Question

In an A.P., T₁₂ = 37, d = 3, find a and S₁₂.

Answer 1

Given:

T₁₂ = 37, d = 3

As T_n = a + (n − 1)d,

T₁₂ = a + (12 − 1)3

37 = a + 11 × 3

37 = a + 33

a = 37 − 33

a = 4

`S_n = n/2 [a + T_n]`

`S_12 = 12/2 [4 + 37]` 

S₁₂ = 6 × 41

S₁₂ = 246

Thus, a = 4 and S₁₂ = 246