{"id":26951,"date":"2021-06-28T18:27:33","date_gmt":"2021-06-28T12:57:33","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26951"},"modified":"2021-06-29T22:59:58","modified_gmt":"2021-06-29T17:29:58","slug":"ncert-solutions-for-class-10-maths-chapter-5-ex-5-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-5-ex-5-2\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 5 Arithmetic Progressions Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
\n\"NCERT
\nSolution:
\n(i) 28
\n(ii) 2
\n(iii) 46
\n(iv) 10
\n(v) 3.5<\/p>\n

Question 2.
\nChoose the correct choice in the following and justify:
\n(i) 30th term of the AP: 10, 7, 4, …, is
\n(A) 97
\n(B) 77
\n(C) -77
\n(D) -87<\/p>\n

(ii) 11th term of the AP: -3, \\(\\frac { -1 }{ 2 }\\) , 2, …, is
\n(A) 28
\n(B) 22
\n(C) -38
\n(D) -48
\nSolution:
\n(i) 10, 7, 4, …,
\na = 10, d = 7 – 10 = – 3, n = 30
\nan<\/sub> = a + (n – 1)d
\n\u21d2 a30<\/sub> = a + (30 – 1) d = a + 29 d = 10 + 29 (-3) = 10 – 87 = – 77
\nTherefore, 30th term of the sequence 10, 7, 4, is – 77 i.e., (C) is the correct choice.<\/p>\n

(ii) We have given the sequence,
\n\"NCERT
\nTherefore, 11th term of the sequence is 22 i.e., (B) is the correct choice.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn the following APs, find the missing terms in the boxes:
\n\"NCERT
\nSolution:
\n\"NCERT<\/p>\n

Question 4.
\nWhich term of the AP: 3, 8, 13, 18, …, is 78?
\nSolution:
\nGiven: 3, 8, 13, 18, ………,
\na = 3, d = 8 – 3 = 5
\nLet nth term is 78
\nan<\/sub> = 78
\na + (n – 1) d = 78
\n\u21d2 3 + (n – 1) 5 = 78
\n\u21d2 (n – 1) 5 = 78 – 3
\n\u21d2 (n – 1) 5 = 75
\n\u21d2 n – 1 = 15
\n\u21d2 n = 15 + 1
\n\u21d2 n = 16
\nHence, a16<\/sub> = 78<\/p>\n

Question 5.
\nFind the number of terms in each of the following APs:
\n(i) 7, 13, 19, …, 205
\n(ii) 18, 15\\(\\frac { 1 }{ 2 }\\), 13, …, -47
\nSolution:
\n(i) We have given the sequence 7,13,19,…. 205 Here, a = 7, d = 13-7 = 6 an = 205 Let n terms are in this AP.
\n\u2234 an<\/sub> = 205
\na + (n – 1)d= 205 (\u2234 an<\/sub> = a + (n – 1) d)
\nor, 7 + (n – 1) x 6 = 205 (\u2234 a = 7 and d = 6)
\nor, (n – 1)6 = 205 – 7
\nor, (n – 1) = \\(\\frac { 198 }{ 6 }\\)
\n\u2234 n = 33 + 1 = 34
\nTherefore, this sequence has 34 terms.<\/p>\n

