Question

sum of n terms of a AP = pn + qn² . Find d

Answer 1

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+p
n=2 T2 =3q +p
common difference =T2-T1
=3q+p-q-p=2q
hence ,
common difference =2q

Answer 2

Hey!
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Ques.→ Sum of n terms of an A. P = pn + qn² . Find d .


Ans. → Given that ,


=> Sn = pn + qn² ---------------( Eq. 1 )


We also know that ,


=> Sn = n /2 [ 2a + ( n - 1 ) d ]-------------( Eq. 2 )


★Equate the values from ( 1 ) and ( 2 )


$= > \frac{n}{2} (2a + (n - 1)d) = pn + qn {}^{2}$


( Multiply the terms )


=> na + n² d/2 - n d/2 = pn + qn²


Now , Compare the coefficient on both sides ,


→ d/2 = q


•°• d = 2q ✔


Thus, the common difference d = 2q ✔


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