Question

if qthterm is 1/p and pth term is 1/q then prove that 1+pq/ 2 where, p≠q​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• $\sf p {}^{th} \: term = \dfrac{1}{q}$

• $\sf q {}^{th} \: term = \dfrac{1}{p}$

$\;\;\underline{\textbf{\textsf{ To Prove:-}}}$

• The sum of first pq term is $S_{pq} = \dfrac{1+pq}{2}$

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

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

Given that,

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

Again,

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

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$\underline{\:\textsf{ Subtracting (1) from (2) :}}$

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

$\underline{\:\textsf{ Now putting the value of d in eq(1) :}}$

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

Here, we have

• First term , a = $\dfrac{1}{pq}$

• Common difference, d = $\dfrac{1}{pq}$

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Now, find the sum

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

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

( Proved)

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Answer 2

if qthterm is 1/p and pth term is 1/q then prove that 1+pq/ 2 where, p≠qif qthterm is 1/p and pth term is 1/q then prove that 1+pq/ 2 where, p≠q