In an A.P., given d = 5, S₉ = 75, find a and a₉.

By formula,

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

Given,

S₉ = 75

$\Rightarrow 75 = \dfrac{9}{2}[2 \times a + (9 - 1) \times 5] \\[1em] \Rightarrow \dfrac{75 \times 2}{9} = 2a + 8 \times 5 \\[1em] \Rightarrow \dfrac{150}{9} = 2a + 40 \\[1em] \Rightarrow 2a + 40 = \dfrac{50}{3} \\[1em] \Rightarrow 2a = \dfrac{50}{3} - 40 \\[1em] \Rightarrow 2a = \dfrac{50- 120}{3} \\[1em] \Rightarrow a = -\dfrac{70}{6} = -\dfrac{35}{3}.$

By formula,

a_n = a + (n - 1)d

Substituting values we get :

$\Rightarrow a_9 = -\dfrac{35}{3} + (9 - 1) \times 5 \\[1em] = -\dfrac{35}{3} + 8 \times 5 \\[1em] = -\dfrac{35}{3} + 40 \\[1em] = \dfrac{-35 + 120}{3} \\[1em] = \dfrac{85}{3}.$

Hence, a = $-\dfrac{35}{3} \text{ and } a_9 = \dfrac{85}{3}$.