Question 12 Marks
The first term of an $A.P.$ is a and the sum of the first $p$ terms is zero$,$ show that the sum of its next $q$ term is $\frac{-\text{a}(\text{p + q})\text{q}}{\text{p}-1}.$
$[$Hint: Required sum $= S_{p + q}- S_p]$
$[$Hint: Required sum $= S_{p + q}- S_p]$
Answer
View full question & answer→Let the common differeence of the given $A.P$ be d.
Given that $S_p = 0$
$\Rightarrow\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]=0$
$\Rightarrow2\text{a}+(\text{p}-1)\text{d}=0$
$\Rightarrow\text{d}=\frac{-2\text{a}}{\text{p}-1}$
Now$,$ sum of next $q$ terms$,$
$=\text{S}_{\text{p + q}}-\text{S}_\text{p}=\text{S}_{\text{p + q}}-0$
$=\frac{\text{p + q}}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$=\frac{\text{p}+\text{q}}{2}[2\text{a}+(\text{p}-1)\text{d}+\text{qd}]$
$=\frac{\text{p + q}}{2}\Big[0+\frac{\text{q}-2\text{a}}{\text{p}-1}\Big]$
$=\frac{-\text{a}(\text{p + q})\text{q}}{\text{p}-1}$
Given that $S_p = 0$
$\Rightarrow\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]=0$
$\Rightarrow2\text{a}+(\text{p}-1)\text{d}=0$
$\Rightarrow\text{d}=\frac{-2\text{a}}{\text{p}-1}$
Now$,$ sum of next $q$ terms$,$
$=\text{S}_{\text{p + q}}-\text{S}_\text{p}=\text{S}_{\text{p + q}}-0$
$=\frac{\text{p + q}}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$=\frac{\text{p}+\text{q}}{2}[2\text{a}+(\text{p}-1)\text{d}+\text{qd}]$
$=\frac{\text{p + q}}{2}\Big[0+\frac{\text{q}-2\text{a}}{\text{p}-1}\Big]$
$=\frac{-\text{a}(\text{p + q})\text{q}}{\text{p}-1}$