{"id":27809,"date":"2021-07-01T18:51:56","date_gmt":"2021-07-01T13:21:56","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27809"},"modified":"2021-07-01T18:51:56","modified_gmt":"2021-07-01T13:21:56","slug":"ncert-solutions-for-class-10-maths-chapter-5-ex-5-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-5-ex-5-3\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 5 Arithmetic Progressions Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFind the sum of the following APs:
\n(i) 2, 7, 12,…… to 10 terms.
\n(ii) -37, -33, -29, …… to 12 terms.
\n(iii) 0.6, 1.7, 2.8, ……, to 100 terms.
\n(iv) \\(\\frac { 1 }{ 15 }\\), \\(\\frac { 1 }{ 12 }\\), \\(\\frac { 1 }{ 10 }\\), …….., to 11 terms.
\nSolution:
\n(i) We have
\n2, 7, 12, … to 10 terms
\nHere, a = 2, d = 7 – 2 = 5 and n = 10
\n\"NCERT
\nTherefore, the sum of 10 terms of the AP 2, 7, 12, ….. is 245<\/p>\n

(ii) We have,
\n– 37, – 33, – 29, … to 12 terms
\nHere, a = – 37, d = – 33 – (- 37) = 4 and n = 12
\n\"NCERT
\nTherefore, the sum of 12 terms of the AP – 37, – 33, – 29,… is – 180.<\/p>\n

(iii) We have,
\n0.6,1.7, 2.8,… to 100 terms
\nHere, a = 0.6, d = 1.7 – 0.6 = 1.1 and n = 100
\n\"NCERT
\nTherefore, the sum of 100 terms of the AP 0.6, 1.7,2.8,. .. is 5505.<\/p>\n

(iv) We have,
\n\"NCERT
\nTherefore, the sum of 11 terms of the AP \\(\\frac { 1 }{ 15 }\\), \\(\\frac { 1 }{ 12 }\\), \\(\\frac { 1 }{ 10 }\\), ….. is \\(\\frac { 33 }{ 20 }\\).<\/p>\n

Question 2.
\nFind the sums given below:
\n(i) 7 + 10\\(\\frac { 1 }{ 2 }\\) + 14 + … + 84
\n(ii) 34 + 32 + 30 + … + 10
\n(iii) -5 + (-8) + (-11) + ….. + (-230)
\nSolution:
\n(i) We have
\n7 + 10\\(\\frac { 1 }{ 2 }\\) + 14 + … + 84
\n\"NCERT
\nTherefore 7 + 10\\(\\frac { 1 }{ 2 }\\) + 14 + … + 84 = 1046\\(\\frac { 1 }{ 2 }\\)<\/p>\n

(ii) We have
\n34 + 32 + 30 + … + 10
\n\"NCERT
\nTherefore, 34 + 32 + 30 + … + 10 = 286<\/p>\n

(iii) We have
\n-5 + (-8) + (-11) + ….. + (-230)
\n\"NCERT
\nTherefore, -5 + (-8) + (-11) + ….. + (-230) is – 8390.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn an AP:
\n(i) given a = 5, d = 3, an<\/sub> = 50, find n and Sn<\/sub>.
\n(ii) given a = 7, a13<\/sub> = 35, find d and S13<\/sub>.
\n(iii) given a12<\/sub> = 37, d = 3, find a and S12<\/sub>.
\n(iv) given a3<\/sub> = -15, S10<\/sub> = 125, find d and a10<\/sub>.
\n(v) given d = 5, S9<\/sub> = 75, find a and a9<\/sub>.
\n(vi) given a = 2, d = 8, Sn<\/sub> = 90, find n and an<\/sub>.
\n(vii) given a = 8, an<\/sub> = 62, Sn<\/sub> = 210, find n and d.
\n(viii) given an<\/sub> = 4, d = 2, Sn<\/sub> = -14, find n and a.
\n(ix) given a = 3, n = 8, S = 192, find d.
\n(x) given l = 28, S = 144, and there are total 9 terms. Find a.
\nSolution:
\n(i) We have
\na = 5, d = 3, an<\/sub> = 50
\nBut, We know that
\n\"NCERT
\nTherefore, the value of n = 16 and Sn<\/sub> = 440.<\/p>\n

(ii) We have a = 7, a13<\/sub> = 35
\nBut, We know that
\na13<\/sub> = a + 12d
\nor 35 = 7 + 12 d
\nor d = \\(\\frac { 28 }{ 12 }\\) = \\(\\frac { 7 }{ 3 }\\)
\n\"NCERT
\nTherefore, the value of d = \\(\\frac { 7 }{ 3 }\\) and S13<\/sub> = 273.<\/p>\n

