{"id":6041,"date":"2023-08-12T04:54:31","date_gmt":"2023-08-12T04:54:31","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6041"},"modified":"2023-08-12T04:54:31","modified_gmt":"2023-08-12T04:54:31","slug":"ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-1\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1"},"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 5.1 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-5-2\" 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-5-3\" 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;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 5.1<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) <\/strong>The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong>It can be observed that<br \/>\nTaxi fare for 1st km = 15<br \/>\nTaxi fare for first 2 km = 15 + 8 = 23<br \/>\nTaxi fare for first 3 km = 23 + 8 = 31<br \/>\nTaxi fare for first 4 km = 31 + 8 = 39<br \/>\nThus, 15, 23, 31, 39 \u2026 forms <strong>an A.P<\/strong>. because every next term is 8 more than the preceding term.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) <\/strong>The amount of air present in a cylinder when a vacuum pump removes 1\/4\u00a0of the air remaining in the cylinder at a time.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Solution &#8211; <\/strong>Let the volume of air in a cylinder, initially, be\u00a0<em>V<\/em>\u00a0litres.<br \/>\nIn each stroke, the vacuum pump removes 1\/4th of air remaining in the cylinder at a time. Or we can say, after every stroke, 1 &#8211; 1\/4 = 3\/4th part of air will remain.<br \/>\nTherefore, volumes will be\u00a0<em>V<\/em>, 3<em>V<\/em>\/4 , (3<em>V<\/em>\/4)<sup>2<\/sup>\u00a0, (3<em>V<\/em>\/4)<sup>3 <\/sup>\u2026 and so on<br \/>\nClearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is <strong>not an A.P<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) <\/strong>The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong>Cost of digging for first metre = 150<br \/>\nCost of digging for first 2 metres = 150 + 50 = 200<br \/>\nCost of digging for first 3 metres = 200 + 50 = 250<br \/>\nCost of digging for first 4 metres =\u00a0250 + 50 = 300<br \/>\nClearly, 150, 200, 250, 300 \u2026 forms <strong>an A.P<\/strong>. with a common difference of 50 between each term.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) <\/strong>The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.<strong><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>We know that if Rs. P\u00a0is deposited at\u00a0<em>r<\/em>% compound interest per annum for n years, the amount of money will be: P(1 + r\/100)<sup>n<br \/>\n<\/sup>Therefore, after each year, the amount of money will be;<br \/>\n10000(1 + 8\/100), 10000(1 + 8\/100)<sup>2<\/sup>, 10000(1 + 8\/100)<sup>3<\/sup>\u2026\u2026<br \/>\nClearly, the terms of this series do not have the common difference between them. Therefore, this is\u00a0<strong>not an A.P.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>2. Write first four terms of the A.P. when the first term a and the common difference are given as follows<\/strong>:<br \/>\n<strong>(i)\u00a0<\/strong><em>a<\/em>\u00a0= 10,\u00a0<em>d<\/em>\u00a0= 10<br \/>\n<strong>(ii)\u00a0<\/strong><em>a<\/em>\u00a0= -2,\u00a0<em>d<\/em>\u00a0= 0<strong><br \/>\n(iii)<\/strong>\u00a0<em>a<\/em>\u00a0= 4,\u00a0<em>d<\/em>\u00a0= \u2013 3<strong><br \/>\n(iv)<\/strong>\u00a0<em>a<\/em>\u00a0= -1\u00a0<em>d<\/em>\u00a0= 1\/2<strong><br \/>\n(v)\u00a0<\/strong><em>a<\/em>\u00a0= \u2013 1.25,\u00a0<em>d<\/em>\u00a0= \u2013 0.25<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\">The general form of an arithmetic progression is a, (a + d), (a + 2d), (a + 3d), \u2026 where <strong>a<\/strong> is the <strong>first term<\/strong> and <strong>d<\/strong> is a <strong>common difference<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(i)\u00a0<i>a<\/i>\u00a0= 10,\u00a0<i>d<\/i>\u00a0= 10<br \/>\n<\/strong>First term, a = 10<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Second term, a + d = 10 + 10 = 20<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Third term, a + 2d = 10 + 20 = 30<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Fourth term, a + 3d = 10 + 30 = 40<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">The first four terms of the AP are 10, 20, 30, and 40.