Question

Can you all answer this question​

Answer 1

If a be the first term of an arithmetic progression and, the sum of its first p terms is zero, show that the sum of its next q terms is : $\rm -\bigg\lgroup\dfrac{p+q}{p-1}\bigg\rgroup ~aq$

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We have :

$\quad$★ First term of AP : a

$\quad$★ Sum of p terms : 0 $\rm [S_p=0]$

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So, now let the common difference of this AP be d.

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We know that :

$\rm :\implies ~ S_n=\dfrac{n}{2}~ \{2a+(n-1)d\}$

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From the above formula, let's get d as we know that $\rm S_p=0$ :

$\rm: \implies~ S_p:\dfrac{p}{2}[2a+(p-1)d]=0$

$\rm:\implies~ 2a+(p-1)d=0\bigg(\dfrac{2}{p}\bigg)$

$\rm :\implies~ 2a+(p-1)d=0$

$\rm :\implies~ d=\dfrac{-2a}{p-1}~~~~~[eq^n.1]$

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Now, let's find the sum of next q terms of the AP using the value of d from the above value.

★ Sum of next q terms = Sum of first p terms + q terms – Sum of p terms

$\rm :\implies ~ S_{(p+q)}-S_p=S_{(p+q)}$

$\rm :\implies~ S_{(p+q)}-0=S_{(p+q)}~~~~~[\because ~ S_p=0]$

$\rm :\implies ~ S_{(p+q)}=S_{(p+q)}$

$:\rm\implies~ S_{(p+q)}=\dfrac{p+q}{2}[2a+(p+q-1)d]$

$\rm :\implies~ S_{(p+q)}=\dfrac{p+q}{2}[2a+(p+q-1)\bigg(\dfrac{-2a}{p-1}\bigg)~~~~~[From ~eq^n.1]$

$\rm :\implies ~ S_{(p+q)}=\dfrac{p+q}{2(p-1)}(2ap-2ap-2a+2a-2aq)$

$\rm :\implies ~ S_{(p+q)}=\dfrac{p+q}{2(p-1)}(-2aq)$

$\rm :\implies S_{(p+q)}=-\bigg\lgroup\dfrac{p+q}{p-1}\bigg\rgroup~aq$

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Henceforth, proved!