Question

# The third term of G.P. is 4. Find the product of its five terms.

Hint: Equate the third term of G.P. with 4. And convert the product of the first five terms in the form of the third term.

According to the question, the third term of the G.P. is 4. Let $a$ and $r$ be the first term and common ratio of the G.P. Then the G.P. is:
$\Rightarrow a,ar,a{r^2},a{r^3},....$
We know that the general term of G.P. is:
${T_r} = a{r^{n - 1}}$
Third term is given as 4. So we have:
$\Rightarrow {T_3} = 4, \\ \Rightarrow a{r^{3 - 1}} = 4, \\ \Rightarrow a{r^2} = 4 .....(i) \\$
The product of first five terms is:
$\Rightarrow$ Product $= a \times ar \times a{r^2} \times a{r^3} \times a{r^4} = {a^5}{r^{10}} = {\left( {a{r^2}} \right)^5}$
Putting the value of $a{r^2}$ from equation $(i)$, we’ll get:
$\Rightarrow$ Product $= {\left( 4 \right)^5} = 1024$
Thus, the product of first terms of G.P. is 1024.

Note: This can be solved by another method as:
If five numbers are in G.P. then the middle number (i.e. third number) is their geometric mean.
Third term is given as 4. So, 4 is the geometric mean of the first five terms of G.P.
And if the geometric mean of five numbers is 4, then their product is ${4^5}$.