If the sum to n terms of an A.P. is $3 n^{2} + 5 n$ while $T_{m} = 164$ , then value of m is

25

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27

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28

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Solution

The correct option is D $27$
$S_{n} = 3 n^{2} + 5 n$
Now
$S_{1} = 3 + 5 = 8$
Hence initial term is 8.
Now the second term is
$T_{2} = S_{2} - S_{1}$
$= (3 (2)^{2} + 5 (2)) - 8$
$= 12 + 10 - 8$
$= 14$ .
Hence common difference is
$d = T_{2} - T_{1}$
$= 14 - 8 = 6$
Now
$T_{n} = a + (n - 1) d$
$= 8 + (n - 1) 6$
$= 6 n + 2$ .
It is given that
$T_{m} = 164$
Or
$164 = 6 m + 2$
$162 = 6 m$
$27 = m$
Hence
$m = 27$ .

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