# Arithmetic ProgressionsExercise – 5.3

1. Find the sum of the following APs:

(i) 2, 7, 12, … , to 10 terms

Here, a = 2 and d = 7 – 2 = 5

The sum of n terms of an AP is given by,

S10 = 10/2[2×2 + (10 – 1)5]

S10 = 5[4 + 45] = 245

(ii) -37, -33, -29, … , to 12 terms

Here, a = -37 and d = -33 – (-37) = 4

The sum of n terms of an AP is given by,

Sn = n/2 [2a+ (n-1)d]

S12 = 12/2[2×-37 + (12 – 1)4]

S12 = 6[ -74 + 44] = -180

(iii) 0.6, 1.7, 2.8, … , to 100 terms

Here, a = 0.6 and d = 1.7 – 0.6 = 1.1

The sum of n terms of an AP is given by,

Sn = n/2 [2a+ (n-1)d]

S100 = 100/2[2×0.6 + (100 – 1)1.1]

S100 = 50[1.2 + 99×1.1] = 50[110.1] = 5505

(iv) 1/15 , 1/12 , 1/10 , … , to 11 terms

Here, a = 1/15 and d = 1/12 – (1/15) = 1/60/157 /15 , 1/12 , 1/10 , ...

The sum of n terms of an AP is given by,

Sn = n/2 [2a+ (n-1)d]

S11 = 11/2[2×(1/15) + (11 – 1)(1/60)]

S11 = 11/2[2/15 + 1/6] = 11/2[9/30] = 33/20

2. Find the sums given below:

(i) 7 + 10 ½ + 14 + … + 84

Here, a = 7 and d = 10 ½ - 7 = 21/2 – 7 = 7/2

Let, the nth term of the AP is 84

Therefore, an = 84

a + (n – 1)d = 84

7 + (n – 1)(7/2) = 84

(n – 1) = 22

n = 23

The sum of n terms of an AP,

[When the last term (an or l) is given.]

S23 = 23/2 [7 + 84]

S23 = 23/2 [91] = 2093/2

S23 = 1046 ½

(ii) 34 + 32 + 30 + … + 10

Here, a = 34 and d = 32 - 34 = -2

Let, the nth term of the AP is 10

Therefore, an = 10

a + (n – 1)d = 10

34 + (n – 1)(-2) = 10

(n – 1)(-2) = -24

n – 1 = 12

n = 13

The sum of n terms of an AP,

Sn = n/2 [a+l]    [When the last term (an or l) is given.]

S13 = 13/2 [34 + 10]

S13 = 13/2 [44] = 13×22

S13 = 286

(iii) -5 + (-8) + (-11) + … + (-230)

Here, a = -5 and d = -8 – (-5) = -3

Let, the nth term of the AP is -230

Therefore, an = -230

a + (n – 1)d = -230

-5 + (n – 1)(-3) = -230

(n – 1)(-3) = -225

n – 1 = 75

n = 76

The sum of n terms of an AP,

Sn = n/2 [a+l]   [When the last term (an or l) is given.]

S76 = 76/2 [-5 - 230]

S76 = 76/2 [-235] = -38×235

S76 = -8930

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 = 7, 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.

(i) Here a = 5, d = 3 and an = 50

Let, the nth term of the AP is 50

Therefore, an = 50

a + (n – 1)d = 50

5 + (n – 1)3 = 50

(n – 1)3 = 45

n – 1 = 15

n = 16

The sum of n terms of an AP,

Sn = n/2 [a+l]   [When the last term (an or l) is given.]

S16 = 16/2 [5 + 50]

S16 = 8 [55] = 440

(ii) Here a = 7 and a13 = 35

Let, the 13th term of the AP is 35

Therefore, a13 = 35

a + (n – 1)d = 35

7 + (13 – 1)d = 35

12d = 28

d = 28/12 = 7/3

The sum of n terms of an AP,

Sn = n/2 [a+l]

S13 = 13/2 [7 + 35]

S13 = 13/2 [42] = 273

(iii) Here d = 3 and a12 = 37

Let, the 12th term of the AP is 37

Therefore, a12 = 37

a + (n – 1)d = 37

a + (12 – 1)3 = 37

a + 11×3 = 37

a = 37 – 33

a = 4

The sum of n terms of an AP,

Sn = n/2 [a+l]

S12 = 12/2 [4 +37]

S16 = 6 [41] = 246

(iv) Here a3 = 15 and S10 = 125

a3 = 15

a + (3 – 1)d = 15

a  + 2d = 15

a = 15 – 2d             ………(1)