(ii) We have given the sequence
\n\"NCERT<\/p>\n

Question 6.
\nCheck, whether -150 is a term of the AP: 11, 8, 5, 2, ….
\nSolution:
\n11, 8, 5, 2, …….
\nHere, a = 11, d = 8 – 11= -3, an<\/sub> = -150
\na + (n – 1) d = an
\n\u21d2 11 + (n – 1) (- 3) = -150
\n\u21d2 (n – 1) (- 3) = -150 – 11
\n\u21d2 -3 (n – 1) = -161
\n\u21d2 n – 1 = \\(\\frac { -161 }{ -3 }\\)
\n\u21d2 n = \\(\\frac { 161 }{ 3 }\\) + 1 = \\(\\frac { 164 }{ 3 }\\) = 53\\(\\frac { 4 }{ 3 }\\)
\nWhich is not an integral number.
\nHence, – 150 is not a term of the AP.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nFind the 31st<\/sup> term of an AP whose 11th term is 38 and the 16th term is 73.
\nSolution:
\na11<\/sub> = 38 and a16<\/sub> = 73
\n\u21d2 a11<\/sub> = a + (11 – 1) d \u21d2 a + 10d = 38 ….. (i)
\n\u21d2 a16<\/sub> = a + (16 – 1 )d \u21d2 a + 15d = 73 …(ii)
\nSubtracting eqn. (i) from (ii), we get
\na + 15d – a – 10d = 73 – 38
\n\u21d2 5d = 35
\n\u21d2 d = 1
\nFrom (i), a + 10 x 7 = 38
\n\u21d2 a = 38 – 70 = – 32
\na31<\/sub> = a + (31 – 1) d = a + 30d = – 32 + 30 x 7 = – 32 + 210 = 178
\nTherefore, the 31th term of this AP is 178.<\/p>\n

Question 8.
\nAn AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
\nSolution:
\nGiven:
\na50<\/sub> = 106
\na50<\/sub> = a + (50 – 1) d
\n\u21d2 a + 49d = 106 … (i)
\nand a3<\/sub> = 12
\n\u21d2 a3<\/sub> = a + (3 – 1 )d
\n\u21d2 a + 2d = 12 … (ii)
\nSubtracting eqn. (ii) from (i), we get
\na + 49d – a – 2d = 106 – 12
\n\u21d2 47d = 94
\n\u21d2 d = \\(\\frac { 94 }{ 47 }\\) = 2
\na + 2d = 12
\n\u21d2 a + 2 x 2 = 12
\n\u21d2 a + 4 = 12
\n\u21d2 a = 12 – 4 = 8
\na29<\/sub> = a + (29 – 1) d = a + 28d = 8 + 28 x 2 = 8 + 56 = 64
\nTherefore, the 29th<\/sup> term of this AP is 64.<\/p>\n

Question 9.
\nIf the 3rd and the 9th term of an AP are 4 and -8 respectively, which term of this AP is zero?
\nSolution:
\nGiven: a3<\/sub> = 4 and a9<\/sub> = – 8
\n\u21d2 a3<\/sub> = a + (3 – 1 )d \u21d2 a + 2d = 4 …(i)
\na9<\/sub> = a + (9 – 1) d \u21d2 a + 8d = -8 ….(ii)
\nSubtracting eqn. (i) from (ii), we get
\na + 8d – a – 2d = -8 – 4
\n\u21d2 6d = -12.
\n\u21d2 d = -2
\nNow,
\na + 2d = 4
\n\u21d2 a + 2(-2) = 4
\n\u21d2 a – 4 = 4
\n\u21d2 a = 4 + 4 = 8
\nLet an<\/sub> = 0
\n\u21d2 a + (n – 1) d = 0
\n\u21d2 8 + (n – 1) (- 2) = 0
\n\u21d2 8 = 2 (n – 1)
\n\u21d2 n – 1 = 4
\n\u21d2 n = 4 + 1 = 5
\nTherefore, 5th term of this AP is 0.<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nThe 17th term of an AP exceeds its 10th term by 7. Find the common difference.
\nSolution:
\nGiven: a17<\/sub> – a10<\/sub> = 7
\n\u21d2 [a + (17 – 1 ) d] – [a + (10 – 1 ) d] = 7
\n\u21d2 (a + 16d) – (a + 9d) = 7
\n\u21d2 7d = 7
\n\u21d2 d = 1
\nTherefore, the common difference of this AP is 1.<\/p>\n

Question 11.
\nWhich term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?
\nSolution:
\n3, 15, 27, 39, …..
\nHere, a = 3, d = 15 – 3 = 12
\nLet an<\/sub> = 132 + a54<\/sub>
\n\u21d2 an<\/sub> – a54<\/sub> = 132
\n\u21d2 [a + (n – 1) d] – [a + (54 – 1) d] = 132
\n\u21d2 a + nd – d – a – 53d = 132
\n\u21d2 12n – 54d = 132
\n\u21d2 12n – 54 x 12 = 132
\n\u21d2 (n – 54)12 = 132
\n\u21d2 n – 54 = 11
\n\u21d2 n = 11 + 54 = 65
\nTherefore, 65th term of this AP be 132 more than its 54th term.<\/p>\n

