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.

Answers (1)

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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