if the sum of p terms of an ap is q and the sum of q terms isp show that sum f p-q terms is (p-q) (1 +2q/p)

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Solution

$Given, S_{p} = q and S_{q} = p q = \frac{p}{2} [2 a + (p - 1) d] . . . . . . (1) p = \frac{q}{2} [2 a + (q - 1) d] . . . . . . (2) Subtracting after multiplying first equation by q and second by p, we get, q^{2} - p^{2} = d [q (p - 1) - p (q - 1)] q^{2} - p^{2} = d [pq - q - pq + p] q^{2} - p^{2} = d [(p - q)] d = - (p + q) . . . . . . . . . . (3) Putting in equation 1 we get, p [2 a - (p + q) (p - 1)] = 2 q 2 a = \frac{2 q}{p} + (p + q) (p - 1) . . . . . . . . . . . . . (4) Now sum of (p - q) term is, S_{p - q} = \frac{p - q}{2} [2 a + (p - q - 1) d] = \frac{p - q}{2} [\frac{2 q}{p} + (p + q) (p - 1) - (p - q - 1) (p + q)] = \frac{p - q}{2} [\frac{2 q}{p} + (p + q) [p - 1 - p + q + 1]] = \frac{p - q}{2} [\frac{2 q}{p} + q (p + q)]$

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