The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Given,

First term (a) = 5

Last term (l) = 45

Let number of terms be n and common difference be d.

By formula,

S = $\dfrac{n}{2}[a + l]$

Substituting values, we get :

$\Rightarrow 400 = \dfrac{n}{2}[5 + 45] \\[1em] \Rightarrow 400 = \dfrac{n}{2} \times 50 \\[1em] \Rightarrow 400 = 25n \\[1em] \Rightarrow n = \dfrac{400}{25} \\[1em] \Rightarrow n = 16.$

Since, nth term is the last term.

⇒ a_n = 45

⇒ a + (n - 1)d = 45

⇒ 5 + (16 - 1)d = 45

⇒ 5 + 15d = 45

⇒ 15d = 40

⇒ d = $\dfrac{40}{15} = \dfrac{8}{3}$.

Hence, no. of terms = 16 and common difference = $\dfrac{8}{3}$.