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 = Sp + q - Sp]
[Hint: Required sum = Sp + q - Sp]
Answer
View full question & answer→Let the common differeence of the given A.P be d.
Given that Sp = 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 Sp = 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,
$=\text{S}_{\text{p + q}}-\text{S}_\text{p}=\text{S}_{\text{p + q}}-0$
$=\frac{\text{p + q}}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$=\frac{\text{p}+\text{q}}{2}[2\text{a}+(\text{p}-1)\text{d}+\text{qd}]$
$=\frac{\text{p + q}}{2}\Big[0+\frac{\text{q}-2\text{a}}{\text{p}-1}\Big]$
$=\frac{-\text{a}(\text{p + q})\text{q}}{\text{p}-1}$