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The third term of a GP is 3 . What is the product of the first five terms?
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The correct answer is:
243
Given, $3^{\text {rd }}$ term of G.P. =3 Let ' $a$ ' be the first term and ' $\mathrm{r}$ ' be the common ratio. $\therefore \mathrm{ar}^{2}=3$
We know, $\mathrm{T}_{1}=\mathrm{a}, \mathrm{T}_{2}=\mathrm{ar}, \mathrm{T}_{3}=\mathrm{ar}^{2}, \mathrm{~T}_{4}=\mathrm{ar}^{3}, \mathrm{~T}_{5}=\mathrm{ar}^{4}$
$\mathrm{T}_{1} \cdot \mathrm{T}_{2} \cdot \mathrm{T}_{3} \cdot \mathrm{T}_{4} \cdot \mathrm{T}_{5}=(\mathrm{a})(\operatorname{ar})\left(\mathrm{ar}^{2}\right)\left(\mathrm{ar}^{3}\right)\left(\mathrm{ar}^{4}\right)$
$=\left(\mathrm{ar}^{2}\right)^{5}$
$=3^{5}$
$=243$
We know, $\mathrm{T}_{1}=\mathrm{a}, \mathrm{T}_{2}=\mathrm{ar}, \mathrm{T}_{3}=\mathrm{ar}^{2}, \mathrm{~T}_{4}=\mathrm{ar}^{3}, \mathrm{~T}_{5}=\mathrm{ar}^{4}$
$\mathrm{T}_{1} \cdot \mathrm{T}_{2} \cdot \mathrm{T}_{3} \cdot \mathrm{T}_{4} \cdot \mathrm{T}_{5}=(\mathrm{a})(\operatorname{ar})\left(\mathrm{ar}^{2}\right)\left(\mathrm{ar}^{3}\right)\left(\mathrm{ar}^{4}\right)$
$=\left(\mathrm{ar}^{2}\right)^{5}$
$=3^{5}$
$=243$
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