NCERT solutions for class 10 Maths chapter 5 Arithmetic Progressions (Exercise 5.3)

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NCERT solutions for class 10 Maths


Chapter 5


Arithmetic Progressions


Exercise 5.3


Ex 5.3 Question 1.


Find the sum of the following APs.

(i) 2, 7, 12 ,…., to 10 terms.
(ii) − 37, − 33, − 29 ,…, to 12 terms
(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms
(iv) 1/15, 1/12, 1/10, …… , to 11 terms

Solutions:

(i) a = 2

d = a2 − a1 = 7−2 = 5

n = 10

Sn = n/2 [2a +(n-1)d]

S10 = 10/2 [2(2)+(10 -1)×5]

= 5[4+(9)×(5)]

= 5 × 49 = 245

(ii) a = −37

d = a2− a1

d= (−33)−(−37)

= − 33 + 37 = 4

n = 12

Sn = n/2 [2a+(n-1)d]

S12 = 12/2 [2(-37)+(12-1)×4]

= 6[-74+11×4]

= 6[-74+44]

= 6(-30) = -180

(iii) a = 0.6

d = a2 − a1 = 1.7 − 0.6 = 1.1

n = 100

Sn = n/2[2a +(n-1)d]

S12 = 50/2 [1.2+(99)×1.1]

= 50[1.2+108.9]

= 50[110.1]

= 5505

(iv) a = 1/5

d = a–a1 = (1/12)-(1/5) = 1/60

n = 11

Sn = n/2 [2a + (n – 1) d]

Sn = 11/2 [2(1/15) + (11 – 1) 1/60]

 

= 11/2(2/15 + 10/60)

= 11/2 (9/30)

= 33/20


Ex 5.3 Question 2.


Find the sums given below:

 

(i) 7 + 10½ + 14 + …………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Solutions:

(i) a = 7

nth term, (an) = 84

d = a–a1 = 10½ – 7 = 7/2

 

a= a(n-1)d

84 = 7+(n – 1)×7/2

77 = (n-1)×7/2

22 = n−1

n = 23

sum of n term –

Sn = n/2 (a + l)

l = 84

Sn = 23/2 (7+84)

Sn  = (23×91/2) = 2093/2

Sn  = 1046½

 

(ii) a = 34

d = a2−a1 = 32−34 = −2

nth term, an= 10

an= a +(n−1)d

10 = 34+(n−1)(−2)

−24 = (n −1)(−2)

12 = n −1

n = 13

sum of n terms –

Sn = n/2 (a +l) 

l = 10

= 13/2 (34 + 10)

= (13×44/2) = 13 × 22

= 286

(iii) a = −5

nth term, an= −230

d = a2−a1 = (−8)−(−5)

⇒d = − 8+5 = −3

ana+(n−1)d

−230 = − 5+(n−1)(−3)

−225 = (n−1)(−3)

(n−1) = 75

n = 76

Sum of n term,

Sn = n/2 (a + l)

= 76/2 [(-5) + (-230)]

= 38(-235)

= -8930


Ex 5.3 Question 3.


In an AP
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5, S9 = 75, find a and a9.
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an = 4, d = 2, Sn = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.

Solutions:

(i)  a = 5

d = 3

an = 50

an = a +(n −1)d

⇒ 50 = 5+(n -1)×3

⇒ 3(n -1) = 45

⇒ n -1 = 15

⇒ n = 16

sum of n terms,

Sn = n/2 (a +an)

Sn = 16/2 (5 + 50) = 440

(ii) a = 7

a13 = 35

an = a+(n−1)d

⇒ 35 = 7+(13-1)d

⇒ 12d = 28

⇒ d = 28/12 = 2.33

Sn = n/2 (a+an)

S13 = 13/2 (7+35) = 273

(iii)a12 = 37

d = 3

an = a+(n −1)d,

⇒ a12 = a+(12−1)3

⇒ 37 = a+33

⇒ a = 4

sum of nth term,

Sn = n/2 (a+an)

Sn = 12/2 (4+37)

