Question

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

Answer 1

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,
$\frac{1}{2n}\left(2a+nd-d\right)=\frac{1}{2m}\left(2a+md-d\right)$
$\Rightarrow m\left(2a+nd-d\right)=n\left(2a+md-d\right)$
$\Rightarrow 2am+mnd-md=2an+mnd-nd$

Solving further, we get,

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

Now,

Thus,

Therefore, the correct option is (C).