Question
If $\text{s}_\text{n}=\text{n}^2\ \text{p}$ and $\text{s}_\text{m}=\text{m}^2\ \text{p},\ \text{m}\not=\text{n},$ in an A.P., prove that $\text{s}_\text{p}=\text{p}^3.$
✓
Answer
Let a be the first term of the AP and d is the common difference. then
$\text{s}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$\text{n}^2\text{p}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$\text{np}=\frac{1}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$2\text{np}=2\text{a}+(\text{n}-1)\text{d}\ .....{(1)}$
Again
$\text{s}_\text{m}=\frac{\text{m}}{2}(2\text{a}+(\text{m}-1)\text{d})$
$\text{m}^2\text{p}=\frac{\text{m}}{2}(2\text{a}+(\text{m}-1)\text{d})$
$\text{mp}=\frac{1}{2}[2\text{a}+(\text{m}-1)\text{d}]\ .....{(2)}$
Now subtract (1) from (2)
$2\text{p}(\text{m}-\text{n})=(\text{m}-\text{n})\text{d}$
$\text{d}=2\text{p}$
Therefore
$2\text{mp}=2\text{a}+(\text{m}-1)\times2\text{p}$
$2\text{a}=2\text{p}$
$\text{a}=\text{p}$
The sum up p terms will be:
$\text{s}_\text{p}=\frac{\text{p}}{2}(2\text{a}+(\text{p}-1)\text{d})$
$=\frac{\text{p}}{2}(2\text{p}+(\text{p}-1).2\text{p})$
$=\frac{\text{p}}{2}(2\text{p}+2\text{p}^2-2\text{p})$
$=\text{p}^3$
$\text{s}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$\text{n}^2\text{p}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$\text{np}=\frac{1}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$2\text{np}=2\text{a}+(\text{n}-1)\text{d}\ .....{(1)}$
Again
$\text{s}_\text{m}=\frac{\text{m}}{2}(2\text{a}+(\text{m}-1)\text{d})$
$\text{m}^2\text{p}=\frac{\text{m}}{2}(2\text{a}+(\text{m}-1)\text{d})$
$\text{mp}=\frac{1}{2}[2\text{a}+(\text{m}-1)\text{d}]\ .....{(2)}$
Now subtract (1) from (2)
$2\text{p}(\text{m}-\text{n})=(\text{m}-\text{n})\text{d}$
$\text{d}=2\text{p}$
Therefore
$2\text{mp}=2\text{a}+(\text{m}-1)\times2\text{p}$
$2\text{a}=2\text{p}$
$\text{a}=\text{p}$
The sum up p terms will be:
$\text{s}_\text{p}=\frac{\text{p}}{2}(2\text{a}+(\text{p}-1)\text{d})$
$=\frac{\text{p}}{2}(2\text{p}+(\text{p}-1).2\text{p})$
$=\frac{\text{p}}{2}(2\text{p}+2\text{p}^2-2\text{p})$
$=\text{p}^3$
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