Question 12.
\nTwo APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
\nSolution:
\nLet a and A be the first term of two APs and d be the common difference.
\nGiven:
\na100<\/sub>\u00a0– A100<\/sub> = 100
\n\u21d2 a + 99d – A – 99d = 100
\n\u21d2 a – A = 100
\n\u21d2 a1000<\/sub> – A1000<\/sub> = a + 999d – A – 999d
\n\u21d2 a – A = 100
\n\u21d2 a1000<\/sub>\u00a0– A1000<\/sub> = 100 [From equation (i)]
\nTherefore, difference of their 1000th term is 100.<\/p>\n

Question 13.
\nHow many three-digit numbers are divisible by 7?
\nSolution:
\nThe three-digit numbers which are divisible by 7 are 105, 112, 119, ………., 994
\nHere, a = 105, d = 112 – 105 = 7 , an<\/sub> = 994
\na + (n – 1) d = 994
\n\u21d2 105 + (n – 1) 7 = 994
\n\u21d2 (n – 1) 7 = 994 – 105
\n\u21d2 7 (n – 1) = 889
\n\u21d2 n – 1 = 127
\n\u21d2 n = 127 + 1 = 128
\nTherefore, the number of three digit numbers which are divisible by 7 is 128.<\/p>\n

\"NCERT<\/p>\n

Question 14.
\nHow many multiples of 4 lie between 10 and 250?
\nSolution:
\nThe multiples of 4 between 10 and 250 be 12, 16, 20, 24,…., 248
\nHere, a = 12, d = 16 – 12 = 4, an<\/sub> = 248
\nan<\/sub> = a + (n – 1) d
\n\u21d2 248 = 12 + (n – 1) 4
\n\u21d2 248 – 12 = (n – 1) 4
\n\u21d2 236 = (n – 1) 4
\n\u21d2 59 = n – 1
\n\u21d2 n = 59 + 1 = 60
\nTherefore, number of terms between 10 and 250 which is multiple of 4 is 60.<\/p>\n

Question 15.
\nFor what value of n, the nth term of two APs: 63, 65, 61,… and 3, 10, 17,… are equal?
\nSolution:
\nFirst AP
\n63, 65, 67,…
\nHere, a = 63, d = 65 – 63 = 2
\nan<\/sub> = a + (n – 1) d = 63 + (n – 1) 2 = 63 + 2n – 2 = 61 + 2n
\nSecond AP
\n3, 10, 17, …
\nHere, a = 3, d = 10 – 3 = 7
\nan<\/sub> = a + (n – 1) d = 3 + (n – 1)7 = 3 + 7n – 7 = 7n – 4
\nNow, an<\/sub>\u00a0= an<\/sub>
\n\u21d2 61 + 2n = 7n – 4
\n\u21d2 61 + 4 = 7n – 2n
\n\u21d2 65 = 5n
\n\u21d2 n = 13
\nTherefore, the 13th term of both APs are same.<\/p>\n

Question 16.
\nDetermine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.
\nSolution:
\nGiven: a3<\/sub> = 16
\n\u21d2 a + (3 – 1)d = 16
\n\u21d2 a + 2d = 16
\nand a7<\/sub> – a5<\/sub> = 12
\n\u21d2 [a + (7 – 1 )d] – [a + (5 – 1 )d] = 12
\n\u21d2 a + 6d – a – 4d = 12
\n\u21d2 2d = 12
\n\u21d2 d = 6
\nSince a + 2d = 16
\n\u21d2 a + 2(6) = 16
\n\u21d2 a + 12 = 16
\n\u21d2 a = 16 – 12 = 4
\na1<\/sub> = a = 4
\na2<\/sub> = a1<\/sub> + d = a + d = 4 + 6 = 10
\na3<\/sub> = a2<\/sub> + d = 10 + 6 = 16
\na4<\/sub> = a3<\/sub> + d = 16 + 6 = 22
\nThus, the required AP is a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>,…,
\nTherefore the required AP is 4, 10, 16, 22.<\/p>\n

