Find the sum of first 21 terms of the AP whose $2^{n d}$ term is 8 and $4^{t h}$ term is $14 .$

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Solution

Given: $a_{2} = 8$ and $a_{4} = 14$ and $n = 21$
We know that,
$a_{2} = a + d = 8 . . . (i)$
and $a_{4} = a + 3 d = 14 . . . (i i)$
Solving the linear equations $(i)$ and $(i i),$ we get
$a + d - a - 3 d = 8 - 14$
$\Rightarrow - 2 d = - 6$
$\Rightarrow d = 3$
Putting the value of $d$ in equation. $(i),$ we get
$a + 3 = 8$
$\Rightarrow a = 8 - 3 = 5$
Now, we have to find the sum of the first $21$ terms.
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$\Rightarrow S_{21} = \frac{21}{2} [2 \times 5 + (21 - 1) (3)]$
$\Rightarrow S_{21} = \frac{21}{2} [10 + 20 \times (3)]$
$\Rightarrow S_{21} = 21 [5 + 10 \times (3)]$
$\Rightarrow S_{21} = 21 [35]$
$\Rightarrow S_{21} = 735$

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