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.

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 = $\dfrac{5d}{3}$ …..(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 - $\dfrac{5d}{3}$ = 4

⇒ $\dfrac{9d - 5d}{3}$ = 4

⇒ 4d = 12

⇒ d = $\dfrac{12}{4}$

⇒ d = 3

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

⇒ a = $\dfrac{5d}{3}$

⇒ a = $\dfrac{5(3)}{3}$

⇒ a = 5

By formula,

$S_n = \dfrac{n}{2}[2a + (n - 1)d]$

where, a = first term, d = common difference

$\Rightarrow S_{9} = \dfrac{9}{2}[2(5) + 8(3)] \\[1em] = \dfrac{9}{2}[10 + 24] \\[1em] = \dfrac{9}{2}[34] \\[1em] = 9 \times 17 \\[1em] = 153.$

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