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

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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.

Complete step-by-step answer:
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 ${4^5}$.
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