= 246

(iv) a3 = 15

S10 = 125

an = a +(n−1)d,

a3 = a+(3−1)d

15 = a+2d …….. (i)

Sum of the nth term,

Sn = n/2 [2a+(n-1)d]

S10 = 10/2 [2a+(10-1)d]

125 = 5(2a+9d)

25 = 2a+9…………….. (ii)

Multiplying equation (i) by (ii),

30 = 2a+4d ……………. (iii)

Subtracting equation (iii) from (ii),

−5 = 5d

d = −1

From equation (i),

15 = a+2(−1)

15 = a−2

a = 17 = First term

a10 = a+(10−1)d

a10 = 17+(9)(−1)

a10 = 17−9 = 8

(v)  d = 5

S9 = 75

Sn = n/2 [2a +(n -1)d]

S9 = 9/2 [2a +(9-1)5]

25 = 3(a+20)

25 = 3a+60

3a = 25−60

a = -35/3

Now,

an = a+(n−1)d

a9 = a+(9−1)(5)

= -35/3+8(5)

= -35/3+40

= (35+120/3) = 85/3

(vi) a = 2

d = 8

Sn = 90

Sn = n/2 [2a +(n -1)d]

90 = n/2 [2a +(n -1)d]

⇒ 180 = n(4+8n -8) = n(8n-4) = 8n2-4n

⇒ 8n2-4n –180 = 0

⇒ 2n2n-45 = 0

⇒ 2n2-10n+9n-45 = 0

⇒ 2n(n -5)+9(n -5) = 0

⇒ (n-5)(2n+9) = 0

n = 5            [ n can only be a positive integer]

∴ a5 = 8+5×4 = 34

(vii) a = 8

an = 62

Sn = 210

Sn = n/2 (a + an)

210 = n/2 (8 +62)

⇒ 35n = 210

⇒ n = 210/35 = 6

Now,

62 = 8+5d

⇒ 5d = 62-8 = 54

⇒ d = 54/5 = 10.8

(viii) an = 4

d = 2

Sn = −14.

an = a+(n−1)d

4 = a+(−1)2

4 = a+2n−2

a+2n = 6

= 6 − 2n …………. (i)

Sn = n/2 (a+an)

-14 = n/2 (a+4)

−28 = (a+4)

From equation (i)

−28 = (6 −2n +4)

−28 = (− 2n +10)

−28 = − 2n2+10n

2n2 −10n − 28 = 0

n2 −5−14 = 0

n2 −7n+2n −14 = 0

(n−7)+2(n −7) = 0

(n −7)(n +2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2 [ n can not be negative nor fractional.]

Therefore, n = 7

From equation (i),

a = 6−2n

a = 6−2(7)

= 6−14

= −8

(ix) a = 3,

n = 8

Sn = 192

Sn = n/2 [2a+(n -1)d]

192 = 8/2 [2×3+(8 -1)d]

192 = 4[6 +7d]

48 = 6+7d

42 = 7d

d = 6

(x)  l = 28

Sn = 144

Sn = n/2 (a + l)

144 = 9/2(a+28)

(16)×(2) = a+28

32 = a+28

a = 4


Ex 5.3 Question 4.


How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Solution:

a = 9

 d = a2a1 = 17−9 = 8

Sn = 636

Sn = n/2 [2a+(n -1)d]

636 = n/2 [2×a+(8-1)×8]

636 = n/2 [18+(n-1)×8]

636 = [9 +4n −4]

636 = (4n +5)

4n2 +5n −636 = 0

4n2 +53n −48n −636 = 0

(4n + 53)−12 (4n + 53) = 0

(4n +53)(n −12) = 0

Either 4n+53 = 0 or n−12 = 0

n = (-53/4) or n = 12          [n cannot be negative or fraction.]

∴n = 12.


Ex 5.3 Question 5.


The 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.

Solution:

a = 5

 l = 45

Sn = 400

Sn = n/2 (a+l)

400 = n/2(5+45)

400 = n/2(50)

 n =16

l = a+(n −1)d

45 = 5 +(16 −1)d

40 = 15d

d = 40/15 = 8/3


Ex 5.3 Question 6.


