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 first 8 terms of the A.P.

Given, a₄ = 22 and a₁₅ = 66.

By formula, a_n = a + (n - 1)d
⇒ a₄ = a + (4 - 1)d
⇒ 22 = a + 3d
⇒ a = 22 - 3d         (Eq 1)

⇒ a₁₅ = a + (15 - 1)d
⇒ 66 = a + 14d
Putting value of a from Eq 1 in above equation
⇒ 66 = 22 - 3d + 14d
⇒ 66 = 22 + 11d
⇒ 11d = 66 - 22
⇒ 11d = 44
⇒ d = 4.

Putting value of d in Eq 1,
⇒ a = 22 - 3(4)
⇒ a = 22 - 12
⇒ a = 10.

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

⇒ S₈ = $\dfrac{8}{2}[2 \times 10 + (8 - 1)4]$
⇒ S₈ = 4[20 + 28]
⇒ S₈ = 4 × 48
⇒ S₈ = 192.

Hence, first term = a = 10, common difference = d = 4 and sum of first 8 terms = S₈ = 192.