{"id":4013,"date":"2021-02-26T12:57:28","date_gmt":"2021-02-26T07:27:28","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=4013"},"modified":"2021-09-15T11:40:39","modified_gmt":"2021-09-15T06:10:39","slug":"ap-ssc-10th-class-maths-solutions-chapter-6-ex-6-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-6-ex-6-2\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 6 Progressions Ex 6.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.2<\/h2>\n

10th Class Maths 6th Lesson Progressions Ex 6.2 Textbook Questions and Answers<\/h3>\n

Question 1.
\nFill in the blanks in the following table, given that ‘a\u2019 is the first term, d the common difference and an the nth term of the A.P:
\n\"AP
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\nFind the i) 30th term of the A.P.: 10, 7, 4,……
\nii) 11th term of the A.P.: -3, –\\(\\frac{1}{2}\\), 2,…
\nAnswer:
\ni) Given A.P. = 10, 7, 4, …….
\na1<\/sub> = 10; d = a2<\/sub> – a1<\/sub> = 7 – 10 = – 3
\nan<\/sub> = a + (n – 1) d
\na30<\/sub> = 10 + (30 – 1) (- 3) = 10 + 29 \u00d7 (- 3) = 10 – 87 = – 77<\/p>\n

ii) Given A.P. = – 3, –\\(\\frac{1}{2}\\), 2,…
\na1<\/sub> = -3; d = a2<\/sub> – a1<\/sub> = –\\(\\frac{1}{2}\\) – (-3) = – 3
\n= –\\(\\frac{1}{2}\\) + 3
\n= \\(\\frac{-1+6}{2}\\)
\n= \\(\\frac{5}{2}\\)
\nan<\/sub> = a + (n – 1) d
\n= -3 + (11-1) \u00d7 \\(\\frac{5}{2}\\)
\n= -3 + 10 \u00d7 \\(\\frac{5}{2}\\)
\n= -3 + 5 \u00d7 5
\n= -3 + 25
\n= 22<\/p>\n

Question 3.
\nFind the respective terms for the following APs.
\ni) a1<\/sub> = 2; a3<\/sub> = 26, find a2<\/sub>.
\nAnswer:
\nGiven: a1<\/sub> = a = 2 …….. (1)
\na3<\/sub> = a + 2d = 26 …….. (2
\nEquation (2) – equation (1)
\n\u21d2 (a + 2d) – a = 26 – 2
\n\u21d2 2d = 24
\nd = \\(\\frac{24}{2}\\) = 12
\nNow a2<\/sub> = a + d = 2 + 12 = 14<\/p>\n

ii) a2<\/sub> = 13; a4<\/sub> = 3, find a1<\/sub>, a3<\/sub>.
\nAnswer:
\nGiven: a2<\/sub> = a + d = 13 ….. (1)
\na4<\/sub> = a + 3d = 3 ….. (2)
\nSolving equations (1) and (2);
\n\"AP
\n\u2234 Substituting d = – 5 in equation (1) we get
\na + (-5) = 13
\n\u2234 a = 13 + 5 = 18 i.e., a1<\/sub> = 18
\na3<\/sub> = a + 2d = 18 + 2(- 5)
\n= 18 – 10 = 8<\/p>\n

iii) a1<\/sub> = 5; a4<\/sub> = 9\\(\\frac{1}{2}\\), find a2<\/sub>, a3<\/sub>.
\nAnswer:
\nGiven: a1<\/sub> = a = 5 ….. (1)
\na4<\/sub> = a + 3d = 9\\(\\frac{1}{2}\\) ….. (2)
\nSolving equations (1) and (2);
\n\"AP
\n\u21d2 3d = 4\\(\\frac{1}{2}\\)
\n\u21d2 3d = \\(\\frac{9}{2}\\)
\n\u21d2 d = \\(\\frac{9}{2 \\times 3}\\) = \\(\\frac{3}{2}\\)
\n\u2234 a2<\/sub> = a + d = 5 + \\(\\frac{3}{2}\\) = \\(\\frac{13}{2}\\)
\na3<\/sub> = a + 2d = 5 + 2 \u00d7 \\(\\frac{3}{2}\\) = 5 + 3 = 8<\/p>\n