The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution:

a = 17

l = 350

d = 9

l = a+(n −1)d

350 = 17+(n −1)9

333 = (n−1)9

(n−1) = 37

n = 38

Sn = n/2 (a+l)

S38 = 13/2 (17+350)

= 19×367

= 6973


Ex 5.3 Question 7.


Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:

d = 7

a22 = 149

S22 = ?

an = a+(n−1)d

a22 = a+(22−1)d

149 = a+21×7

149 = a+147

a = 2

Sum of n terms,

Sn = n/2(a+an)

S22 = 22/2 (2+149)

= 11×151

= 1661


Ex 5.3 Question 8.


Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:

a2 = 14

a3 = 18

d = a3a2 = 18−14 = 4

a2 = a+d

14 = a+4

a = 10

Sum of n terms

Sn = n/2 [2a + (n – 1)d]

S51 = 51/2 [2×10 (51-1) 4]

= 51/2 [2+(20)×4]

= 51 × 220/2

= 51 × 110

= 5610


Ex 5.3 Question 9.


If 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.

Solution:

S7 = 49

S17 = 289

Sn = n/2 [2a + (n – 1)d]

S77/2 [2a +(n -1)d]

S7 = 7/2 [2a + (7 -1)d]

49 = 7/2 [2a +6d]

7 = (a+3d)

a + 3d = 7 …………. (i)

S17 = 17/2 [2a+(17-1)d]

289 = 17/2 (2a +16d)

17 = (a+8d)

a +8d = 17 ……….(ii)

Subtracting equation (i) from equation (ii),

5d = 10

d = 2

From equation (i),

a+3(2) = 7

a+ 6 = 7

a = 1

Sn = n/2[2a+(n-1)d]

n/2[2(1)+(n – 1)×2]

n/2(2+2n-2)

n/2(2n)

n2


Ex 5.3 Question 10.


Show that a1a… , an , … form an AP where an is defined as below

(i) an = 3+4n
(ii) an = 9−5n
Also find the sum of the first 15 terms in each case.

Solutions:

(i) an = 3+4n

a1 = 3+4(1) = 7

a2 = 3+4(2) = 3+8 = 11

a3 = 3+4(3) = 3+12 = 15

a4 = 3+4(4) = 3+16 = 19

d =?

a2 − a1 = 11−7 = 4

a3 − a2 = 15−11 = 4

a4 − a3 = 19−15 = 4

Hence , a = 7

d= 4

Sn = n/2[2a+(n -1)d]

S15 = 15/2[2(7)+(15-1)×4]

= 15/2[(14)+56]

= 15/2(70)

= 15×35

= 525

(ii) an = 9−5n

a1 = 9−5×1 = 9−5 = 4

a2 = 9−5×2 = 9−10 = −1

a3 = 9−5×3 = 9−15 = −6

a4 = 9−5×4 = 9−20 = −11

a2 − a1 = −1−4 = −5

a3 − a2 = −6−(−1) = −5

a4 − a3 = −11−(−6) = −5

Hence, a = 4

d = -5

Sn = n/2 [2a +(n-1)d]

S15 = 15/2[2(4) +(15 -1)(-5)]

= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31)

= -465


Ex 5.3 Question 11.


If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the10th and the nth terms.

Solution:

Sn = 4nn2

a = S1 = 4(1) − (1)2 = 4−1 = 3

S2= 4(2)−(2)2 = 8−4 = 4

a2 = S2 − S1 = 4−3 = 1

d = a2a = 1−3 = −2

an = a+(n−1)

= 3+(n −1)(−2)

= 3−2n +2

= 5−2n

a3 = 5−2(3) = 5-6 = −1

a10 = 5−2(10) = 5−20 = −15

Sum of first two terms = 4.

Second term = 1.

3rd  term = -1

10th term = -15

nth terms = 5 − 2n.


Ex 5.3 Question 12.


Find the sum of first 40 positive integers divisible by 6.

Solution:

a = 6

d = 6

S40 = ?

