Question

If the sumk of n terms of an A.P. is $(pn+q{n}^2),$ Where pa and q are constants, find the common difference.

Answer 1

Here $S_n=pn+q{n}^2$

Repalcing n by n-1 in $S_n$

$\therefore S_{n-1}=p(n-1)+q(n-1{)}^2$

= $pn-p+q{n}^2-2pn+q$

We know that $a_n=S_n-S_{n-1}$

= $pn+q{n}^2-(pn-p+q{n}^2-2qn+q)$

= $pn+q{n}^2-pn-p+q{n}^2-2qn+q)$

= p + 2qn -q

Replacing n by (n-1) in $a_n$

$\therefore a_{n-1}=p+2q(n-1)-q$

= $p+2qn-2q-q=p+2qn-3q$

We know that d = a_n - a_{n-1}

= $p+2qn-q-(p+2qn-3q)$

=$p+2qn-q-p-2qn+3q=2q$