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The third term of a G.P is 4. The product of the first five terms is
A .\[{4^3}\]
B. \[{4^5}\]
C .\[{4^4}\]
D. None of these

Last updated date: 19th Apr 2024
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Hint- Proceed the solution of this question by considering the general term of GP in our mind such that their multiplication can itself form such a number which are either known or can be found.

Complete step-by-step solution -
Let the common ratio be r and the terms be ${\text{a,ar,a}}{{\text{r}}^2}{\text{,a}}{{\text{r}}^3}{\text{,a}}{{\text{r}}^4}....$and so on in G.P.
Here a is the 1st number, ${\text{ar}}$ be the 2nd number, ${\text{a}}{{\text{r}}^2}$be the third number and so on.
In the question, it is given that the third term of GP is equal to 4.
⇒${\text{a}}{{\text{r}}^2}$=4 ... (1)
Therefore, the product of the first five term is given by,
⇒\[{\text{a}} \times {\text{ar}} \times {\text{a}}{{\text{r}}^2} \times {\text{a}}{{\text{r}}^3} \times {\text{a}}{{\text{r}}^4} = {{\text{a}}^5} \times {{\text{r}}^{10}}\]

On further simplifying
⇒\[{\text{a}} \times {\text{ar}} \times {\text{a}}{{\text{r}}^2} \times {\text{a}}{{\text{r}}^3} \times {\text{a}}{{\text{r}}^4} = {\left( {{\text{a}} \times {{\text{r}}^2}} \right)^5}\]
From equation (1), substitute ${\text{a}}{{\text{r}}^2}$=4; we get
⇒\[{\text{a}} \times {\text{ar}} \times {\text{a}}{{\text{r}}^2} \times {\text{a}}{{\text{r}}^3} \times {\text{a}}{{\text{r}}^4} = {\left( 4 \right)^5}\]
Thus, the product of the first five term is \[{\left( 4 \right)^5}\]
Hence option B is correct.

Note- In a G.P. as we know that, each term is multiplied by the common ratio \[{\text{r}}\]. To get the second term, the first term is multiplied by \[{\text{r}}\]. We get the third term by multiplying the first term by \[{{\text{r}}^2}\]Similarly, we will get the fourth term by multiplying the first term by \[{{\text{r}}^3}\] and so on. Hence 3rd term is the geometric mean of 2nd and 4th term as well as 1st and 5th term of GP. Hence multiplication of the first 5 numbers can be written in exponential form of their geometrical form.