The first and the last terms of an AP are 5 and 45 respectively.

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 $\frac{8}{3}$.

Leave a comment