Answer
384.6k+ views
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.
\[ \Rightarrow 512\]
This is our final answer.
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.
\[ \Rightarrow 512\]
This is our final answer.
If \[{5^{th}}\] term of a G.P. is 2, then the product of its 9 terms is 512.
Note:
Here note that the fifth term is having fourth power of 2 and not fifth power. We need not to find all nine terms separately; only finding the product is enough because that product will then be written in the form of the term that is known. Terms in a G.P. are having a common ratio in between. That’s why the power of r is increasing as the terms are increasing.
Note:
Here note that the fifth term is having fourth power of 2 and not fifth power. We need not to find all nine terms separately; only finding the product is enough because that product will then be written in the form of the term that is known. Terms in a G.P. are having a common ratio in between. That’s why the power of r is increasing as the terms are increasing.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)