\"AP<\/p>\n

iv) a1<\/sub> = -4; a6<\/sub> = 6, find a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>.
\nAnswer:
\nGiven: a1<\/sub> = a = -4 ….. (1)
\na6<\/sub> = a + 5d = 6 ….. (2)
\nSolving equations (1) and (2);
\n(-4) + 5d = 6
\n\u21d2 5d = 6 + 4
\n\u21d2 5d = 10
\n\u21d2 d = \\(\\frac{10}{5}\\)
\nNow
\n\u2234 a2<\/sub> = a + d = -4 + 2 = -2
\na3<\/sub> = a + 2d = -4 + 2 \u00d7 2 = -4 + 4 = 0
\na4<\/sub> = a + 3d = -4 + 3 \u00d7 2 = -4 + 6 = 2
\na5<\/sub> = a + 4d = -4 + 4 \u00d7 2 = -4 + 8 = 4<\/p>\n

v) a2<\/sub> = 38; a6<\/sub> = -22, find a1<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>.
\nAnswer:
\nGiven: a2<\/sub> = a + d = 38 ….. (1)
\na6<\/sub> = a + 5d = -22 ….. (2)
\nSubtracting (2) from (1) we get
\n\"AP
\nNow substituting, d = – 15 in equation (1), we get
\na + (- 15) = 38 \u21d2 a = 38 + 15 = 53
\nThus,
\na1<\/sub> = a = 53;
\na3<\/sub> = a + 2d = 53 + 2 \u00d7 (- 15) = 53 – 30 = 23;
\na4<\/sub> = a + 3d = 53 + 3 \u00d7 (- 15) = 53 – 45 = 8;
\na5<\/sub> = a + 4d = 53 + 4 \u00d7 (- 15) = 53 – 60 = – 7<\/p>\n

Question 4.
\nWhich term of the AP:
\n3, 8, 13, 18,…, is 78?
\nAnswer:
\nGiven: 3, 8, 13, 18, ……
\nHere a = 3; d = a2<\/sub> – a1<\/sub> = 8 – 3 = 5
\nLet \u201878\u2019 be the nth term of the given A.P.
\n\u2234 an<\/sub> = a + (n – 1) d
\n\u21d2 78 = 3 + (n – 1) 5
\n\u21d2 78 = 3 + 5n – 5
\n\u21d2 5n = 78 + 2
\n\u21d2 n = \\(\\frac{80}{2}\\) = 16
\n\u2234 78 is the 16th term of the given A.P.<\/p>\n

Question 5.
\nFind the number of terms in each of the following APs:
\ni) 7, 13, 19, ….., 205
\nAnswer:
\nGiven: A.P: 7, 13, 19, ……….
\nHere a1<\/sub> = a = 7; d = a2<\/sub> – a1<\/sub> = 13 – 7 = 6
\nLet 205 be the nth<\/sup> term of the given A.P.
\nThen, an<\/sub> = a + (n – 1) d
\n205 = 7 + (n- 1)6
\n\u21d2 205 = 7 + 6n – 6
\n\u21d2 205 = 6n + 1
\n\u21d2 6n = 205 – 1 = 204
\n\u2234 n = \\(\\frac{204}{6}\\) = 34
\n\u2234 34 terms are there.<\/p>\n

