Answer :

The given series is A.P whose first term is ‘a’ and common difference is ‘d’.

We know that,



[ Sn = qn2]


2qn = 2a + (n – 1)d


2qn – (n – 1)d = 2a …(i)


and


[ Sm = qm2]


2qm = 2a + (m – 1)d


2qm – (m – 1)d = 2a …(ii)


Solving eq. (i) and (ii), we get


2qn – (n – 1)d = 2qm – (m – 1)d


2qn – 2qm = (n – 1)d – (m – 1)d


2q(n – m) = d[n – 1 – (m – 1)]


2q(n – m) = d[n – 1 – m + 1]


2q(n – m) = d(n – m)


2q = d


Putting the value of d in eq. (i), we get


2qn – (n – 1)(2q) = 2a


2qn – 2qn + 2q = 2a


2q = 2a


q = a


a = q and d = 2q. So,





Sq = q2 + q2(q – 1)


Sq = q2 + q3 – q2


Sq = q3


Hence, the correct option is (c)

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