In a sequence the sum of the nth terms is 3n^2+5n.If its mth term is 164, then what is the value of m?

1 Answer
Nov 15, 2017

m=27.

Explanation:

Let S_n and T_n denote the sum of the first n terms and the

n^(th) term of the sequence in question, respectively.

Then, S_n=T_1+T_2+...+T_(n-1)+T_n.

:. S_n={T_1+T_2+...+T_(n-1)}+T_n.

But, T_1+T_2+...+T_(n-1)=S_(n-1).

rArr S_n=S_(n-1)+T_n............(ast).

Since, S_n=3n^2+5n, :. S_(n-1)=3(n-1)^2+5(n-1).

Therefore, by (ast), we have,

T_n=S_n-S_(n-1),

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

=3n^2+5n-(3n^2-6n+3+5n-5),

:. T_n=6n+2.

Now, given that T_m=164 rArr 6m+2=164.

:. m=(164-2)/6=27.