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Arithmetic Progressions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSArithmetic Progressions3 Marks
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 $\text{s}_\text{p}=\text{p}^3.$
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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: $\text{s}_\text{p}=\frac{\text{p}}{2}(2\text{a}+(\text{p}-1)\text{d})$ $=\frac{\text{p}}{2}(2\text{p}+(\text{p}-1).2\text{p})$ $=\frac{\text{p}}{2}(2\text{p}+2\text{p}^2-2\text{p})$ $=\text{p}^3$

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