The first term of an A.P. is 5, the last term is 45 and the sum of its term is 1000. Find the number of terms and the common difference of the A.P.

Given, a = 5, l = 45 and S = 1000.

$\Rightarrow S = \dfrac{n}{2}(a + l) \\[1em] \Rightarrow 1000 = \dfrac{n}{2}(5 + 45) \\[1em] \Rightarrow 1000 = \dfrac{n}{2} \times 50 \\[1em] \Rightarrow n = \dfrac{1000 \times 2}{50} \\[1em] \Rightarrow n = 40.$

Given, l = a₄₀ = 45

⇒ a + (40 - 1)d = 45
⇒ 5 + 39d = 45
⇒ 39d = 40
⇒ d = $\dfrac{40}{39}$.

Hence, n = 40 and d = $\dfrac{40}{39}$.