Question

1. if the sum of n terms of an ap is np+1/2n(n-1)Q, where P and Q are constant , find the common difference.

2. if the sum of n term of an AP is pn²+qn², where P and q are constant , find the common difference.

Answer 1

S(n) = np+(1/2)(n)(n-1)q 

The first term, a[1] = S(1), so substitute n=1 
p + (1/2)(1)(0)q = p 

S(2), substitute n=2 
= 2p+(1/2)(2)q = 2p+q 

Now a[2] = S(2) - S(1) 
= p+q 

Common difference: a[2] - a[1] 
(p+q)-p = q

☆2nd question :--

Formula of sum = Sn = n/2 ( 2a + ( n-1)d ) 
where n terms , a first term , d common difference . 
Sn = pn^2 + qn² = n ( p + qn ) = n/2 ( 2p + 2qn ) = n/2 ( 2p +2q + (n-1 ) 2q ) 
then we can see that 
a = 2p+2q ; 
d = 2q 
hence , common diff. be 2q.

Answer 2

Np + n/2(n-1)q

N/2(2p + (n-1)q

Compare with n/2(2a + (n-1)d

We get a = p

D = q