In an A.P., given a = 7, a₁₃ = 35, find d and S₁₃.

By formula,

a_n = a + (n - 1)d

Given,

⇒ a₁₃ = 35 and a = 7.

⇒ 7 + (13 - 1)d = 35

⇒ 7 + 12d = 35

⇒ 12d = 28

⇒ d = $\dfrac{28}{12} = \dfrac{7}{3}$

By formula,

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

Substituting values we get :

$\Rightarrow S_{13} = \dfrac{13}{2}[2 \times 7 + (13 - 1) \times \dfrac{7}{3}] \\[1em] = \dfrac{13}{2} \times [14 + 12 \times \dfrac{7}{3}] \\[1em] = \dfrac{13}{2} \times [14 + 28] \\[1em] = \dfrac{13}{2} \times 42 \\[1em] = 13 \times 21 \\[1em] = 273.$

Hence, d = $\dfrac{7}{3}$ and S₁₃ = 273.