\"AP<\/p>\n

ii) 18, 15\\(\\frac{1}{2}\\), 13, …, -47
\nAnswer:
\nGiven: A.P: 18, 15\\(\\frac{1}{2}\\), 13, …….
\nHere a1<\/sub> = a = 18;
\nd = a2<\/sub> – a1<\/sub> = 15\\(\\frac{1}{2}\\) – 18 = -2\\(\\frac{1}{2}\\) = –\\(\\frac{5}{2}\\)
\nLet ‘-47’ be the nth term of the given A.P.
\nan<\/sub> = a + (n – 1) d
\n\"AP
\n\u21d2 -94 = 36 – 5n + 5
\n\u21d2 5n = 94 + 41
\n\u21d2 n = \\(\\frac{135}{5}\\) = 27
\n\u2234 27 terms are there.<\/p>\n

Question 6.
\nCheck whether, -150 is a term of the AP: 11, 8, 5, 2…
\nAnswer:
\nGiven: A.P. = 11, 8, 5, 2…
\nHere a1<\/sub> = a = 11;
\nd = a2<\/sub> – a1<\/sub> = 8 – 11 = -3
\nIf possible, take – 150 as the nth<\/sup> term of the given A.P.
\nan<\/sub> = a + (n – 1) d
\n\u21d2 -150 = 11 + (n – 1) \u00d7 (-3)
\n\u21d2 -150 = 11 – 3n + 3
\n\u21d2 14 – 3n = – 150
\n\u21d2 3n= 14 + 150 = 164
\n\u2234 n = \\(\\frac{164}{3}\\) = 54\\(\\frac{2}{3}\\)
\nHere n is not an integer.
\n\u2234 -150 is not a term of the given A.P.<\/p>\n

Question 7.
\nFind the 31st<\/sup> term of an A.P. whose 11th<\/sup> term is 38 and the 16th<\/sup> term is 73.
\nAnswer:
\nGiven: An A.P. whose
\n\"AP
\n\u21d2 -5d = -35
\n\u21d2 d = \\(\\frac{-35}{-5}\\) = 7
\nSubstituting d = 7 in the equation (1)
\nwe get,
\na + 10 x 7 = 38
\n\u21d2 a + 70 = 38
\n\u21d2 a = 38 – 70 = -32
\nNow, the 31st<\/sup> term = a + 30d
\n= (-32) + 30 \u00d7 7
\n= -32 + 210 = 178<\/p>\n

\"AP<\/p>\n

Question 8.
\nIf the 3rd and the 9th<\/sup> terms of an A.P are 4 and -8 respectively, which term of this A.P is zero?
\nAnswer:
\nGiven: An A.P. whose
\n\"AP
\nSubstituting d = -2 in equation (1) we get
\na + 2 \u00d7 (-2) = 4
\n\u21d2 a – 4 = 4
\n\u21d2 a = 4 + 4 = 8
\nLet nth<\/sup> term of the given A.P be equal to zero.
\nan = a + (n – 1)d
\n\u21d2 0 = 8 + (n – 1) \u00d7 (-2)
\n\u21d2 0 = 8 – 2n + 2
\n\u21d2 10 – 2n = 0
\n\u21d2 2n = 10 and n = \\(\\frac{10}{2}\\) = 5
\n\u2234 The 5th<\/sup> term of the given A.P is zero.<\/p>\n

Question 9.
\nThe 17th<\/sup> term of an A.P exceeds its 10 term by 7. Find the common difference.
\nAnswer:
\nGiven an A.P in which a17<\/sub> = a10<\/sub> + 7
\n\u21d2 a17<\/sub> – a10<\/sub> = 7
\nWe know that an<\/sub> = a + (n – 1)d
\n\"AP
\n\u21d2 d = \\(\\frac{7}{7}\\) = 1<\/p>\n