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(ii)\u00a0<i>a<\/i>\u00a0= &#8211; 2,\u00a0<i>d<\/i>\u00a0= 0<br \/>\n<\/strong>First term, a = &#8211; 2<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Second term, a + d = &#8211; 2 + 0 = &#8211; 2<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Third term, a + 2d = &#8211; 2 + 0 = &#8211; 2<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Fourth term, a + 3d = &#8211; 2 + 0 = &#8211; 2<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">The first four terms of the AP are &#8211; 2, &#8211; 2, &#8211; 2, and -2.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(iii) <i>a<\/i>\u00a0= 4,\u00a0<i>d<\/i>\u00a0= &#8211; 3<br \/>\n<\/strong>First term, a = 4<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Second term, a + d = 4 + (- 3) = 1<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Third term, a + 2d = 4 + 2(- 3) = 4 &#8211; 6 = &#8211; 2<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Fourth term, a + 3d = 4 + 3(- 3) = 4 &#8211; 9 = &#8211; 5<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">The first four terms of the AP are 4, 1, &#8211; 2, &#8211; 5.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(iv)\u00a0<i>a<\/i>\u00a0= &#8211; 1,\u00a0<i>d<\/i>\u00a0= 1\/2<br \/>\n<\/strong>First term,\u00a0a = &#8211; 1<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Second term,\u00a0a + d = &#8211; 1 + 1\/2 = &#8211; 1\/2<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Third term,\u00a0a + 2d = &#8211; 1 + 2(1\/2) = &#8211; 1 + 1 = 0<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Fourth term,\u00a0a + 3d = &#8211; 1 + 3(1\/2) = &#8211; 1 + 3\/2 = 1\/2<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">The first four terms of the AP are &#8211; 1, &#8211; 1\/2, 0, 1\/2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(v)\u00a0<i>a<\/i>\u00a0= &#8211; 1.25,\u00a0<i>d<\/i>\u00a0= &#8211; 0.25<br \/>\n<\/strong>First term,\u00a0a = &#8211; 1.25<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Second term,\u00a0a + d\u00a0= -1.25 + (-0.25)\u00a0= &#8211; 1.25 &#8211; 0.25\u00a0= &#8211; 1.5<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Third term,\u00a0a + 2d\u00a0= &#8211; 1.25 + 2 (- 0.25)\u00a0= &#8211; 1.25 &#8211; 0.50\u00a0= &#8211; 1.75<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Fourth term,\u00a0a + 3d\u00a0= &#8211; 1.25 + 3 (-0.25)\u00a0= &#8211; 1.25 &#8211; 0.75\u00a0= &#8211; 2<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">The first four terms of the AP are &#8211; 1.25, &#8211; 1.5, &#8211; 1.75, and &#8211; 2.00.\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. For the following A.P.s, write the first term and the common difference.<br \/>\n(i) <\/strong>3, 1, \u2013 1, \u2013 3 \u2026<br \/>\n<strong>(ii) <\/strong>-5, \u2013 1, 3, 7 \u2026<br \/>\n<strong>(iii) <\/strong>1\/3, 5\/3, 9\/3, 13\/3 \u2026.<br \/>\n<strong>(iv) <\/strong>0.6, 1.7, 2.8, 3.9 \u2026<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solutions &#8211;\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\">The general form of an arithmetic progression is a, (a + d), (a + 2d), (a + 3d),\u2026 where <strong>a<\/strong> is the <strong>first term<\/strong> and <strong>d<\/strong> is a <strong>common difference<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(i) 3, 1, \u2013 1, \u2013 3 \u2026<br \/>\n<\/strong>First term,\u00a0<em>a<\/em>\u00a0= 3<br \/>\nCommon difference,\u00a0<em>d<\/em>\u00a0= Second term \u2013 First term<br \/>\n\u21d2\u00a0 1 \u2013 3 = -2<br \/>\n\u21d2\u00a0 d = -2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(ii) -5, \u2013 1, 3, 7 \u2026<br \/>\n<\/strong>First term,\u00a0<em>a<\/em>\u00a0= -5<br \/>\nCommon difference,\u00a0<em>d<\/em>\u00a0= Second term \u2013 First term<br \/>\n\u21d2\u00a0( \u2013 1)-( \u2013 5) = \u2013 