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The product of first nine terms of a GP is, in general, equal to which one of the following?
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Verified Answer
The correct answer is:
The 9 th power of the 5 th term
Let a be the first term and $\mathrm{r}$, the common ratio First nine terms of a GP are a, ar, ar $^{2}, \ldots .$ ar $^{8}$.
$$
\begin{aligned}
\therefore \quad \mathrm{P} &=\mathrm{a} . \mathrm{ar} \cdot \mathrm{ar}^{2} \ldots . \mathrm{ar}^{8}=\mathrm{a}^{9} \cdot \mathrm{r}^{1+2+\ldots+8} \\
&=\mathrm{a}^{9} \cdot \mathrm{r}^{\frac{8.9}{2}}=\mathrm{a}^{9} \mathrm{r}^{36}=\left(\mathrm{ar}^{4}\right)^{9}=\left(\mathrm{T}_{5}\right)^{9}
\end{aligned}
$$
$=9$ th power of the 5 th term
$$
\begin{aligned}
\therefore \quad \mathrm{P} &=\mathrm{a} . \mathrm{ar} \cdot \mathrm{ar}^{2} \ldots . \mathrm{ar}^{8}=\mathrm{a}^{9} \cdot \mathrm{r}^{1+2+\ldots+8} \\
&=\mathrm{a}^{9} \cdot \mathrm{r}^{\frac{8.9}{2}}=\mathrm{a}^{9} \mathrm{r}^{36}=\left(\mathrm{ar}^{4}\right)^{9}=\left(\mathrm{T}_{5}\right)^{9}
\end{aligned}
$$
$=9$ th power of the 5 th term
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