Question

18^th term of an A.P. is equal to 4 times its 4^th term and the 6^th term exceeds twice the 2^nd term by 4. Find the sum of the first 9 terms of this A.P.

Answer 1

We know that,

a_n = a + (n − 1)d

Given,

18^th term of an A.P. is equal to 4 times its 4^th term

a₁₈ = a + 17d

a₄ = a + 3d

⇒ a + 17d = 4(a + 3d)

⇒ a + 17d = 4a + 12d

⇒ a - 4a + 17d − 12d = 0

⇒ -3a + 5d = 0

⇒ 5d = 3a

⇒ a = `(5d)/3`​ .....(1)

Given,

6th term exceeds twice 2nd term by 4

a₆ = a + 5d

a₂ = a + d

⇒ a + 5d = 2(a + d) + 4

⇒ a + 5d = 2a + 2d + 4

⇒ a + 5d − 2a − 2d = 4

⇒ 3d − a = 4

⇒ 3d − a = 4 ...(2)

Substituting value of a from equation (1) in equation (2), we get:

⇒ 3d − `(5d)/3`​ = 4

⇒ `(9d − 5d)/3 = 4`

⇒ 4d = 12

⇒ d = `12/4`

⇒ d = 3

Substituting value of d in equation (1), we get :

⇒ a = `(5d)/3`

⇒ a = `(5(3))/3`

⇒ a = 5

By formula,

`S_n ​= n/2 ​[2a+(n−1)d]`

where, a = first term, d = common difference

⇒ `S_9 = 9/2[2(5)+8(3)]`

= `9/2[10+24]`

= `9/2[34]`

= 9 × 17

=153.

Hence, the sum of the first 9 terms of this A.P. = 153.