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  4. 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

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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

Sequences and Series

Solution:

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} $