(iii) We have a12<\/sub> = 37, d = 3
\nBut, We know that
\na12<\/sub> = a + 11d
\nor 37 = a + 11 x 3
\nor 37 = a \u00f7 33
\nor a = 37 – 33 = 4
\nAgain we know that
\n\"NCERT
\nTherefore, the value of a = 4 and S12<\/sub> = 246.<\/p>\n

(iv) We have a3<\/sub> = -15, S10<\/sub> = 125
\nBut, We know that
\n\"NCERT
\nMultiplying equation (i) by 10 then subtracting from equation (ii), we get,
\n\"NCERT
\nPutting the value of d = – 1 in equation, (i) we get,
\na + 2(-1) = 15
\na – 2 = 15 \u21d2 a = 17
\nNow, a = a+ (n – 1)d
\na10<\/sub> = 17 + (10 – 1) (- 1)
\na10<\/sub> = 17 – 9 = 8
\nTherefore, d = – 1 and a10<\/sub> = 8<\/p>\n

(v) We have d = 5, S9<\/sub> = 75
\nBut, We know that
\n\"NCERT
\nTherefore, the value of a = \\(\\frac { -35 }{ 3 }\\) and a9<\/sub> = \\(\\frac { 85 }{ 3 }\\).<\/p>\n

(vi) We have a = 2, d = 8, Sn<\/sub> = 90
\nBut, We know that
\n\"NCERT
\nBut number of terms cannot be in negative
\n\u2234 n = \\(\\frac { -9 }{ 2 }\\) is neglected.
\nSo, n = 5
\nAgain we know that
\nan<\/sub> = a + (n – 1)d
\n\u2234 a5<\/sub> = 2 + (5 – 1) x 8
\n= 2 + 4 x 8
\nor a5<\/sub> = 2 + 32 = 34
\nTherefore, the value of n = 5 and a5<\/sub> = 34.<\/p>\n

(vii) We have a = 8, an<\/sub> = 62, Sn<\/sub> = 210
\nBut, We know that
\nan<\/sub> = a + (n – 1)d
\n62 = 8 + (n – 1) d
\n\u2234 (n – 1)d = 62 – 8 = 54
\nor a5<\/sub> = 2 + 32 = 34 … (i)
\nAgain
\n\"NCERT
\nTherefore, the value of n = 6 and d = \\(\\frac { 54 }{ 5 }\\).<\/p>\n

(viii) We have an<\/sub> = 4, d = 2, Sn<\/sub> = – 14
\nBut, We know that
\nan<\/sub> = a + (n – 1)d
\nor 4 = a+ (n – 1)2
\nor 4 = a + 2n – 2
\n\u2234 a + 2n = 6
\n\u2234 a = 6 – 2n … (i)
\nAgain, we know that
\nSn<\/sub> = \\(\\frac { n }{ 2 }\\)[2a + (n – 1)d]
\nor – 14 = \\(\\frac { n }{ 2 }\\)[2 x a + (n – 1) x 2]
\nor – 14 x 2 = n (2a + 2n – 2)
\nor – 28 = n[2(6 – 2n) + 2n – 2]
\n[Putting the value of a from equation (i)]
\nor – 28 = n (12 – 4n + 2n – 2)
\nor – 28 = n (- 2n +10)
\nor – 28 = – 2n\u00b2 + 10n
\nor 2n\u00b2 – 10n – 28 = 0
\nor n\u00b2 – 5n – 14 = 0
\nor n\u00b2 – 7n + 2n – 14 = 0
\nor n (n – 7) + 2 (n – 7) = 0
\nor (n + 2) (n – 7) = 0
\n\u2234 n = 7 and n= -2
\nBut, number of terms cannot be in negative
\n\u2234 n = – 2 is neglected.
\nSo, n = 7
\nPutting the value of n in equation (i), we get,
\na = 6 – 2n = 6 – 2 x 7
\n\u2234 a = 6 – 14 = – 8
\nTherefore, the value of a = – 8 and n = 7.<\/p>\n

(ix) We have a = 3, n = 8, S = 192
\nBut, We know that
\n\"NCERT
\nTherefore, the value of d = 6.<\/p>\n

(x) We have l = 28, S = 144, and n = 9
\nBut, We know that
\nan<\/sub> = l
\nor a + (n – 1)d = 28
\nor a + (9 – 1)d = 28
\nor a + 8d = 28 … (i)
\nAgain we know that
\n\"NCERT
\nMultiply equation (i) by 2 then subtract equation (ii) from equation (i) we get,
\n\"NCERT
\nPutting the value of d in equation (i)
\na + 8 d = 28
\nor a + 8 x 3 = 28
\n\u2234 a = 28 – 24 = 4
\nTherefore, the value of a = 4.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nHow many terms of AP: 9, 17, 25, ….. must be taken to give a sum of 636?
\nSolution:
\nWe have, 9,17,25 Here,
\na = 9 and d = 17 – 9 = 8
\nLet n terms of this AP must be taken to give a sum of 636.
\nWe know that
\n\"NCERT<\/p>\n