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S10 = 10/2 [2a + (10 – 1)d]

125 = 5[2a + 9d]

2a + 9d = 25

Substituting the value of ‘a’ from Equation (1),

2(15 – 2d) + 9d = 25

30 – 4d + 9d = 25

5d = -5

d = -1

Substituting the value of ‘d’ from Equation (1),

a = 15 – 2(-1) = 17

an = a + (n – 1)d

a10 = 17 + (10 – 1)(-1) = 17 – 9 = 8

a10 = 8

(v) Here d = 5 and S9 = 75

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S9 = 9/2 [ 2a + (9 – 1)5]

75 = 9/2 [ 2a + 40]

75 = 9a + 180

a = -105/9 = -35/3

an = a + (n – 1)d

a9 = -35/3 + (9 – 1)5 = -35/3 + 40 = 85/3

a9 = 85/3

(vi) Here a = 2, d = 8 and Sn = 90

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

90 = n/2 [ 2×2 + (n – 1)8]

90 = n/2 [4 + 8n – 8]

90 = n/2 [8n – 4]

90 = 4n2 – 2n

2n2 – n – 45 = 0

2n2 – 10n + 9n  - 45 = 0

2n(n – 5) + 9(n – 5) = 0

(2n + 9)(n – 5) = 0

n – 5 = 0             [2n + 9 ≠ 0, n ≠ -9/2]

n = 5

an = a + (n – 1)d

a5 = 2 + (5 – 1)8 = 2 + 32 = 34

(vii) Here a = 8, an = 62 and Sn = 210

The sum of n terms of an AP,

Sn = n/2 [a + an]

210 = n/2 [8 + 62]

210 = 35n

n = 210/35 = 6

an = a + (n – 1)d

62 = 8 + (6 – 1)d

54 = 5d

d = 54/5

(viii) Here an = 4, d = 2 and Sn = -14

an = a + (n – 1)d

4 = a + (n – 1)2

4 = a + 2n – 2

a = 6 – 2n          ………(1)

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

-14 = n/2 [2a + (n – 1)2]

-14 = n[a + n - 1]

Substituting the value of ‘a’ from Equation (1),

-14 = n[6 – 2n + n – 1]

-14 = n[5 – n]

-14 = 5n – n2

n2 – 5n – 14 = 0

n2 - 7n + 2n – 14 = 0

n(n – 7) + 2(n – 7) = 0

(n – 7)(n + 2) = 0

n – 7 = 0   [ n + 2 ≠ 0, n ≠ -2]

Substituting the value of ‘d’ from Equation (1),

a = 6 – 2(7) = -8

(ix) Here a = 3, n = 8 and S = 192

The sum of n terms of an AP,

Sn = n/2 [a + an]

192 = 8/2 [3 + an]

192 = 4 [3 + an]

3 + an = 192/4 = 48

an = 45

an = a + (n – 1)d

45 = 3 + (8 – 1)d

42 = 7d

d = 42/7

d = 6

(x) Here l = 28, S = 144 and n = 9

The sum of n terms of an AP,

Sn = n/2 [a + l]

144 = 9/2[a + 28]

144×2/9 = a + 28

32 = a + 28

a = 4

4. How many terms of the A.P.: 9, 17, 25, …36 4a + 28

+ 6(1

+ 9 + ... must be taken to given a sum 636?

Here a = 9, d = 17 – 9 = 8 and Sn = 636

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

636 = n/2 [2×9 + (n – 1)8]

636 = n[9 + 4n – 4]

4n2 + 53n – 48n – 636 = 0

n(4n + 53) -12 (4n + 53) = 0

(n – 12) (4n + 53) = 0

n – 12 = 0          [4n + 53 ≠ 0, n ≠ -53/4]

n = 12

So to get the sum 636, 12 terms of A.P.: 9, 17, 25, … should be taken.

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.

Here a = 5, an = 45 and Sn = 400

The sum of n terms of an AP,

Sn = n/2 [a + an]

400 = n/2 [5 + 45]

400 = 25n

n = 400/25 = 16

an = a + (n – 1)d

45 = 5 + (16 – 1)d

40 = 15d

d = 40/15 = 8/3

Hence, the number of terms is 16 and the common difference is 8/3.

6. The 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?

Here a = 17, an = 350 and d = 9

an = a + (n – 1)d

350 = 17 + (n – 1)9

350 = 17 + 9n – 9

350 = 8 + 9n

n = 342/9 = 38

The sum of n terms of an AP,

Sn = n/2 [a + an]

S38 = 38/2 [17 + 350]

400 = 19 × 367

= 6973

Hence, it has 38 terms and their sum is 6973.

