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If the sum of the first p terms of an A.P. is v and the sum of the first v terms is $p$ ; then show that the sum of the first $(p+v)$ terms is $-(p+v)$ .

Answers (1)

Given, sum of first $p$ terms, $S_{p}=v$

$\Rightarrow S_{P}=\frac{p}{2}[2a+(p-1)d]=v$ ______(a)

Sum of first $v$ terms, $S_{q}=p$

$\Rightarrow S_{q}=\frac{V}{2}\left [ 2a+(V-1)d \right ]=P$ ______(b)

To prove, $S_{(p+v)}=-\left \{ P+V \right \}$

$S_{(p+v)}=\frac{(P+v)}{2}\left [ 2a+(p+v-1)d \right ]$ ________(c)

Let's find value of $'a'$ and $'d'$ in terms of $P$ and $V$ .

$\Rightarrow$ eq (a) can be written as $2a+(P-1)d=\frac{2v}{p}$ _____(i)

$\Rightarrow$ eq (b) can be written as $2a+(v-1)d=\frac{2p}{v}$ _______(ii)
Substarct eq. (i) from (ii)

$(2a+(v-1)d)-(2a+(p-1)d)=\frac{2p}{v}-\frac{2v}{p}$

$2a+(v-1)d-2a-(p-1)d=\frac{2p}{v}-\frac{2v}{p}$

$\Rightarrow (v-p)d=2\left ( \frac{p^{2}-v^{2}}{pv} \right )$

$\Rightarrow (v-p)d=2\left [ \frac{(p-v)(p+q)}{pv} \right ]$

$\Rightarrow d=\frac{-2(p+v)}{pv}$

$\Rightarrow$ now put the value 'd' in eq (i)

$\Rightarrow 2a+(p-1)\left [ \frac{-2(p+v)}{pv} \right ]=\frac{2v}{p}$

$\Rightarrow a+\frac{\left [ -(p-1)(p+v) \right ]}{pv}=\frac{v}{p}$

$\Rightarrow a=\frac{v}{p}+\frac{(p-1)(p+v)}{pv}$

$a=\frac{v^{2}+p^{2}-v+pv-p}{pv}$

$a=\frac{p^{2}+v^{2}-p-v+pv}{pv}$

now, put values of 'a' and 'd' in eq. (c)

$S_{(p+v)}=\frac{(p+v)}{2}\left [ 2\frac{(p^{2}+v^{2})-p-v+pv}{pv}-\frac{2(p+v-1)(p+v)}{pv} \right ]$

$=\frac{(pv)}{2}\left [ 2\left \{ \frac{p^{2}+v^{2}-p-v+pv-p^{2}-2pv+p+v}{pv} \right \} \right ]$

$=(p+v)\left \{ -\frac{pv}{pv} \right \}$

$S_{(p+v)}=-(p+v)$

hence proved

Posted by

Safeer PP

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