Question:
The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find the common difference and the number of terms.
Solution:
Suppose there are n terms in the AP.
Here, a = 5, l = 45 and Sn = 400
$S_{n}=400$
$\Rightarrow \frac{n}{2}(5+45)=400 \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$
$\Rightarrow \frac{n}{2} \times 50=400$
$\Rightarrow n=\frac{400 \times 2}{50}=16$
Thus, there are 16 terms in the AP.
Let d be the common difference of the AP.
$\therefore a_{16}=45$
$\Rightarrow 5+(16-1) \times d=45 \quad\left[a_{n}=a+(n-1) d\right]$
$\Rightarrow 15 d=45-5=40$
$\Rightarrow d=\frac{40}{15}=\frac{8}{3}$
Hence, the common difference of the AP is $\frac{8}{3}$.