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

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25

26

27

28

answer is C.

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Detailed Solution

We are given the sum of n terms as

S_{n} = 3 n^{2} + 5 n

.
For n = 1, we will get the first term of the AP.

S_{1} = 3 + 5 = 8

Hence first term is 8.
Now the second term will be,

T_{2} = S_{2} - S_{1}

\Rightarrow T_{2} = (3 (2)^{2} + 5 (2)) - 8

\Rightarrow T_{2}

= 12 + 10 - 8

\Rightarrow T_{2} = 14

So, the common difference is

d = T_{2} - T_{1}

d = 14 - 8

d = 6

We know that

n^{th}

term of an AP is given as,

t_{n} = a + (n - 1) d

.
We are given that

m^{th}

term of the AP is 164.
So, we get

T_{m} = 164

\Rightarrow 164 = 8 + (m - 1) 6

\Rightarrow 164 - 8 = (m - 1) 6

\Rightarrow 156 = (m - 1) 6

\Rightarrow 26 = m - 1

\Rightarrow 26 + 1 = m

\Rightarrow m = 27

Therefore, the value of m is 27.
Hence, the correct option is 3.

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