Q. 21

# If in an A.P., S<

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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