Question

The sum of first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.

Answer 1

Let a and a₈ be the first and eighth terms of A.P.

Let common difference be d.

∴ a + a₈ = 32   ...(given)

⇒ a + [a + (8 − 1) d] = 32      ...[∵ T_n =a + (n − 1)d]

⇒ a + (a + 7d) = 32

⇒ a + 7d = 32 − a   ...(i)

Also, a . a₈ = 60   ...(given)

⇒ a · [a + (8 − 1) d] = 60

⇒ a (a + 7d) = 60

⇒ a (32 − a) = 60   [from eq. (i)]

⇒ 32a − a² = 60

⇒ a² − 32a + 60 = 0

⇒ a² − 30a − 2a + 60 = 0

⇒ a (a − 30) − 2 (a − 30) = 0

⇒ (a − 30) (a − 2) = 0

⇒ a = 2, 30

For a = 2, from eq. (i)

2 + 7d = 32 − 2

7d = 28

d = 4

For a = 30, from eq. (i)

30 + 7d = 32 − 30

7d = − 28

d = −4

for (a, d) = (2, 4)

a₈ = 2 + 7 × 4 = 30

∴ a + a₈ = 32 and a . a₈ = 60

for (a, d) = (30, −4)

a₈ = 30 + 7 (−4) = 2

∴ a + a₈ = 32 and a · a₈ = 60

Taking (a, d) = (2, 4)

`S_20 = 20/2[2 xx 2 + (20 - 1) xx 4]`   `...[∵ S_n = n/2[2a + (n - 1)d]]`

= 10 [4 + 19 × 4]

= 40 × 20 = 800

Taking (a, d) = (30, −4)

`S_20 = 20/2[30 xx 2 + (20 - 1)(-4)]`     `...[∵ S_n = n/2[2a + (n - 1)d]]`

= 10 [60 + 19(−4)]

= 10 (60 − 76)

= 10 × (− 16)

= −160