If Sn=n2 a+(n/4)(n-1) d is the sum of first n terms of an A.P., then common difference is equal to

Question

If Sn=n2 a+(n/4)(n-1) d is the sum of first n terms of an A.P., then common difference is equal to (A) a +2 d (B) 2 a + d (C) (a+d/2) (D) 2 a +( d /2)

Tardigrade Admin · Accepted Answer

Let d1 be the common difference, so, S 1= a = T 1 S 2=4 a +( d /2) ⇒ T 1+ T 2=4 a +( d /2) ⇒ T 2=3 a +( d /2)= a + d 1 ⇒ d 1= T 2- T 1=2 a +( d /2)

Q. If $S_{n} = n^{2} a + \frac{n}{4} (n - 1) d$ is the sum of first $n$ terms of an A.P., then common difference is equal to

$a + 2 d$

$2 a + d$

$\frac{a + d}{2}$

$2 a + \frac{d}{2}$

Let $d_{1}$ be the common difference,
so, $S_{1} = a = T_{1}$
$S_{2} = 4 a + \frac{d}{2} \Rightarrow T_{1} + T_{2} = 4 a + \frac{d}{2}$
$\Rightarrow T_{2} = 3 a + \frac{d}{2} = a + d_{1}$
$\Rightarrow d_{1} = T_{2} - T_{1} = 2 a + \frac{d}{2}$

Q. If Sn​=n2a+4n​(n−1)d is the sum of first n terms of an A.P., then common difference is equal to

a+2d

2a+d

2a+d​

2a+2d​