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jaswasri2006 @jaswasri2006

March 2021 1 73 Report

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Stera

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Let us consider the first term and common difference of the AP be a and d respectively

Given ,

$\sf p {}^{th} \: term = \dfrac{1}{q} \\ \\ \implies \sf a + (p - 1)d = \dfrac{1}{q} \dashrightarrow(1)$

And

$\sf q {}^{th} \: term = \dfrac{1}{p} \\ \\ \sf \implies a + (q - 1)d = \dfrac{1}{p} \dashrightarrow(2)$

Subtracting (1) from (2)

$\sf\implies a + (q-1)d - \{a +(p-1)d \} = \dfrac{1}{p} - \dfrac{1}{q}\\\\ \sf\implies (q-1)d - (p-1)d = \dfrac{q - p}{pq} \\\\ \sf\implies d(q - 1 - p + 1) = \dfrac{q-p}{pq} \\\\ \sf\implies d(q - p) = \dfrac{q - p}{pq} \\\\ \sf\implies d = \dfrac{1}{pq}$

Now putting the value of ‘d’ in (1)

$\sf\implies a + (p - 1)\times\dfrac{1}{pq} = \dfrac{1}{q} \\\\ \sf\implies a = \dfrac{1}{q} - \dfrac{p-1}{pq} \\\\ \sf\implies a = \dfrac{p-p+1}{pq} \\\\ \sf\implies a = \dfrac{1}{pq}$

Thus ,

first term , a = 1/pq

common difference , d = 1/pq

Thus , sum upto pq terms is :

$\sf S_{pq} = \dfrac{pq}{2}\left\{2\times \dfrac{1}{pq} + (pq-1)\times\dfrac{1}{pq} \right\} \\\\ \sf\implies S_{pq} = \dfrac{pq}{2}\left\{ \dfrac{2 + pq - 1}{pq} \right\} \\\\ \sf\implies S_{pq} = \dfrac{pq}{2}\left\{\dfrac{1+pq}{pq} \right\} \\\\ \sf\implies S_{pq} = \dfrac{1+pq}{2}$

Where p≠q (proved)

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