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Q.

If $S_{r}$ denotes the sum of $r$ terms of an AP and $\frac{S_{a}}{a^{2}} = \frac{S_{b}}{b^{2}} = c$ . Then $S_{c}$ is

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a

c^{3}

b

\frac{c}{{ab}^{}}

c

abc

d

a + b + c

answer is A.

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Detailed Solution

Given that,

\frac{S_{a}}{a^{2}} = \frac{S_{b}}{b^{2}} = c

and

S_{r}

denotes the sum of

r

terms.
We have to find the value of

S_{c} .

We know,

(S_{n}) = \frac{n}{2} (2 a + (n - 1) d)

.

\frac{S_{a}}{a^{2}} = \frac{S_{b}}{b^{2}}

\Rightarrow \frac{\frac{a}{2} (2 a_{1} + (a - 1) d)}{a^{2}} = \frac{\frac{b}{2} (2 a_{1} + (b - 1) d)}{b^{2}}

\Rightarrow \frac{2 a_{1}}{a} + \frac{(a - 1) d}{a} = \frac{2 a_{1}}{b} + \frac{(b - 1) d}{b} or

\Rightarrow \frac{2 a_{1}}{a} + d - \frac{d}{a} = \frac{2 a_{1}}{b} + d - \frac{d}{b} or,

\Rightarrow 2 a_{1} (\frac{1}{a} - \frac{1}{b}) = d (\frac{1}{a} - \frac{1}{b})

So, \frac{a}{2} (\frac{2 a_{1} + (a - 1) d}{a^{2}}) = c

2 a_{1} \frac{(1 + a - a)}{2 a} = c

\frac{a_{1} a}{a} = c

c = a_{1}

S_{c} = \frac{c}{2} (2 a_{1} + (c - 1) d)

\Rightarrow S_{c} = \frac{c}{2} (2 a_{1} + (c - 1) 2 a_{1})

\Rightarrow S_{c}

=

\frac{c}{2} (2 c + (c - 1) 2 c)

\Rightarrow S_{c} = \frac{c}{2} \times 2 c (1 + c - 1)

\Rightarrow S_{c} = c^{3}

Therefore,

S_{c} = c^{3}

.
Hence, option 1 is correct.

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