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  4. If sum to n terms of an A.P. is 5n2 + 6n and rth term is 401, then the value of r is

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Q. If sum to $n$ terms of an $A.P$. is $5n^2 + 6n$ and $r^{th}$ term is $401$, then the value of $r$ is

Sequences and Series

Solution:

$T_{n} = S_{n} -S_{n-1} $
i.e., $T_{n}= 5n^{2} + 6n -5\left(n-1\right)^{2} -6\left(n-1\right)$
$= 10n +1 $
$ \therefore T_{r} = 10r +1 = 401$
$\Rightarrow 10r= 400$
$\Rightarrow r= 40$