\"NCERT<\/p>\n

Question 17.
\nFind the 20th term from the last term of the AP: 3, 8, 13, …, 253.
\nSolution:
\nGiven: AP is 3, 8, 13,…….. ,253
\nOn reversing the given A.P., we have
\n253, 248, 243 ,………, 13, 8, 3.
\nHere, a = 253, d = 248 – 253 = -5
\na20<\/sub> = a + (20 – 1)d = a + 19d = 253 + 19 (-5) = 253 – 95 = 158<\/p>\n

Question 18.
\nThe sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
\nSolution:
\nWe have,
\nor a4<\/sub> + a8<\/sub> – 24
\na + 3d + a + 7d = 24
\nor 2a + 10d = 24 … (i)
\nand a6<\/sub> + a10<\/sub>
\nor a + 5d + a + 9d = 44
\nor 2a + 14d = 44 … (ii)
\nSubtracting equation (ii) from equation (i), we
\n2a + 10d – (2a + 14d) = 24 – 44
\nor 2a + 10d – 2a – 14d = – 20
\nor 10d – 14d = – 20
\nor – 4d = – 20
\n\u2234 d = \\(\\frac { -20 }{ -4 }\\)
\nPutting the value of d in equation (i), we get
\n2a + 10d = 24
\nor 2a + 10 x 5 = 24
\nor 2a + 50 = 24
\nor 2a = 24 – 50
\nor 2 a = – 26
\n\u2234 a = \\(\\frac { -26 }{ 2 }\\) = – 13
\nTherefore, the sequence can be,
\n– 13, – 13 + 5, – 13 + 2 x 5,… or -13, -8, -3,…
\n\u2234 The first three terms of this AP are -13, -8, and-3<\/p>\n

Question 19.
\nSubba Rao started work in 1995 at an annual salary of \u20b9 5000 and received an increment of \u20b9 200 each year. In which year did his income reach \u20b9 7000 ?
\nSolution:
\nWe have,
\nStarting salary of Subba Rao is \u20b9 5000 and annual increment = \u20b9 500
\nTherefore, the sequence of Subba Rao’s salary is 5000, 5200, 5400,…
\nWe have, a + (n – 1) d = 7000
\n\u21d2 5000 + (n – 1) 200 = 7000
\n\u21d2 (n – 1) 200 = 7000 – 5000
\n\u21d2 (n – 1) 200 = 2000
\n\u21d2 (n- 1) = 10
\n\u21d2 n = 11
\n\u21d2 1995 + 11 = 2006
\nTherefore, after 11 years from 1995, salary of Subba Roa each \u20b9 7000.
\n\u2234 19995 + 11 = 2006
\nSo, in year 2006, salary of Subba Rao will reach will \u20b9 7000.<\/p>\n

\"NCERT<\/p>\n

Question 20.
\nRamkali saved \u20b9 5 in the first week of a year and then increased her weekly saving by \u20b9 1.75. If in the nth week, her weekly saving become \u20b9 20.75, find n.
\nSolution:
\nRamkali savings in the first week of the year be \u20b9 5.
\nand weekly saving increased by \u20b9 1.75
\nTherefore, the sequence of weekly savings be 5, 6.75,8.50,…
\nHere, a = 5 and d = 6.75 – 5 = 1.75
\nLet nth<\/sup> term of this sequence be 20.75
\nan<\/sub> = 20.75
\na + (n – 1) d – 20.75
\n\u21d2 5 + (n – 1) 1.75 = 20.75
\n\u21d2 (n – 1) x 1.75 = 20.75 – 5
\n\u21d2 (n – 1) 1.75 = 15.75
\n\u21d2 n – 1 = 9
\n\u21d2 n = 9 + 1
\n\u21d2 n = 10
\nHence, in 10th week Ramkali\u2019s saving will be \u20b9 20.75.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2 Question 1. Fill in the blanks in the following table, given that a is the first term, d the …<\/p>\n

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2 Question 1. 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