NCERT Solution Class X Mathematics Arithmetic Progressions Question 3 (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.

Answer:

(i) Given that, a = 5, d = 3, an = 50

As an = a + (n − 1)d,

∴ 50 = 5 + (n − 1)3

45 = (n − 1)3

15 = n − 1

n = 16

(ii) Given that, a = 7, a13 = 35

As an = a + (n − 1) d,

∴ a13 = a + (13 − 1) d

35 = 7 + 12 d

35 − 7 = 12d

28 = 12d

(iii)Given that, a12 = 37, d = 3

As an = a + (n − 1)d,

a12 = a + (12 − 1)3

37 = a + 33

a = 4

(iv) Given that, a3 = 15, S10 = 125

As an = a + (n − 1)d,

a3 = a + (3 − 1)d

15 = a + 2d              (i)

On multiplying equation (1) by 2, we obtain

30 = 2a + 4d (iii)

On subtracting equation (iii) from (ii), we obtain

−5 = 5d

d = −1

From equation (i),

15 = a + 2(−1)

15 = a − 2

a = 17

a10 = a + (10 − 1)d

a10 = 17 + (9) (−1)

a10 = 17 − 9 = 8

(v)Given that, d = 5, S9 = 75

(vi) Given that, a = 2, d = 8, Sn = 90

90 = n [2 + (n − 1)4]

90 = n [2 + 4n − 4]

90 = n (4n − 2) = 4n2 − 2n

4n2 − 2n − 90 = 0

4n2 − 20n + 18n − 90 = 0

4n (n − 5) + 18 (n − 5) = 0

(n − 5) (4n + 18) = 0

Either n − 5 = 0 or 4n + 18 = 0

n = 5 or n =−(18/4) = −9/2

However, n can neither be negative nor fractional.

Therefore, n = 5

an = a + (n − 1)d

a5 = 2 + (5 − 1)8

= 2 + (4) (8)

= 2 + 32 = 34

(vii) Given that, a = 8, an = 62, Sn = 210

n = 6

an = a + (n − 1)d

62 = 8 + (6 − 1)d

62 − 8 = 5d

54 = 5d

d = 54/5

(viii) Given that, an = 4, d = 2, Sn = −14

an = a + (n − 1)d

4 = a + (n − 1)2

4 = a + 2n − 2

a + 2n = 6

a = 6 − 2n (i)

−28 = n (a + 4)

−28 = n (6 − 2n + 4) {From equation (i)}

−28 = n (− 2n + 10)

−28 = − 2n2 + 10n

2n2 − 10n − 28 = 0

n2 − 5n −14 = 0

n2 − 7n + 2n − 14 = 0

n (n − 7) + 2(n − 7) = 0

(n − 7) (n + 2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we obtain

a = 6 − 2n

a = 6 − 2(7)

= 6 − 14

= −8

(ix)Given that, a = 3, n = 8, S = 192

192 = 4 [6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

(x)Given that, l = 28, S = 144 and there are total of 9 terms.

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