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nth Term of an AP

If the sum of...

Question

If the sum of an terms of an A.P. is $3 n^{2} + 5 n$ and its $m^{t h}$ term is 164, find the value of m.

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Solution

Here, $S_{n} = 3 n^{2} + 5 n a n d a_{m} = 164.$

Replacing $n b y (n - 1) i n S_{n},$ We have

$S_{n - 1} = 3 (n - 1)^{2} + 5 (n - 1)$

= $3 (n^{2} - 2 n + 1) + 5 n - 5$

= $3 n^{2} - 6 n + 3 + 5 n - 5 = 3 n^{2} - n - 2$

We know that $a_{n} = S_{n} - S_{n - 1}$

$∴ a_{n} = 3 n^{2} + 5 n - (3 n^{2} - n - 2)$

= $3 n^{2} + 5 n - 3 n^{2} + n + 2 = 6 n + 2$
$∴ 164 = 6 m + 2$
$6 m = 162$
$m = 27$

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