# Q : 5    The first term of an AP is $\small 5$, the last term is $\small 45$ and the sum is $\small 400$. Find the number of terms and the common difference.

G Gautam harsolia

It is given that
$\small a=5,a_n=45,S_n=400,$
Now, we know that
$a_n = a+(n-1)d$
$45 = 5+(n-1)d$
$(n-1)d= 40 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow 400 = \frac{n}{2}\left \{ 2\times(5) +(n-1)d\right \}$
$\Rightarrow 800 = n\left \{ 10+40 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$\Rightarrow 800 = n\left \{ 50 \right \}$
$\Rightarrow n = 16$
Now, put this value in (i) we will get
$d = \frac{40}{15}= \frac{8}{3}$
Therefore, value of n and d are $16 \ and \ \frac{8}{3}$ respectively

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