Question 10.
\nTwo APs have the same common difference. The difference between their 100th<\/sup> terms is 100, what is the difference between their 1000th<\/sup> terms?
\nAnswer:
\nLet the first A.P be:
\na, a + d, a + 2d, ……..
\nSecond A.P be:
\nb, b + d, b + 2d, b + 3d, ………
\nAlso, general term, an<\/sub> = a + (n – 1)d
\nGiven that, a100<\/sub> – b100<\/sub> = 100
\n\u21d2 a + 99d – (b + 99d) = 100
\n\u21d2 a – b = 100
\nNow the difference between their 1000th<\/sup> terms,
\n\"AP
\n\u2234 The difference between their 1000th<\/sup> terms is (a – b) = 100.
\nNote: If the common difference for any two A.Ps are equal then difference between nth<\/sup> terms of two A.Ps is same for all natural values of n.<\/p>\n

\"AP<\/p>\n

Question 11.
\nHow many three-digit numbers are divisible by 7?
\nAnswer:
\nThe least three digit number is 100.
\n\"AP
\n\u2234 The least 3 digit number divisible by 7 is 100 + (7 – 2) = 105
\nThe greatest 3 digit number is 999
\n\"AP
\n\u2234 The greatest 3 digit number divisible by 7 is 999 – 5 = 994.
\n\u2234 3 digit numbers divisible by 7 are
\n105, 112, 119,….., 994.
\na1<\/sub> = a = 105; d = 7; an<\/sub> = 994
\nan<\/sub> = a + (n – 1)d
\n\u21d2 994 = 105 + (n – 1)7
\n\u21d2 (n – 1)7 = 994 – 105
\n\u21d2 (n – 1)7 = 889
\n\u21d2 n – 1 = \\(\\frac{889}{7}\\) = 127
\n\u2234 n = 127 + 1 = 128
\n\u2234 There are 128, 3 digit numbers which are divisible by 7.
\n(or)
\n\\(\\frac{\\text { last number – first number }}{7}\\)
\n\\(\\frac{999-100}{7}\\)
\n\u2243 128.4 = 128 numbers divisible by 7.<\/p>\n

Question 12.
\nHow many multiples of 4 lie between 10 and 250?
\nAnswer:
\nGiven numbers: 10 to 250
\n\"AP
\n\u2234 Multiples of 4 between 10 and 250 are
\nFirst term: 10 + (4 – 2) = 12
\nLast term: 250 – 2 = 248
\n\u2234 12, 16, 20, 24, ….., 248
\na = a1<\/sub> = 12; d = 4; an<\/sub> = 248
\nan<\/sub> = a + (n – 1)d
\n248 = 12 + (n – 1) \u00d7 4
\n\u21d2 (n – 1)4 = 248 – 12
\n\u21d2 n – 1 = \\(\\frac{236}{4}\\) = 59
\n\u2234 n = 59 + 1 = 60
\nThere are 60 numbers between 10 and 250 which are divisible by 4.<\/p>\n

\"AP<\/p>\n

Question 13.
\nFor what value of n, are the nth<\/sup> terms of two APs: 63, 65, 67, ….. and 3, 10, 17,… equal?
\nAnswer:
\nGiven : The first A.P. is 63, 65, 67, ……
\nwhere a = 63, d = a2<\/sub> – a1<\/sub>,
\n\u21d2 d = 65 – 63 = 2
\nand the second A.P. is 3, 10, 17, …….
\nwhere a = 3; d = a2<\/sub> – a1<\/sub> = 10 – 3 = 7
\nSuppose the nth terms of the two A.Ps are equal, where an<\/sub> = a + (n – 1)d
\n\u21d2 63 + (n – 1)2 = 3 + (n – 1)7
\n\u21d2 63 + 2n – 2 = 3 + 7n – 7
\n\u21d2 61 + 2n = 7n – 4
\n\u21d2 7n – 2n = 61 + 4
\n\u21d2 5n = 65
\n\u21d2 n = \\(\\frac{65}{5}\\) = 13
\n\u2234 13th<\/sup> terms of the two A.Ps are equal.<\/p>\n

