In an A.P., given a₃ = 15, S₁₀ = 125, find d and a₁₀.

By formula,

a_n = a + (n - 1)d

Given,

⇒ a₃ = 15

⇒ a + (3 - 1)d = 15

⇒ a + 2d = 15

⇒ a = 15 - 2d ………(1)

By formula,

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

Given,

⇒ S₁₀ = 125

$\Rightarrow \dfrac{10}{2}[2 \times a + (10 - 1)d] = 125 \\[1em] \Rightarrow 5[2a + 9d] = 125 \\[1em] \Rightarrow 2a + 9d = \dfrac{125}{5} \\[1em] \Rightarrow 2a + 9d = 25$

Substituting value of a from equation (1) in above equation :

$\Rightarrow 2(15 - 2d) + 9d = 25 \\[1em] \Rightarrow 30 - 4d + 9d = 25 \\[1em] \Rightarrow 5d = 25 - 30 \\[1em] \Rightarrow 5d = -5 \\[1em] \Rightarrow d = \dfrac{-5}{5} = -1.$

Substituting value of d in equation (1), we get :

⇒ a = 15 - 2d = 15 - 2(-1) = 15 + 2 = 17.

a₁₀ = 17 + (10 - 1)(-1) = 17 - 9 = 8.

Hence, d = -1 and a₁₀ = 8.