The 4th term of an A.P. is 22 and 15th term is 66. Find the first term and the common difference. Hence, find the sum of the series upto 8 terms.

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

The 4th term of an A.P. is 22.

⇒ a₄ = a + (4 - 1)d

⇒ 22 = a + 3d

⇒ a + 3d = 22 …..(1)

Given,

The 15th term of an A.P. is 66.

⇒ a₁₅ = a + (15 - 1)d

⇒ 66 = a + 14d

⇒ a + 14d = 66 …..(2)

Subtracting Equation 1 from Equation 2, we get:

⇒ a + 14d - (a + 3d) = 66 - 22

⇒ a + 14 d - a - 3d = 44

⇒ 11d = 44

⇒ d = $\dfrac{44}{11}$

⇒ d = 4.

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

⇒ a + 3(4) = 22

⇒ a + 12 = 22

⇒ a = 22 - 12

⇒ a = 10.

We know that,

Sum of n terms of an A.P. is given by,

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

⇒ S₈ = $\dfrac{8}{2}$ [2(10) + (8 - 1)4]

= 4[20 + (7)4]

= 4[20 + 28]

= 4 × (48)

= 192.

Hence, a = 10, d = 4, S₈ = 192.