Question

If the sum of n terms of an A.P. is (pn+qn^2), where p and q are constants, find the common difference...​

Answer 1

Answer:

••hey mate••

••here is ur answer••

➡️Given ,

Sn=pn +qn^2

we know ,Tn =Sn-S (n-1)

where Tn is nth term of AP

so,

Tn =pn+qn^2-q (n-1)^2-p (n-1)=q (n^2-n^2+2n-1)+p (n-n+1)=q (2n-1)+p

hence ,

Tn =q (2n-1)+p

put n=1

T1 = q+pn=2

T2 =3q +p

common difference =T2-T1=3q+p-q-p=2q

hence ,common difference =2q

》hope it helps u deaR..《

✌✌

☆plz like and mark brainliest ☆

Answer 2

$sn = (pn + {qn}^{2} )$

$s1 = a1 = (p1 + q \times {1}^{2} )$

$a1 = (p + q)$

$a2 = p + 3q$

$d = 2q$