If sum of $n$ terms of a series is given by $S_{n} = 3 n^{2} + 3 n$ , then $6^{t h}$ term of the series is

Open in App

Solution

Given : $S_{n} = 3 n^{2} + 3 n$
Now,
$T_{n} = S_{n} - S_{n - 1} \Rightarrow T_{n} = 3 n^{2} + 3 n - (3 (n - 1)^{2} + 3 (n - 1)) \Rightarrow T_{n} = 3 n^{2} + 3 n - (3 (n^{2} - 2 n + 1) + 3 n - 3) \Rightarrow T_{n} = 3 n^{2} + 3 n - (3 n^{2} - 3 n) \Rightarrow T_{n} = 6 n ∴ T_{6} = 36$

Suggest Corrections

Similar questions

n

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

n^{t h}

n

S_{n} = n^{2} - n,

10^{t h}

n

S_{n} = n^{2} - n,

10^{t h}

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

6^{t h}

n

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

Join BYJU'S Learning Program