Question

if the sum of n terms of an A. P is 2mn + pn2 where m and p are constant. Find the common difference​

Answer 1

$\sf \bold {Question\::}$

If the sum of n terms of an A. P is 2mn + pn² where m and p are constant. Find the common difference.

$\sf \bold {Given\::}\: \sf S_{n} =\:2mn\:+\:pn^{2}$

$\sf \bold{To\:find\::}\:\:common difference = d (let)\\\\\sf \bold {Formula\:used\::}\:\sf S_{n}\:=\:(\frac{n}{2})(\:2a\:+\:(n-1)d\:)$

Here

$\bullet$ n = nth term

$\bullet$ a = 1st term

$\bullet$ d = common difference

$\bullet\sf S_{n} = sum\:of\:n\:terms\\\\\\\sf \bold {Solution\::}\\\\As\:we\:know,\\\\\implies\sf S_{n}\:=\:(\frac{n}{2})(\:2a\:+\:(n-1)d\:)\\\\\implies \sf S_{n}\:=\: (\dfrac{n}{2} )(\:2a\:+\: nd - d)\\\\\\\implies \sf S_{n}\:=\:\dfrac{(n\times 2a)+(n\times nd)\:-\:(n \times d)}{2}\\\\\\\implies \sf S_{n}\:=\:\dfrac{2an\:+dn^{2} \:-\:nd}{2}\\\\\\\implies \sf S_{n}\:=\:\dfrac{(2an-\:nd)\:+dn^{2} \:}{2}\\\\\\\implies \sf S_{n}\:=\:\dfrac{(2an-\:nd) }{2}+\:\dfrac{dn^{2}}{2} .....(equation 1)$

Also ,we are given that,

$\bullet$ $\sf S_{n} =\:2mn\:+\:pn^{2}$......(equation 2)

$\sf Putting\:value\:of\:S_{n} , from\: equation\:1\:to\:equation\:2\\\sf We\:get\:,\\\\\sf \dfrac{(2an-\:nd) }{2}+\:\dfrac{dn^{2}}{2}\:=\:2mn\:+\:pn^{2}$

On comparison, we get

$\sf \dfrac{dn^{2}}{2}\:=\:pn^{2}$

{as coefficient of n² will be same}

$\implies \sf \dfrac{d}{2}\:=\:p\\\\\implies \sf d\:=\:p\times2\\\\\implies \sf d\:=\:2p$

$\sf \bold {ANSWER\::}\: common\: difference, \:\sf \bold{d\:=\:2p}$