1+5 = 4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(iii) <\/strong><strong>1\/3, 5\/3, 9\/3, 13\/3, \u2026<br \/>\n<\/strong>First term, <em>a<\/em>\u00a0= 1\/3<br \/>\nCommon difference,\u00a0<em>d<\/em>\u00a0= Second term \u2013 First term<br \/>\n\u21d2\u00a05\/3 \u2013 1\/3 = 4\/3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(iv) <\/strong><strong>0.6, 1.7, 2.8, 3.9, \u2026<br \/>\n<\/strong>First term,\u00a0<em>a<\/em>\u00a0= 0.6<br \/>\nCommon difference,\u00a0<em>d<\/em>\u00a0= Second term \u2013 First term<br \/>\n\u21d2 1.7 \u2013 0.6 =\u00a01.1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>4. Which of the following are APs? If they form an A.P. find the common difference\u00a0<em>d<\/em>\u00a0and write three more terms.<br \/>\n<\/strong><strong>(i) <\/strong>2, 4, 8, 16 \u2026<br \/>\n<strong>(ii) <\/strong>2, 5\/2, 3, 7\/2 \u2026.<br \/>\n<strong>(iii)<\/strong> -1.2, -3.2, -5.2, -7.2 \u2026<br \/>\n<strong>(iv)<\/strong> -10, \u2013 6, \u2013 2, 2 \u2026<br \/>\n<strong>(v)<\/strong> 3, 3 +\u00a0\u221a2, 3\u00a0+ 2\u221a2, 3\u00a0+ 3\u221a2<br \/>\n<strong>(vi)<\/strong> 0.2, 0.22, 0.222, 0.2222 \u2026.<br \/>\n<strong>(vii)<\/strong> 0, \u2013 4, \u2013 8, \u2013 12 \u2026<br \/>\n<strong>(viii)<\/strong> -1\/2, -1\/2, -1\/2, -1\/2 \u2026.<br \/>\n<strong>(ix)<\/strong> 1, 3, 9, 27 \u2026<br \/>\n<strong>(x)\u00a0<\/strong><em>a<\/em>, 2<em>a<\/em>, 3<em>a<\/em>, 4<em>a<\/em>\u00a0\u2026<br \/>\n<strong>(xi)\u00a0<\/strong><em>a<\/em>,\u00a0<em>a<\/em><sup>2<\/sup>,\u00a0<em>a<\/em><sup>3<\/sup>,\u00a0<em>a<\/em><sup>4<\/sup>\u00a0\u2026<br \/>\n<strong>(xii)<\/strong> \u221a2, \u221a8, \u221a18, \u221a32\u00a0\u2026<br \/>\n<strong>(xiii)<\/strong> \u221a3, \u221a6, \u221a9, \u221a12\u00a0\u2026<br \/>\n<strong>(xiv)<\/strong> 1<sup>2<\/sup>, 3<sup>2<\/sup>, 5<sup>2<\/sup>, 7<sup>2<\/sup>\u00a0\u2026<br \/>\n<strong>(xv)<\/strong> 1<sup>2<\/sup>, 5<sup>2<\/sup>, 7<sup>2<\/sup>, 7<sup>3<\/sup>\u00a0\u2026<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(i) 2, 4, 8, 16 \u2026<br \/>\n<\/strong>Here, the common difference is;<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub> = 4 \u2013 2 = 2<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub> = 8 \u2013 4 = 4<br \/>\n<em>a<sub>4<\/sub><\/em>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub> = 16 \u2013 8 = 8<br \/>\n<em>(a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) \u2260 (<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nSo, 2, 4, 8, 16, &#8230; are <strong>not in A.P.<\/strong>, because the common difference is not equal.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) \u00a02, 5\/2 ,3, 7\/2, &#8230;<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0=\u00a05\/2-2 = 1\/2<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0=\u00a03-5\/2 = 1\/2<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0=\u00a07\/2-3 = 1\/2<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) = (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) =\u00a0 (<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\n2, 5\/2 ,3, 7\/2 forms <strong>an A.P.<\/strong> and <strong>common difference<\/strong> is <strong>1\/2<br \/>\n<\/strong>The next three terms are;<br \/>\nFifth term <em>a<\/em><sub>5<\/sub> = a\u2081 + 4d = 2 + 4 (1\/2) = 4<br \/>\nSixth term <em>a<\/em><sub>6 <\/sub>= a\u2081 + 5d = 2 + 5 \u00d7 1\/2 = 2 + 5\/2 = (4 + 5)\/2 = 9\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Seventh term <em>a<\/em><sub>7\u00a0 <\/sub>= a + 6d = 2 + 6 \u00d7 1\/2 = 5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) &#8211; 1.2, &#8211; 3.2, &#8211; 5.2, &#8211; 7.2, &#8230;<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= (-3.2)-(-1.2) = -2<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0= (-5.2)-(-3.2) = -2<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= (-7.2)-(-5.2) = -2<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) = (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) =\u00a0 (<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore,\u00a0<strong><em>d<\/em>\u00a0= -2<\/strong> and the given series are in <strong>A.P<\/strong>.