In-an-A-P-the-sum-of-p-terms-is-q-and-the-sum-of-q-terms-is-p-Prove-that-the-sum-of-p-q-terms-is-p-q-

Question Number 21502 by 786786AM last updated on 25/Sep/17

In an A . P, the sum of p terms is q and the sum of q terms is p .

Prove that the sum of (p + q) terms is - (p + q) .

Answered by $@ty@m last updated on 25/Sep/17

A T Q

S_{p} = q

\frac{p}{2} [2 a + (p - 1) d] = q

\Rightarrow 2 a + (p - 1) d = \frac{2 q}{p} - - (1)

S_{q} = p

\frac{q}{2} [2 a + (q - 1) d] = p

\Rightarrow 2 a + (q - 1) d = \frac{2 p}{q} - - (2)

S u b t r a c t i n g (2) f r o m (1)

(p - 1 - q + 1) d = \frac{2 q}{p} - \frac{2 p}{q}

(p - q) d = \frac{2 (q^{2} - p^{2})}{p q}

d = \frac{- 2 (p + q)}{p q} - - (3)

∴ f r o m (1),

2 a + (p - 1) \frac{- 2 (p + q)}{p q} = \frac{2 q}{p}

2 a = \frac{2 (p - 1) (p + q)}{p q} + \frac{2 q}{p} - - - (4)

∴ S_{p + q} = \frac{p + q}{2} [2 a + (p + q - 1) d]

\Rightarrow S_{p + q} = \frac{p + q}{2} [\frac{2 (p - 1) (p + q)}{p q} + \frac{2 q}{p} + (p + q - 1) \frac{- 2 (p + q)}{p q}]

f r o m (3) & (4)

= \frac{2 (p + q)}{2 p q} [p^{2} - p + p q - q + q^{2} - {(p + q)}^{2} + p + q]

= \frac{p + q}{p q} \times (- p q)

= - (p + q)

Q . E . D .

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