112<br \/>\nThird number = 112 + 7 = 119<br \/>\nTherefore, 105, 112, 119, \u2026<br \/>\nAs we know, the largest possible three-digit number is 999.<br \/>\nWhen we divide 999 by 7, the remainder will be 5.<br \/>\nTherefore, 999 &#8211; 5 = 994 is the maximum possible three-digit number that is divisible by 7.<br \/>\nNow the series is as follows.<br \/>\n105, 112, 119, \u2026, 994<br \/>\nLet 994 be the\u00a0nth term of this A.P.<br \/>\n<em>a<\/em> = 105<br \/>\n<em>d\u00a0<\/em>= 7<br \/>\n<em>a<sub>n<\/sub>\u00a0<\/em>= 994<br \/>\n<em>n\u00a0<\/em>= ?<br \/>\n<em>a<sub>n<\/sub>\u00a0<\/em>=\u00a0<em>a + (n \u2212 1)d<br \/>\n<\/em>994 = 105 + (<em>n <\/em>\u2212 1)7<br \/>\n889 = (<em>n <\/em>\u2212 1)7<br \/>\n(<em>n<\/em>\u22121) = 127<br \/>\n<em>n\u00a0<\/em>= 128<br \/>\nTherefore, 128 three-digit numbers are divisible by 7.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>14. How many multiples of 4 lie between 10 and 250?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\n12 is the minimum number that is divisible by 4 between 10 and 250.<br \/>\n<em>a<\/em> = 12<\/span><br \/>\n<span style=\"color: #000000;\"><em>d<\/em> = 4\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">When we divide 250 by 4, the remainder will be 2.<br \/>\nTherefore, 250 \u2212 2 = 248 is divisible by 4.<br \/>\nThe series is as follows, now;<br \/>\n12, 16, 20, 24, \u2026, 248<br \/>\n<em>a<\/em><sub>n<\/sub> =\u00a0248<br \/>\na<sub>n<\/sub> = a + (n \u2212 1)d<br \/>\n248 = 12 + (<em>n <\/em>&#8211; 1) \u00d7 4<br \/>\n236\/4 = n &#8211; 1<br \/>\n59 \u00a0= n &#8211; 1<br \/>\nn\u00a0= 60<br \/>\nTherefore, there are 60 multiples of 4 between 10 and 250.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>15. For what value of\u00a0<em>n<\/em>, are the\u00a0<em>n<\/em><sup>th<\/sup>\u00a0terms of two APs 63, 65, 67, and 3, 10, 17, \u2026 equal?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\nTwo APs as; 63, 65, 67,\u2026 and 3, 10, 17,\u2026.<br \/>\n<strong>Taking First AP,<br \/>\n<\/strong>63, 65, 67, \u2026<br \/>\n<em>a<\/em> = 63<br \/>\n<em>d<\/em> = a<sub>2<\/sub>\u2212 a<sub>1<\/sub> = 65 \u2212 63 = 2<br \/>\n<em>a<sub>n<\/sub>\u00a0<\/em>=\u00a0<em>a + (n \u2212 1)d<br \/>\n<\/em><em>a<sub>n<\/sub><\/em>= 63 + (<em>n <\/em>\u2212 1)2<br \/>\n<em>a<sub>n <\/sub><\/em>= 63 + 2<em>n <\/em>\u2212 2<br \/>\n<em>a<sub>n<\/sub><\/em> = 61 + 2<em>n<\/em>\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212; (i)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Taking second AP,<br \/>\n<\/strong>3, 10, 17, \u2026<br \/>\n<em>a<\/em> = 3<br \/>\n<em>d<\/em> = <em>a<sub>2<\/sub>\u00a0\u2212\u00a0a<sub>1<\/sub>\u00a0<\/em>= 10 \u2212 3 = 7<br \/>\n<em>a<sub>n<\/sub>\u00a0=\u00a0a + (n \u2212 1)d<br \/>\n<\/em><em>a<sub>n<\/sub>\u00a0=\u00a0<\/em>3 + (<em>n <\/em>\u2212 1)7<br \/>\n<em>a<sub>n<\/sub><\/em> = 3 + 7<em>n <\/em>\u2212 7<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0= 7<em>n <\/em>\u2212 4\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nAccording to the question, n<sup>th<\/sup> terms of both APs are equal<br \/>\n61 + 2<em>n<\/em> = 7<em>n <\/em>\u2212 4<br \/>\n61 + 4 = 5<em>n<br \/>\n<\/em>5<em>n<\/em>\u00a0= 65<br \/>\n<em>n<\/em>\u00a0= 13<br \/>\nTherefore, 13<sup>th<\/sup>\u00a0terms of both these A.P.s are equal to each other.