<br \/>\nHence, next three terms are;<br \/>\nFifth term <em>a<\/em><sub>5<\/sub>\u00a0= a<sub>\u2081<\/sub>\u00a0+ 4d = &#8211; 1.2 + 4(- 2) = -1.2 &#8211; 8 = &#8211; 9.2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sixth term <em>a<\/em><sub>6<\/sub>\u00a0= a<sub>\u2081<\/sub>\u00a0+ 5d = &#8211; 1.2 + 5(- 2) = &#8211; 1.2 &#8211; 10 = &#8211; 11.2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Seventh term <em>a<\/em><sub>7<\/sub>\u00a0= a<sub>\u2081<\/sub>\u00a0+ 6d = &#8211; 1.2 + 6(- 2) = &#8211; 1.2 &#8211; 12 = &#8211; 13.2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) &#8211; 10, &#8211; 6, &#8211; 2, 2, &#8230;<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= (-6)-(-10) = 4<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0= (-2)-(-6) = 4<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= (2 -(-2) = 4<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) = (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) =\u00a0 (<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore,\u00a0<strong><em>d<\/em>\u00a0= 4<\/strong> and the given numbers are in <strong>A.P<\/strong>.<br \/>\nHence, next three terms are;<br \/>\nFifth Term <em>a<\/em><sub>5<\/sub> = a<sub>\u2081<\/sub> + 4d = &#8211; 10 + 16 = 6<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sixth Term <em>a<\/em><sub>6 <\/sub>= a<sub>\u2081<\/sub> + 5d = &#8211; 10 + 20 = 10<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Seventh Term <em>a<\/em><sub>7<\/sub> = a<sub>\u2081<\/sub> + 6d = &#8211; 10 + 24 = 14<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(v) 3, 3 + \u221a2, 3 + 2\u221a2, 3 + 3\u221a2, &#8230;<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= 3+\u221a2-3 = \u221a2<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0= (3+2\u221a2)-(3+\u221a2) = \u221a2<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= (3+3\u221a2) \u2013 (3+2\u221a2) = \u221a2<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) = (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) =\u00a0 (<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore,\u00a0<strong><em>d<\/em>\u00a0=\u00a0\u221a2\u00a0<\/strong>and the given series forms <strong>an<\/strong> <strong>A.P<\/strong>.<br \/>\nHence, next three terms are;<br \/>\nFifth term <em>a<\/em><sub>5<\/sub>\u00a0= a\u2081\u00a0+ 4d = 3 + 4 \u00d7 \u221a2 = 3 + 4\u221a2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sixth term <em>a<\/em><sub>6<\/sub>\u00a0= a\u2081\u00a0+ 5d = 3 + 5 \u00d7 \u221a2 = 3 + 5\u221a2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Seventh term <em>a<\/em><sub>7<\/sub>\u00a0= a\u2081\u00a0+ 6d = 3 + 6 \u00d7 \u221a2 = 3 + 6\u221a2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(vi) 0.2, 0.22, 0.222, 0.2222, &#8230;<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0=\u00a00.22-0.2 = 0.02<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0=\u00a00.222-0.22 = 0.002<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0=\u00a00.2222-0.222 = 0.0002<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) \u2260 (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) <em>\u2260 <\/em>(<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore, and the given series <strong>doesn\u2019t forms an A.P.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(vii) 0, -4, -8, -12 \u2026<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0=\u00a0(-4)-0 = -4<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0=\u00a0(-8)-(-4) = -4<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0=\u00a0(-12)-(-8) = -4<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) = (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) <em>= <\/em>(<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore,\u00a0<strong><em>d<\/em>\u00a0=\u00a0-4\u00a0<\/strong>and the given series forms <strong>an A.P<\/strong>.