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

Let nth be the last term so,
a = 5, l = a_n = 45, S_n = 400.

By formula a_n = a + (n - 1)d
⇒ 45 = 5 + (n - 1)d
⇒ 45 - 5 = (n - 1)d
⇒ (n - 1)d = 40      (Eq 1)

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

$\Rightarrow 400 = \dfrac{n}{2}[2 \times 5 + (n - 1)d]$

Putting value of (n - 1)d from Eq 1 in above equation,

$\Rightarrow 400 = \dfrac{n}{2}[10 + 40] \\[1em] \Rightarrow \dfrac{n}{2} \times 50 = 400 \\[1em] \Rightarrow 25n = 400 \\[1em] \Rightarrow n = 16$

Putting value of n in Eq 1 we get,

$\Rightarrow (16 - 1)d = 40 \\[1em] \Rightarrow 15d = 40 \\[1em] \Rightarrow d = \dfrac{40}{15} = \dfrac{8}{3}. \\[1em]$

Hence, number of terms = 16 and common difference = $\dfrac{8}{3}.$