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

26

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27

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28

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25

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Solution

The correct option is A $27$
Since $S_{n} = {3 n}^{2} + 5 n$
Replace $n$ by $(n - 1)$ ; we get
$S_{n - 1} = {3 (n - 1)}^{2} + 5 (n - 1)$
$= (n - 1) [3 (n - 1) + 5]$
$= (n - 1) [3 n - 3 + 5]$
$= (n - 1) (3 n + 2)$
Now, $T_{n} = S_{n} - S_{n - 1} = 6 n + 2$
Substituting $S_{n}$ and $S_{n - 1}$ ; we get
$T_{n} = 6 n + 2 = 164$
$n = 27$

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

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

3 n^{2} + 5 n

T_{m} = 164

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

n

3 n^{2} + 5 n

m^{t h}

164,

m

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