<br \/>\nHence, next three terms are;<br \/>\nFifth term <em>a<\/em><sub>5<\/sub>\u00a0= a\u2081\u00a0+ 4d = 0 + 4(- 4) = &#8211; 16<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sixth term <em>a<\/em><sub>6<\/sub>\u00a0= a\u2081\u00a0+ 5d = 0 + 5(- 4) = &#8211; 20<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Seventh term <em>a<\/em><sub>7<\/sub>\u00a0= a\u2081\u00a0+ 6d = 0 + 6(- 4) = &#8211; 24<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(viii) -1\/2, -1\/2, -1\/2, -1\/2 \u2026<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= (-1\/2) \u2013 (-1\/2) = 0<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0= (-1\/2) \u2013 (-1\/2) = 0<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= (-1\/2) \u2013 (-1\/2) = 0<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) = (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) <em>= <\/em>(<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore,\u00a0<strong><em>d<\/em>\u00a0= 0<\/strong> and the given series forms <strong>an A.P<\/strong>.<br \/>\nHence, next three terms are;<br \/>\nFifth term <em>a<\/em><sub>5<\/sub>\u00a0= a\u2081\u00a0+ 4d = &#8211; 1\/2\u00a0+ 4 (0) = &#8211; 1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sixth term <em>a<\/em><sub>6<\/sub>\u00a0= a\u2081\u00a0+ 5d = &#8211; 1\/2\u00a0+ 5 (0) = &#8211; 1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Seventh term <em>a<\/em><sub>7<\/sub>\u00a0= a\u2081\u00a0+ 6d = &#8211; 1\/2\u00a0+ 6 (0) = &#8211; 1\/2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ix) 1, 3, 9, 27 \u2026.<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0=\u00a03-1 = 2<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0=\u00a09-3 = 6<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0=\u00a027-9 = 18<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) \u2260 (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) <em>\u2260 <\/em>(<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore, and the given series <strong>doesn\u2019t forms an A.P.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(x)\u00a0<em>a<\/em>, 2<em>a<\/em>, 3<em>a<\/em>, 4<em>a<\/em>\u00a0\u2026<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0=\u00a02<em>a<\/em>\u2013<em>a\u00a0<\/em>=\u00a0<em>a<br \/>\n<\/em><em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0=\u00a03<em>a<\/em>-2<em>a<\/em>\u00a0=\u00a0<em>a<br \/>\n<\/em><em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0=\u00a04<em>a<\/em>-3<em>a<\/em>\u00a0=\u00a0<em>a<br \/>\n(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) = (a<sub>3<\/sub>\u00a0\u2013\u00a0a<sub>2<\/sub>) = (a<sub>2<\/sub>\u00a0\u2013\u00a0a<sub>1<\/sub>)<br \/>\n<\/em>Therefore,\u00a0<strong><em>d<\/em>\u00a0=\u00a0<em>a<\/em>\u00a0<\/strong>and the given series forms <strong>an A.P.<br \/>\n<\/strong>Hence, next three terms are;<br \/>\nFifth term <em>a<\/em><sub>5<\/sub>\u00a0= a\u2081\u00a0+ 4d = a + 4a = 5a<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sixth term <em>a<\/em><sub>6<\/sub>\u00a0= a\u2081\u00a0+ 5d = a + 5a = 6a<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Seventh term <em>a<\/em><sub>7<\/sub>\u00a0= a\u2081\u00a0+ 6d = a + 6a = 7a<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(xi)\u00a0<em>a<\/em>,\u00a0<em>a<\/em><sup>2<\/sup>,\u00a0<em>a<\/em><sup>3<\/sup>,\u00a0<em>a<\/em><sup>4<\/sup>\u00a0\u2026<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0=\u00a0<em>a<\/em><sup>2<\/sup>\u2013<em>a<\/em>\u00a0= a(<em>a <\/em>&#8211; 1)<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0=\u00a0<em>a<\/em><sup>3\u00a0<\/sup>\u2013<sup>\u00a0<\/sup><em>a<\/em><sup>2\u00a0<\/sup>=\u00a0<em>a<\/em><sup>2<\/sup>(<em>a <\/em>&#8211; 1)<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0=\u00a0<em>a<\/em><sup>4<\/sup>\u00a0\u2013\u00a0<em>a<\/em><sup>3\u00a0<\/sup>=\u00a0<em>a<\/em><sup>3<\/sup>(<em>a <\/em>&#8211; 1)<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) \u2260 (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) <em>\u2260 <\/em>(<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore, and the given series <strong>doesn\u2019t forms an A.P.