Question 5.
\nThe first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
\nSolution:
\n\"NCERT<\/p>\n

Question 6.
\nThe first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
\nSolution:
\n\"NCERT<\/p>\n

Question 7.
\nFind the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
\nSolution:
\n\"NCERT<\/p>\n

Question 8.
\nFind the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
\nSolution:
\n\"<\/p>\n

Question 9.
\nIf the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
\nSolution:
\nWe have
\nS7<\/sub> and S17<\/sub> = 289
\nBut, we know that
\n\"NCERT
\nAgain, we know that
\n\"NCERT
\nSubtracting equation (ii) from equation (i), we get,
\n3d – 8d = 7 – 17
\nor – 5d = – 10
\nPutting the value of d in equation (i), we get,
\na + 3d = 7
\nor a + 3 x 2 = 7
\n\u2234 a = 7 – 6 = 1
\nTherefore, sum of n terms of this AP is
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nShow that a1<\/sub>, a2<\/sub>, ……. an<\/sub>,…… form an AP where an is defined as below:
\n(i) an<\/sub> = 3 + 4n
\n(ii) an<\/sub> = 9 – 5n
\nAlso find the sum of the first 15 terms in each case.
\nSolution:
\n(i) We have
\na = 3 + 4n
\nPut n = 1 in this equation
\n\u2234 a1<\/sub> = 3 + 4 x 1 = 7
\nNow, put n = 2 in this equation
\na2<\/sub> = 3 + 4 x 2 = 11
\nAgain, put n = 3 in this equation
\na3<\/sub> = 3 + 4 x 3 = 15
\nTherefore, the required AP is 7, 11, 15, ….
\nHere, a = 7, d = 11 – 7 = 4
\nWe know that
\n\"NCERT
\nTherefore, sum of 15 terms of the sequence an<\/sub> = 3 + 4n is 525.<\/p>\n

(ii) We have,
\n\u2234 an<\/sub> = 9 – 5n
\na1<\/sub> = 9 – 5 x 1 = 4
\na2<\/sub> = 9 – 5 x 2 = – 1
\nand a3<\/sub>= 9 – 5 x 3 = – 6
\n\u2234 AP is 4,-1,-6,…
\n\u2234 Common difference (d) = – 1 – 4 = – 5
\n\"NCERT
\nTherefore, sum of 15 terms of the sequence an<\/sub> = 9 – 5n is – 465.<\/p>\n

Question 11.
\nIf the sum of the first n terms of an AP is 4n – n2<\/sup>, what is the first term (that is S1<\/sub>)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
\nSolution:
\nSn<\/sub> = 4n – n\u00b2
\nPut n = 1, 2, 3, …
\nS1<\/sub> = 4(1) – 1
\nS1<\/sub> = 3,
\nS2<\/sub>=4,
\nS3<\/sub>=3,
\nThe second term is
\na2<\/sub> = S2<\/sub> – S1<\/sub>
\na2<\/sub> = 4 – 3 \u21d2 a2<\/sub> = 1;
\nThird term,
\na2<\/sub>= S3<\/sub> – S2<\/sub>
\na3<\/sub> = 3 – 4 \u21d2 a2<\/sub> = – 1
\nThird term
\na2<\/sub> = S3<\/sub> – S2<\/sub>
\na3<\/sub> = 3 – 4 \u21d2 a3<\/sub> = – 1;
\n\u21d2 – 60 + 45 = – 15
\nand nth term is
\nan<\/sub> = (4n – n2<\/sup>) – [(4n – 4) – (n – 1)\u00b2]
\n= 4n – n2<\/sup> – 4n + 4 + n2<\/sup> + 1 – 2n
\n= 5 – 2n
\nnth term is (5 – 2n).<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nFind the sum of the first 40 positive integers divisible by 6.
\nSolution:
\nThe positive integers divisible by 6 are
\n6,12,18, …….
\nHere, a = 6 and d = 12 – 6 = 6
\n\u2234 Sn<\/sub> = \\(\\frac { 40 }{ 2 }\\)[2 x 6 + (40 – 1) x 6]
\n= 20 (12 + 39 x 6) = 20 (12 + 234)
\n\u2234 S40<\/sub> = 20 x 246 = 4920<\/p>\n

