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Question Number 47628 by 786786AM last updated on 12/Nov/18

$$\mathrm{If}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}{\mathrm{n}}^{\mathrm{th}}\mathrm{term}\mathrm{of}\mathrm{G}.\mathrm{P}.,\mathrm{are}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{respectively},$$$$\mathrm{p}\mathrm{is}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{n}\mathrm{terms}.\mathrm{Prove}\mathrm{that}{\mathrm{p}}^2={\left(\mathrm{ab}\right)}^{\mathrm{n}}.$$$$$$$$$$

Answered by math1967 last updated on 12/Nov/18

$$P=a.ar.....{ar}^{n-1}[r=commonratio]$$$$P=a^n.r^{1+2+.....+(n-1)}$$$$P=a^n.r^{\frac{(n-1)(n-1+1)}{2}}$$$$P=a^n.r^{\frac{n(n-1)}{2}}$$$$P^2=a^{2n}.r^{n(n-1)}$$$$P^2={(a.{ar}^{n-1})}^n$$$$\therefore P^2={\left(ab\right)}^n[\because b={ar}^{n-1}]$$

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