Question

Q.10. If for an A.P., tn= 6n + 1, then show that
Sn = 3n square + 4n​

Answer 1

Answer:

In any AP

nth term is given by

$T_n \: or \: a_n = 6n + 1$

So,

$a_1 = 6 + 1 = 7$

From this we get,

First term of the AP,

a = 7

We only need first term and last term of an ap to find sum of its first n terms.

Now,

We know that,

sum of first n terms of the AP is given by

$S_n = \dfrac{n}{2} (a + a_n)$

$= \dfrac{n}{2}(7 + 6n + 1) \\ \\ = \frac{n}{2}(6n + 8) \\ \\ = 3 {n}^{2} + 4n$

Hence the required result is proved.