If in an A.P. S_n = n²p and S_m = m²p, where S_r denotes the sum of r terms of the A.P., then S_p is equal to

(a) $\frac{1}{2} p^{3}$

(b) m n p

(c) p³

(d) (m + n) p²

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Solution

In the given problem, we are given an A.P whose and

We need to find

Now, as we know,

Where, first term = a

Common difference = d

Number of terms = n

So,

Similarly,

Equating (1) and (2), we get,
$\frac{1}{2 n} (2 a + n d - d) = \frac{1}{2 m} (2 a + m d - d)$
$\Rightarrow m (2 a + n d - d) = n (2 a + m d - d)$
$\Rightarrow 2 a m + m n d - m d = 2 a n + m n d - n d$

Solving further, we get,

Further, substituting (3) in (1), we get,

Now,

Thus,

Therefore, the correct option is (C).

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Similar questions

If in an A.P., $S_{n} = n^{2} p$ and $S_{m} = m^{2} p$ , where $S_{r}$ denotes the sum of r terms of the A.P., then $S_{p}$ is equal to

S_{n} = n^{2} p

S_{m} = m^{2} p

m \neq n

S_{p}

S_{n} = n^{2} p

S_{m} = m^{2} p, m \neq n

S_{p} = p^{3}

S_{n} = n^{2} p

S_{m} = m^{2} p

m \neq n

S_{n} = p^{3}

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