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

Here d = 7, an = 149 and n = 22

an = a + (n – 1)d

149 = a + (22 – 1)7

149 = a + 147

a = 2

The sum of n terms of an AP,

Sn = n/2 [a + an]

S22 = 22/2 [2 + 149]

S22 = 11 × 151

= 1661

Hence, the sum of first 22 terms of this AP is 1661

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

Here a2 = 14, a3 = 2 and n = 51

an = a + (n – 1)d

a2 = a + (2 – 1)d

14 = a + d

a = 14 – d          ………(1)

and a3 = a + (3 – 1)d

18 = a + 2d

Substituting the value of ‘a’ from Equation (1),

18 = 14 – d + 2d

d = 4

Substituting the value of ‘d’ from Equation (1),

a = 14 – 4 = 10

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S51 = 51/2 [2×10 + (51 – 1)4]

S51 = 51/2 [220] = 5610

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.

Here S7 = 49 and S17 = 289

The sum of n terms of an AP,

Sn = n/2 [2a+ (n – 1)d]

S7 = 7/2 [2a + (7 – 1)d]

49 = 7/2 [2a + 6d]

49 = 7(a + 3d)

7 = a + 3d

a = 7 – 3d     ………….(1)

and S17 = 17/2 [2a + (17 – 1)d]

289 = 17/2 [2a + 16d]

289 = 17(a + 8d)

17 = a + 8d

Substituting the value of ‘a’ from Equation (1),

17 = 7 – 3d + 8d

5d = 10

d = 2

Substituting the value of ‘d’ from Equation (1),

a = 7 – 3 × 2 = 1

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

= n/2 [2×1 + (n – 1)2]

= n/2 [2 + 2n – 2] = n2

10. Show that a1, a2, … , 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.

(i) an = 3 + 4n

Putting n = 1,

a1 = 3 + 4×1 = 7

Putting n = 2,

a2 = 3 + 4×2 = 11

Similarly, a3 = 3 + 4×3 = 15

a4 = 3 + 4×4 = 19

difference in consecutive terms:

a2 – a1 = 11 – 7 = 4

a3 – a2 = 15 – 11 = 4

a4 – a3 = 19 – 15 = 4

The difference between consecutive terms is the same, hence a1, a2, … , an, … is an A.P.

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S15 = 15/2 [2×7 + (15 – 1)4]

S15 = 15/2 [70] = 525

(ii) an = 9 - 5n

Putting n = 1,

a1 = 9 - 5×1 = 4

Putting n = 2,

a2 = 9 - 5×2 = -1

Similarly, a3 = 9 - 5×3 = -6

a4 = 9 - 5×4 = -11

difference in consecutive terms:

a2 – a1 = -1 – 4 = -5

a3 – a2 = -6 – (-1) = -5

a4 – a3 = -11 – (-6) = -5

The difference between consecutive terms is the same, hence a1, a2, … , an, … is an  A.P.

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S15 = 15/2 [2×4 + (15 – 1)×-5]

S15 = 15/2 [-62] = -465

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 second term? Similarly, find the 3rd, the 10th and the nth terms.

The sum of n terms of an AP,

Sn = 4n – n2

Putting n = 1,

First term = a1 = S1 = 4×1 – 12 = 3

Putting n = 2,

sum of first two terms = a1 + a2 = S2 = 4×2 – 22 = 4

a1 + a2 = 4

3 + a2 = 4              [First term a1 = 3 ]

a2 = 1

Hence, the second term is 1

The common differences d = a2 – a1 = 1 – 3 = - 2

Therefore, the tenth term = a10 = a + 9d = 3 + 9×-2 = -16

Similarly, the nth term = an = a + (n – 1)d = 3 + (n – 1)×-2 = 5 – 2n

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

the first 40 positive integers divisible by 6: 6, 12, 18, … , 240

Here a = 6, d = 12 – 6 = 6 and n = 40

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S40 = 40/2 [2×6 + (40 – 1)6]

= 20 [12 + 234]

= 20 [246]

= 4920

Hence, the sum of the first 40 positive integers divisible by 6 is 4920.

13. Find the sum of the first 15 multiples of 8.

The first 15 multiples of 8: 8, 16, 24, … , 120

Here, a = 8, d = 16 - 8 = 8 and n = 15

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S15 = 15/2 [2×8 + (15 – 1)8]

= 15/2 [16 + 112]

= 15/2 (128)

= 960

Hence, the sum of the first 15 multiples of 8 is 960.