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>16. Determine the A.P. whose third term is 16 and the 7<sup>th<\/sup>\u00a0term exceeds the 5<sup>th<\/sup>\u00a0term by 12.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211;\u00a0 <\/strong>Given :<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0= 16<br \/>\n<em>a<\/em> + (3 \u2212 1)<em>d<\/em>\u00a0= 16<br \/>\n<em>a <\/em>+ 2<em>d<\/em> = 16\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<br \/>\nIt is given that, 7<sup>th<\/sup>\u00a0term exceeds the 5<sup>th<\/sup>\u00a0term by 12.<br \/>\n<em>a<\/em><sub>7<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>5<\/sub>\u00a0= 12<br \/>\n[<em>a <\/em>+ (7 \u2212 1)<em>d<\/em>] \u2212 [<em>a\u00a0<\/em>+ (5 \u2212 1)<em>d<\/em>] = 12<br \/>\n(<em>a <\/em>+ 6<em>d<\/em>) \u2212 (<em>a <\/em>+ 4<em>d<\/em>) = 12<br \/>\n2<em>d<\/em>\u00a0= 12<br \/>\n<em>d<\/em>\u00a0= 6<br \/>\nFrom equation\u00a0(i), we get,<br \/>\n<em>a <\/em>+ 2(6) = 16<br \/>\n<em>a <\/em>+ 12 = 16<br \/>\n<em>a<\/em>\u00a0= 4<br \/>\nTherefore, A.P. will be4, 10, 16, 22, \u2026<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>17. Find the 20<sup>th<\/sup>\u00a0term from the last term of the A.P. 3, 8, 13, \u2026, 253.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\nd = 8 &#8211; 3 = 5<br \/>\nTherefore, we can write the given AP in reverse order as;<br \/>\n253, 248, 243, \u2026, 13, 8, 5<br \/>\nNow for the new AP,<br \/>\na = 253<br \/>\nd = 248 \u2212 253 = \u22125<br \/>\nn\u00a0= 20<br \/>\nTherefore, using nth term formula, we get,<br \/>\n<em>a<\/em><sub>20<\/sub>\u00a0=\u00a0<em>a <\/em>+ (20 \u2212 1)<em>d<br \/>\n<\/em><em>a<\/em><sub>20<\/sub> = 253 + (19)(\u22125)<br \/>\n<em>a<\/em><sub>20<\/sub> = 253 \u2212 95<br \/>\n<em>a<\/em>\u00a0= 158<br \/>\nTherefore, 20<sup>th<\/sup>\u00a0term from the last term of the AP 3, 8, 13, \u2026, 253<strong>.<\/strong>is 158.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>18. The sum of 4<sup>th<\/sup>\u00a0and 8<sup>th<\/sup>\u00a0terms of an A.P. is 24 and the sum of the 6<sup>th<\/sup>\u00a0and 10<sup>th<\/sup>\u00a0terms is 44. Find the first three terms of the A.P<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\n<em>a<sub>4<\/sub>\u00a0+ a<sub>8 <\/sub><\/em>= 24<em><br \/>\n<\/em><em>[a <\/em>+ (4 \u2212 1)<em>d] + [a + (8 \u2212 1)d] = <\/em>24<br \/>\n2<em>a + 10d = <\/em>24<br \/>\n<em>a<\/em> + <em>5d<\/em> = 12\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (i)<br \/>\n<em>a<sub>6<\/sub>\u00a0+ a<sub>10<\/sub>\u00a0<\/em>= 44<br \/>\n<em>(a <\/em>+ 5<em>d) + (a + 9d) = <\/em>44<br \/>\n2<em>a<\/em> + 14<em>d<\/em> = 44\u00a0 \u00a0 <\/span><br \/>\n<span style=\"color: #000000;\"><em>a<\/em> + 7<em>d<\/em> = 22\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (ii)<br \/>\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get,<br \/>\n2d\u00a0= 22 \u2212 12<br \/>\n2d\u00a0= 10<br \/>\n<em>d<\/em>\u00a0= 5<br \/>\nFrom equation\u00a0(i), we get,<br \/>\n<em>a <\/em>+ 5<em>d<\/em>\u00a0= 12<br \/>\n<em>a <\/em>+ 5(5) = 12<br \/>\n<em>a <\/em>+ 25 = 12<br \/>\n<em>a<\/em>\u00a0= \u221213<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0=\u00a0<em>a <\/em>+ <em>d<\/em> = \u2212 13 + 5 = \u22128<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0=\u00a0<em>a<\/em><sub>2<\/sub>+ <em>d<\/em> = \u2212 8 + 5 = \u22123<br \/>\nTherefore, the first three terms of this A.P. are \u221213, \u22128, and \u22123.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 \u00a0<\/strong>As, salary of Subba Rao is increasing by a fixed amount in every year hence this will form an AP with <\/span><br \/>\n<span style=\"color: #000000;\">First term (<em>a<\/em>) = 5000 <\/span><br \/>\n<span style=\"color: #000000;\">common difference (<em>d<\/em>)\u00a0= 200<\/span><br \/>\n<span style=\"color: #000000;\"><em>a<sub>n<\/sub>\u00a0<\/em>= 7000<\/span><br \/>\n<span style=\"color: #000000;\"><em>a + (n \u2013 1)d<\/em> = 7000<\/span><br \/>\n<span style=\"color: #000000;\">5000 + (n \u2013 1) 200 = 7000<\/span><br \/>\n<span style=\"color: #000000;\">n \u2013 1 = 2000\/200<\/span><br \/>\n<span style=\"color: #000000;\">n = 10 + 1<\/span><br \/>\n<span style=\"color: #000000;\">n = 11<\/span><br \/>\n<span style=\"color: #000000;\">Hence, the salary will be 7000 in 11<sup>th<\/sup> year.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the\u00a0<em>n<\/em><sup>th<\/sup>\u00a0week, her weekly savings become Rs 20.75, find\u00a0<em>n.<\/em><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.<br \/>\nFirst term, <em>a\u00a0<\/em>= 5<br \/>\nCommon difference, <em>d<\/em> = 1.75<br \/>\n<em>a<sub>n\u00a0<\/sub><\/em>= 20.75<br \/>\nn = ?<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n <\/em>\u2212 1)<em>d<br \/>\n<\/em>20.75 = 5 + (<em>n<\/em> &#8211; 1)\u00d71.75<br \/>\n15.75 = (<em>n<\/em> &#8211; 1)\u00d71.75<br \/>\n(<em>n<\/em> &#8211; 1) = 15.75\/1.75<br \/>\n(<em>n<\/em> &#8211; 1) = 1575\/175<br \/>\n(<em>n<\/em> &#8211; 1) = 9<br \/>\n<em>n<\/em> &#8211; 1 = 9<br \/>\n<em>n<\/em>\u00a0= 10<br \/>\nHence,\u00a0<em>n<\/em>\u00a0is 10.<\/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 5.2 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":[1045,1074,1076,1044,1049,1048],"class_list":["post-6042","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-5-2-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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