Sn = n/2 [2a +(n – 1)d]

n = 40

S40 = 40/2 [2(6)+(40-1)6]

= 20[12+(39)(6)]

= 20(12+234)

= 20×246

= 4920


Ex 5.3 Question 13.


Find the sum of first 15 multiples of 8.

Solution:

a = 8

d = 8

S15 = ?

Sn = n/2 [2a+(n-1)d]

S15 = 15/2 [2(8) + (15-1)8]

= 15/2[6 +(14)(8)]

= 15/2[16 +112]

= 15(128)/2

= 15 × 64

= 960


Ex 5.3 Question 14.


Find the sum of the odd numbers between 0 and 50.

Solution:

a = 1

d = 2

l = 49

l = a+(n−1) d

49 = 1+(n−1)2

48 = 2(n − 1)

n − 1 = 24

n = 25

Sn = n/2(a +l)

S25 = 25/2 (1+49)

= 25(50)/2

=(25)(25)

= 625


Ex 5.3 Question 15.


A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 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.

Solution:

a = 200

d = 50

Sn = 30

Sn = n/2[2a+(n -1)d]

S30= 30/2[2(200)+(30 – 1)50]

= 15[400+1450]

= 15(1850)

= 27750

Therefore, the contractor has to pay as penalty = ₹ 27750


Ex 5.3 Question 16.


A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Solution:

Let the cost of 1st prize = ₹ x.

Cost of 2nd prize = ₹ x − 20

And cost of 3rd prize = ₹ x − 40

a = x

d = −20

S7 = 700

Sn = n/2 [2a + (n – 1)d]

7/2 [2a + (7 – 1)d] = 700

[2a + (6) (-20)] / 2 = 100

 

a + 3(−20) = 100

a −60 = 100

a = 160

Therefore, the seven  prizes was ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, and ₹ 40.


Ex 5.3 Question 17.


In 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, e.g., 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?

Solution:

Each section of each class will plant tree = 3 x class

Total number of tree planted by class I = 3 x 1 = 3

Total number of tree planted by class I = 3 x 2 = 6

Total number of tree planted by class I = 3 x 3 = 9

a = 3

d = 6−3 = 3

n = 12

Sn = n/2 [2a +(n-1)d]

S12 = 12/2 [2(3)+(12-1)(3)]

= 6(6+33)

= 6(39)

= 234.


Ex 5.3 Question 18.


A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 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? (Take π = 22/7)

Solution:

cercumference of semi-circle = ½(2πr) = πr

Radii of semi – circle = 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm

∴ The length of first sprial l1=π (0.5) cm

The length of second sprial l2=π (1.0) cm

l= π (1.5) cm

l= π (2.0) cm

a = 0.5π

d = 1.0π – 0.5π = 0.5 π

n= 13

Sn = n/2 [2a + (n – 1)d]

S13 = 13/2 [2(0.5π) + (13 – 1)0.5π]

6.5 [π + 6π]

=13/2 (7π)

13/2 × 7 × 22/7

= 143 cm.


Ex 5.3 Question 19.


200 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. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution: 

a = 20

d = a2a1 = 19−20 = −1

Sn = 200

Sn = n/2 [2a +(n -1)d]

S12 = 12/2 [2(20)+(n -1)(-1)]

400 = n (40−n+1)

400 = (41-n)

400 = 41nn2

n2−41+ 400 = 0

n2−16n−25n+400 = 0

n(n −16)−25(n −16) = 0

(−16)(n −25) = 0

Either (n −16) = 0 or n−25 = 0

n = 16 or n = 25

For nth term formula,

an = a+(n−1)d

a16 = 20+(16−1)(−1)

a16 = 20−15

a16 = 5

25th term

a25 = 20+(25−1)(−1)

a25 = 20−24

= −4 (which is not possible)

Therefore, 200 logs can be placed in 16 rows and 5 loga are in the top row.


Ex 5.3 Question 20.


In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

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?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2×5+2×(5+3)]

Solution:

a = 10

d = 16−10 = 6

S10 =?

S10 = 12/2 [2(20)+(n -1)(-1)]

= 5[20+54]

= 5(74)

= 370