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

The odd numbers between 0 and 50: 1, 3, 5, … , 49

Here, a = 1, d = 3 – 1 = 2 and n = 25

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S25 = 25/2 [2×1 + (25 – 1)2]

= 25/2 [2 + 48]

= 25/2 (50)

= 625

Hence, the sum of odd numbers between 0 and 50 is 625.

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

The amount paid as penalty is in the following arithmetic progression:

₹200, ₹250, ₹300, ₹350, …

Here a = 200, d = 250 – 200 = 50 and n = 30

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S30 = 30/2 [2×200 + (300 – 1)50]

= 15[400 + 1450]

= 15(1850)

= 27750

Hence, a contractor will have to pay an amount of ₹27750 as penalty.

16. A sum of700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is20 less than its preceding prize, find the value of each of the prizes.

Considered first prize = ₹x

Total prizes = 7

Total Amount = ₹700

Therefore, the series of 7 prizes are as follows:

x + (x – 20) + (x – 40) + (x – 60) + (x – 80) + (x – 100) + (x – 120) = 700

7x – 420 = 700

7x = 1120

x = 1120/7 = 160

Hence, the amount of 7 prizes is: 160, 140, 120, 100, 80, 60 and 40.

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 1 will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are there sections of each class. How many trees will be planted by the students?

One section of the class will plant 1 tree and each class has three sections. Therefore,

Total number of trees planted by class I students = 3 × 1 = 3

Total number of trees planted by class II students = 3 × 2 = 6

Total number of trees planted by class III students = 3 × 3 = 9

Similarly, the total number of trees planted by the three sections of each class,

3, 6, 9, … , 36

Here a = 3, d =  6 – 3 = 3 and n = 12

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S12 = 12/2 [2×3 + (12 – 1)3]

= 6[6 + 33]

= 6(39)

= 234

Hence, the total number of trees planted by the students of this school = 234

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 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in Figure 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

[ Hint: Length of successive semicircles is  l1, l2, l3, l4,… with centres at A, B, A, B,… respectively.]

Perimeter of semicircle = ½ (2πr)

radii of semicircles: 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …

Therefore, the length of the first spiral l1 = π(0.5) cm

length of second spiral l2 = π(1.0) cm

Similarly, the length of spirals l1, l2, l3, l4, … are as follows:

π(0.5) cm, π(1.0) cm, π(1.5) cm, π(2.0) cm, …

Here a = 0.5π, d = 1.0π – 0.5π = 0.5π and n = 13

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S13 = 13/2 [2(0.5π) + (13 – 1)0.5π]

= 6.5[π + 6π]

= 6.5(7 × 22/7)

= 6.5 × 22 = 143 cm

Hence, the total length of this spiral made of thirteen consecutive semicircles is 143 cm.

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 (see Figure 5.5). In how many rows are the 200 logs placed and how many logs are in the top row?

Bottom row logs = 20

logs in next row = 19

logs in next row = 18

Similarly, the A.P. of logs is,

20, 19, 18, 17, …

Here  a = 20, d = 19 – 20 = -1 and Sn = 200

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

200 = n/2 [2×20 + (n – 1)×-1]

400 = n[40 – n + 1]

400 = 41n – n2

n2 – 41n + 400 = 0

n2 – 16n – 25n + 400 = 0

n(n – 16) – 25(n – 16) = 0

(n – 16) (n – 25) = 0

n – 16 = 0, n – 25 = 0

n = 16, n = 25

If n = 16, a16 = a + 15d = 20 + 15(-1) = 5

If n = 25, a25 = a + 24d = 20 + 24(-1) = -4 जो कि संभव नहीं है।

Hence, these 200 logs are placed in 16 rows and the top row has 5 logs.

20. In 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 figure 5.6).

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) ]

Distance travel by first picking up the potato and putting it in the bucket = 2×5 = 10

Distance travel by second picking up the potato and putting it in the bucket = 2 × (5 +3) = 16

Similarly, the distance covered in pick up each potato is A.P. following as:

10, 16, 22, 28, …

Here a = 10, d = 16 – 10 = 6 and n = 10

The sum of n terms of an AP,

Sn = n/2 [2a+ (n-1)d]

S10 = 10/2 [2×10 + (10 – 1)6]

= 5[20 + 54]

= 5(74)

= 370

The total distance the competitor has to run 370 m.