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(xii) \u221a2, \u221a8, \u221a18, \u221a32\u00a0\u2026<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= \u221a8-\u221a2\u00a0\u00a0= 2\u221a2-\u221a2\u00a0= \u221a2<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>\u00a0= \u221a18-\u221a8\u00a0= 3\u221a2-2\u221a2\u00a0= \u221a2<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= 4\u221a2-3\u221a2\u00a0= \u221a2<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) = (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) <em>= <\/em>(<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore,\u00a0<strong><em>d<\/em>\u00a0=\u00a0\u221a2\u00a0<\/strong>and the given series forms <strong>an A.P.<br \/>\n<\/strong>Hence, next three terms are;<br \/>\nFifth term <em>a<\/em><sub>5<\/sub>\u00a0= a\u2081\u00a0+ 4d = \u221a2 + 4\u221a2 = 5\u221a2 = \u221a25 \u00d7 2 = \u221a50<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sixth term <em>a<\/em><sub>6<\/sub>\u00a0= a\u2081\u00a0+ 5d = \u221a2 + 5\u221a2 = 6\u221a2 = \u221a36 \u00d7 2 = \u221a72<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Seventh term <em>a<\/em><sub>7<\/sub>\u00a0= a\u2081\u00a0+ 6d = \u221a2 + 6\u221a2 = 7\u221a2 = \u221a49 \u00d7 2 = \u221a98<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(xiii)\u00a0\u221a3, \u221a6, \u221a9, \u221a12\u00a0\u2026<br \/>\n<\/strong><em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub> = \u221a6-\u221a3 = \u221a3\u00d7\u221a2-\u221a3 = \u221a3(\u221a2 &#8211; 1)<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub> = \u221a9-\u221a6 = 3-\u221a6 = \u221a3(\u221a3 &#8211; \u221a2)<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>3<\/sub> = \u221a12 \u2013 \u221a9 = 2\u221a3 \u2013 \u221a3\u00d7\u221a3 = \u221a3(2 &#8211; \u221a3)<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) \u2260 (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) <em>\u2260 <\/em>(<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore, and the given series <strong>doesn\u2019t forms an A.P.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(xiv) 1<sup>2<\/sup>, 3<sup>2<\/sup>, 5<sup>2<\/sup>, 7<sup>2<\/sup>\u00a0\u2026<br \/>\n<\/strong>1, 9, 25, 49 \u2026..<br \/>\nHere,<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= 9\u22121 = 8<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>2\u00a0<\/sub>= 25\u22129 = 16<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= 49\u221225 = 24<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) \u2260 (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) <em>\u2260 <\/em>(<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore, and the given series <strong>doesn\u2019t forms an A.P.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(xv) 1<sup>2<\/sup>, 5<sup>2<\/sup>, 7<sup>2<\/sup>, 73 \u2026<br \/>\n<\/strong>1, 25, 49, 73 \u2026<br \/>\nHere,<br \/>\n<em>a<\/em><sub>2<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>1<\/sub>\u00a0= 25\u22121 = 24<br \/>\n<em>a<\/em><sub>3<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>2\u00a0<\/sub>= 49\u221225 = 24<br \/>\n<em>a<\/em><sub>4<\/sub>\u00a0\u2212\u00a0<em>a<\/em><sub>3<\/sub>\u00a0= 73\u221249 = 24<br \/>\n<em>(a<sub>4<\/sub>\u00a0\u2013\u00a0a<sub>3<\/sub>) = (a<\/em><sub>3<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>2<\/sub>) <em>= <\/em>(<em>a<\/em><sub>2<\/sub>\u00a0\u2013\u00a0<em>a<\/em><sub>1<\/sub>)<br \/>\nTherefore,\u00a0<strong><em>d<\/em>\u00a0= 24\u00a0<\/strong>and the given series forms <strong>an A.P<\/strong>.<br \/>\nHence, next three terms are;<br \/>\nFifth term <em>a<\/em><sub>5<\/sub>\u00a0= a\u2081\u00a0+ 4d = 1 + 4 \u00d7 24 = 1 + 96 = 97<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sixth term <em>a<\/em><sub>6<\/sub>\u00a0= a\u2081\u00a0+ 5d = 1 + 5 \u00d7 24 = 1 + 120 = 121<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Seventh term <em>a<\/em><sub>7<\/sub>\u00a0= a\u2081\u00a0+ 6d = 1 + 6 \u00d7 24 = 1 + 144 = 145<\/span><\/p>\n<p><span style=\"font-family: Georgia, Palatino;\"><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><\/span><\/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 5.1 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; 5.2 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