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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.

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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

Posted by

Gautam harsolia

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