Question

The sum of first $n$ terms of $A.P.$ is $5n-2n^2$. Find the $A.P.$ i.e., $a$ and $d$.

Answer 1

Given that the sum of first n terms an A.P is $5n-2n^2$

$S_n=5n-2n^2$

First term, $a_1=S_1=5\left(1\right)-2(1{)}^2=3$

Sum of first two terms ,$S_2=5\left(2\right)-2(2{)}^2=2$

Second term ,$a_2=S_2-S_1=2-3=-1$

Common difference $d=a_2-a_1=-1-3=-4$

$a_n=a+(n-1)d$

$a_n=3+(n-1)(-4)$

$a_n=3-4n+4$

$a_n=7-4n$

$nth$ term of an A.P$=7-4n$

Therefore A.P is

$3,-1,-5,-\mathrm{9........}$

$\therefore a=3,d=-4$