If the sum of p terms of an AP is $3 p^{2} + 4 p$ . Find its $n^{t h}$ term.

Open in App

Solution

Given sum of first $p$ terms is $3 p^{2} + 4 p$
So sum of $n$ terms is $3 n^{2} + 4 n$
Sum of $(n - 1)$ terms is $3 (n - 1)^{2} + 4 (n - 1)$
$n^{t h}$ term is the difference between the sum of $n$ terms and sum of $(n - 1)$ terms
$⟹ n^{t h}$ term is $3 n^{2} + 4 n - 3 (n - 1)^{2} - 4 (n - 1) = 3 (n^{2} - (n - 1)^{2}) + 4 (n - (n - 1)) = 3 (n + n - 1) (n - n + 1) + 4 (n - n + 1)$
$= 3 (2 n - 1) + 4 = 6 n + 1$

Suggest Corrections

Similar questions

p

A P

3 p^{2} + 4 p

{3 n}^{2} + 5 n

n^{t h}

n^{t h}

A P

a_{n} = 2 n + 1

(\frac{5 n^{2}}{2} + \frac{3 n}{2})

(\frac{3 n^{2}}{2} + \frac{5 n}{2})

S_{n}

5 n^{2} + 3 n

Join BYJU'S Learning Program