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Q : 3      In an AP:

             (vii) given  \small a=8,a_n=62,S_n=210, find \small n and \small d.

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It is given that
\small a=8,a_n=62,S_n=210,
Now, we know that
a_n = a+(n-1)d
62 = 8+(n-1)d
(n-1)d= 54 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 210 = \frac{n}{2}\left \{ 2\times(8) +(n-1)d\right \}
\Rightarrow 420 = n\left \{ 16+54 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow 420 = n\left \{ 70 \right \}
\Rightarrow n = 6
Now, put this value in (i) we will get
d = \frac{54}{5}
Therefore, value of n and d are 6 \ and \ \frac{54}{5} respectively

Posted by

Gautam harsolia

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