The 4th term of an A.P. is 22 and 15th term is 66

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

Answer

Let the first term be ‘a’ and the common difference d.
T_n = a + (n – 1) d
∴ T₄ = a + (4 – 1) d, T₁₅ = a + (15 – 1) d
22 = a + 3d      66 = a + 14d
a + 14d = 66 and a + 3d = 22
Solving a = 10, d = 4
S_n = n/2 [2a + (n – 1) d
S₈ = 8/2 [2 × 10 + (8 – 1) 4
= 4 [20 + 28]
∴ S₈ = 192