Question 13.
\nFind the sum of the first 15 multiples of 8.
\nSolution:
\nThe first 15 multiples of 8 are 8,16, 24…
\nHere, a = 8,d = 16-8 = 8
\nSn<\/sub> = \\(\\frac { n }{ 2 }\\)[2a + (n – 1)d]
\n\u21d2 S15<\/sub> = \\(\\frac {15 }{ 2 }\\)[2 x 8 + (15 – 1)8]
\n\u21d2 S15<\/sub> = \\(\\frac { 15 }{ 2 }\\)[16 + 112]
\n\u21d2 S15<\/sub> = \\(\\frac { 15 }{ 2 }\\) x 128
\n\u21d2 S15<\/sub> = 15 x 64 = 960
\n\u2234 The sum of first 15 multiples of 8 is 960.<\/p>\n

Question 14.
\nFind the sum of the odd numbers between 0 and 50.
\nSolution:
\nLet odd numbers between 0 and 50 be 1, 3, 5, 7,…….., 49.
\n\"NCERT<\/p>\n

Question 15.
\nA contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:
\n\u20b9 200 for the first day, \u20b9 250 for the second day, \u20b9 300 for the third day, etc. the penalty for each succeeding day being \u20b9 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
\nSolution:
\nWe have given that penalty for delay of completion beyond a certain date are 200, 250, 300, ….
\nIt is the form of an AP, where a = 200 d = 50
\nSn<\/sub> = \\(\\frac { n }{ 2 }\\)[2a + (n – 1)d]
\nS30<\/sub>= \\(\\frac { 30 }{ 2 }\\) [2 x 200 + (30 – 1)50]
\n= 15 [400 + 1450]
\n= (15 x 1850) = 27750
\nThe contractor has to pay \u20b9 27750 as penalty.<\/p>\n

\"NCERT<\/p>\n

Question 16.
\nA sum of \u20b9 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is \u20b9 20 less than its preceding prize, find the value of each of the prizes.
\nSolution:
\nLet 1st prize be of \u20b9 a
\n2nd prize be \u20b9 (a – 20) and
\n3rd prize be \u20b9 (a – 20 – 20) = \u20b9 (a – 40)
\n\"NCERT<\/p>\n

Question 17.
\nIn a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, example a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
\nSolution:
\nLet the trees be planted 1, 2, 3, 4, 5 , …… 12
\nHere, a = 1, d = 1, n = 12
\nTotal number of trees planted by each section
\nS12<\/sub> = \\(\\frac { 12 }{ 2 }\\) [2a + (n – 1) d] = 6 [2 x 1 + (12 – 1) x 1]
\n= 6 [2 + 11] = 6 x 13 = 78
\nTotal number of trees planted by 3 sections = 78 x 3 = 234<\/p>\n

\"NCERT<\/p>\n

Question 18.
\nA spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?
\n(Take \u03c0 = \\(\\frac { 22 }{ 7 }\\))
\n[Hint:Length of successive semicircles is l1<\/sub>, l2<\/sub>, l3<\/sub>, l4<\/sub>, … with centres at A, B, respectively.]
\nSolution:
\nLength of successive semi circles is l1<\/sub>, l2<\/sub>, l3<\/sub>, l4<\/sub>, …
\nSeries is 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm… we have to found ln<\/sub>(total length)
\nHere, l1<\/sub> = 0.5 \u03c0
\nd = 1.0 – 0.5 = 0.5\u03c0
\nn = 13 (consective semi circles)
\nSum of ln<\/sub> terms
\n\"NCERT
\n\u2234 Total length of spiral of 143 cm.<\/p>\n

Question 19.
\n200 logs are stacked in the following manner 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Figure). In how many rows are the 200 logs placed and how many logs are in the top row?
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 20.
\nIn a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig.) A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
\n[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 x 5 + 2 x (5 + 3)]
\n\"NCERT
\nSolution:
\nDistance covered by the girl pick up the first potato = 10m
\nThe distance covered by the girl to pick up the second potato = 8 x 2 = 16m
\nand the distance covered by the girl to pick up the third potato = 11 x 2 = 22m
\nSo, we get the following series ;
\n10, 16, 22, 28, … upto 10th<\/sup> term.
\nWe know that, [where a = 10, d = 6]
\nTotal distance We know that
\n\u2234 Sn<\/sub> = \\(\\frac { n }{ 2 }\\)[2a + (n – 1)d]
\n= \\(\\frac { 10 }{ 2 }\\)[2 x 10 + (10 – 1)6]
\n= 5 [20 + 54]
\n= 5 [74] = 370 m
\nHence, total distance covered by the competitor is 370.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 Question 1. Find the sum of the following APs: (i) 2, 7, 12,…… to 10 terms. (ii) -37, -33, …<\/p>\n

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 Question 1. Find the sum of the following APs: (i) 2, 7, 12,…… to 10 terms. 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