Question

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

Answer 1

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.