Answer :

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

We know that,

[∵ S_{n} = qn^{2}]

⇒ 2qn = 2a + (n – 1)d

⇒ 2qn – (n – 1)d = 2a …(i)

and

[∵ S_{m} = qm^{2}]

⇒ 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,

⇒ S_{q} = q^{2} + q^{2}(q – 1)

⇒ S_{q} = q^{2} + q^{3} – q^{2}

⇒ S_{q} = q^{3}

Hence, the correct option is **(c)**

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