In an A.P., given a₁₂ = 37, d = 3, find a and S₁₂.

By formula,

a_n = a + (n - 1)d

Given,

⇒ d = 3

⇒ a₁₂ = 37

⇒ a + 3(12 - 1) = 37

⇒ a + 3(11) = 37

⇒ a + 33 = 37

⇒ a = 4.

By formula,

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

Substituting values we get :

$\Rightarrow S_{12} = \dfrac{12}{2}[2 \times 4 + (12 - 1) \times 3] \\[1em] = 6 \times [8 + 11 \times 3] \\[1em] = 6 \times [8 + 33] \\[1em] = 6 \times 41 \\[1em] = 246.$

Hence, a = 4 and S₁₂ = 246.