# If the sum of first n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.

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### 2 Answers

For an arithmetic progression with the first term as a and the common difference as d, the sum of the first n terms is given by (n/2)( 2a + (n-1)*d). The nth term is equal to a + (n-1)*d.

Here we won't need to use any of these expressions.

Instead we use the fact that the sum of the first m terms is given as 3m^2 + 5m and the mth term is 164.

Now, the difference of the sum of the first m terms and the sum of the first (m-1) terms is equal to the mth term.

So tm = 3m^2 + 5m - 3(m-1)^2 - 5(m - 1)

=> tm = 3m^2 + 5m - 3m^2 - 3 + 6m - 5m + 5

=> tm = 6m + 2

Now we have tm = 164

=> 6m + 2 = 164

=> 6m = 162

=> m = 162/6

=> m = 27

Therefore m = 27.

**The required value of m is 27.**

The sum of the n terms Sn = 3n^2+5n.

Therefore the the sum of the n-1 terms = Sn-1 = 3(n-1)^2+5(n-1).

Therefore the nth term of the AP is given by:

an = Sn-Sn-1,

=> Sn-Sn-1= 3n^2+5n-3(n-1)^2+5(n-1).

=> an = 3{n^2-(n-1)^2+5(n-(n-1)).

=> an = 3(n+n-1)(n-n+1) +5(n-n+1)

=> an = 3(2n-1)+5 = 6n-3+5

=> **an = 6n+2**.

Therefore am = 6m+2.

But a m = mth term is given to be 164.

=> am = 6m+2= 144.

=> 6m= 164-2.

=> 6m = 162.

=> 6m/6 = 162 /6 = 27.

=> m= 27.