Question 14.
\nDetermine the AP whose third term is 16 and the 7th<\/sup> term exceeds the 5th<\/sup> term by 12.
\nAnswer:
\nGiven : An A.P in which
\na3<\/sub> = a + 2d = 16 …… (1)
\nand a7<\/sub> = a5<\/sub> + 12
\ni.e., a + 6d = a + 4d + 12
\n\u21d2 6d – 4d = 12
\n\u21d2 2d = 12
\n\u21d2 d = \\(\\frac{12}{2}\\) = 6
\nSubstituting d = 6 in equation (1) we get
\na + 2 \u00d7 6 = 16
\n\u21d2 a = 16 – 12 = 4
\n\u2234 The series\/A.P is
\na, a + d, a + 2d, a + 3d, …….
\n\u21d2 4, 4 + 6, 4 + 12, 4 + 18, ……
\n\u21d2 A.P.: 4, 10, 16, 22, …….<\/p>\n

Question 15.
\nFind the 20th<\/sup> term from the end of the AP: 3, 8, 13,…, 253.
\nAnswer:
\nGiven: An A.P: 3, 8, 13, …… , 253
\nHere a = a1<\/sub> = 3
\nd = a2<\/sub> – a1<\/sub> = 8 – 3 = 5
\nan<\/sub> = 253, where 253 is the last term
\nan<\/sub> = a + (n – l)d
\n\u2234 253 = 3 + (n – 1)5
\n\u21d2 253 = 3 + 5n – 5
\n\u21d2 5n = 253 + 2
\n\u21d2 n = \\(\\frac{255}{5}\\) = 51
\n\u2234 The 20th<\/sup> term from the other end would be
\n1 + (51 – 20) = 31 + 1 = 32
\n\u2234 a32<\/sub> = 3 + (32 – 1) \u00d7 5
\n= 3 + 31 \u00d7 5
\n= 3 + 155 = 158<\/p>\n

\"AP<\/p>\n

Question 16.
\nThe sum of the 4th<\/sup> and 8th<\/sup> terms of an AP is 24 and the sum of the 6th<\/sup> and 10th<\/sup> terms is 44. Find the first three terms of the AP.
\nAnswer:
\nGiven an A.P in which a4<\/sub> + a8<\/sub> = 24
\n\u21d2 a + 3d + a + 7d = 24
\n\u21d2 2a + 10d = 24
\n\u21d2 a + 5d = 12 ……. (1)
\nand a6<\/sub> + a10<\/sub> = 44
\n\u21d2 a + 5d + a + 9d = 44
\n\u21d2 2a + 14d = 44
\n\u21d2 a + 7d = 22 ……. (2)
\nAlso a + 5d = 12
\n\u21d2 a + 5(5) = 12
\n\u21d2 a + 25 = 12
\n\u21d2 a = 12 – 25 = -13
\n\u2234 The A.P is a, a + d, a + 2d, ……
\ni.e., – 13, (- 13 + 5), (-13 + 2 \u00d7 5)…
\n\u21d2 -13, -8, -3, …….<\/p>\n

Question 17.
\nSubba 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?
\nAnswer:
\nGiven: Salary of Subba Rao in 1995 = Rs. 5000
\nAnnual increment = Rs. 200
\ni.e., His salary increases by Rs. 200 every year.
\n\"AP
\nClearly 5000, 5200, 5400, forms an A.P in which a = 5000 and d = 200.
\nNow suppose that his salary reached Rs. 7000 after x – years.
\ni.e., an<\/sub> = 7000
\nBut, an<\/sub> = a + (n – 1)d
\n7000 = 5000 + (n – 1)200
\n\u21d2 7000 – 5000 = (n – 1)200
\n\u21d2 n – 1 = \\(\\frac{2000}{200}\\) = 10
\n\u21d2 n = 10 + 1
\n\u2234 In 11th<\/sup> year his salary reached Rs. 7000.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.2 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.2 10th Class Maths 6th Lesson Progressions Ex 6.2 Textbook Questions and Answers Question 1. Fill in the blanks in the following table, given …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4013"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=4013"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4013\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=4013"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=4013"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=4013"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}