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# If ${5^{th}}$ term of a G.P. is 2, then the product of its 9 terms is,A. 256B. 512C. 1024D. None of these

Last updated date: 13th Jun 2024
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Hint: Given is only the fifth term of the G.P. and asked to find the product of 9 terms. We will consider the first term as a and the r be the common ratio of the whole G.P. then we can simply equate the fifth term as $a{r^4} = 2$ . Then remaining 9 terms will be multiplied such that their product will be in the form of $a \times ar \times a{r^2} \times a{r^3} \times a{r^4}.... \times a{r^8}$ and we will try to adjust this in the form of fifth term. So let’s solve it!

Complete Step by Step Solution:
Given that
${5^{th}}$ term of a G.P. is 2
Let first term as a and the r be the common ratio of the whole G.P.
Then fifth term will be $a{r^4} = 2$
But they are asked to find the product of 9 terms of the G.P.
So we can write the product as $a \times ar \times a{r^2} \times a{r^3} \times a{r^4}.... \times a{r^8}$
So ,
$\Rightarrow a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}$
Now we can write the terms with base a separately and those with r separately.
$\Rightarrow aaaaaaaaa \times r{r^2}{r^3}{r^4}{r^5}{r^6}{r^7}{r^8}$
Now adding the powers of the bases separately,
$\Rightarrow {a^{1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1}} \times {r^{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}}$
Now we can write as,
$\Rightarrow {a^9} \times {r^{36}}$
But we have to write this in the form of power 4,
$\Rightarrow {a^9} \times {r^{9 \times 4}}$
Now we will take the common power out,
$\Rightarrow {\left( {a{r^4}} \right)^9}$
Putting the value of fifth term in above bracket,
$\Rightarrow {\left( 2 \right)^9}$
Then the value of the ninth power of 2 is our answer.
$\Rightarrow 521$