# Q : 8      Find the sum of first $\small 51$ terms of an AP whose second and third terms are $\small 14$ and $\small 18$               respectively.

G Gautam harsolia

It is given that
$\small a_{2}=14,a_3=18,n = 51$
And $d= a_3-a_2= 18-14=4$
Now,
$a_2 = a+d$
$a= 14-4 = 10$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow S_{51}= \frac{51}{2}\left \{ 2\times(10) +(51-1)4\right \}$
$\Rightarrow S_{51}= \frac{51}{2}\left \{ 20 +200\right \}$
$\Rightarrow S_{51}= \frac{51}{2}\left \{ 220\right \}$
$\Rightarrow S_{51}= 51 \times 110$
$\Rightarrow S_{51}=5304$
Therefore, there are 51 terms  and their sum is 5610

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