# 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?

Nov 15, 2017

$m = 27.$

#### Explanation:

Let ${S}_{n} \mathmr{and} {T}_{n}$ denote the sum of the first $n$ terms and the

${n}^{t h}$ term of the sequence in question, respectively.

Then, ${S}_{n} = {T}_{1} + {T}_{2} + \ldots + {T}_{n - 1} + {T}_{n} .$

$\therefore {S}_{n} = \left\{{T}_{1} + {T}_{2} + \ldots + {T}_{n - 1}\right\} + {T}_{n} .$

But, ${T}_{1} + {T}_{2} + \ldots + {T}_{n - 1} = {S}_{n - 1} .$

$\Rightarrow {S}_{n} = {S}_{n - 1} + {T}_{n} \ldots \ldots \ldots \ldots \left(\ast\right) .$

Since, ${S}_{n} = 3 {n}^{2} + 5 n , \therefore {S}_{n - 1} = 3 {\left(n - 1\right)}^{2} + 5 \left(n - 1\right) .$

Therefore, by $\left(\ast\right) ,$ we have,

${T}_{n} = {S}_{n} - {S}_{n - 1} ,$

$= 3 {n}^{2} + 5 n - \left\{3 {\left(n - 1\right)}^{2} + 5 \left(n - 1\right)\right\} ,$

$= 3 {n}^{2} + 5 n - \left(3 {n}^{2} - 6 n + 3 + 5 n - 5\right) ,$

$\therefore {T}_{n} = 6 n + 2.$

Now, given that ${T}_{m} = 164 \Rightarrow 6 m + 2 = 164.$

$\therefore m = \frac{164 - 2}{6} = 27.$