Answer
438k+ views
Hint: First of all, consider the first three terms of the G.P. as variables then use the condition of the terms to be in A.P. after multiplying the extreme and middle terms with the given constants. So, use this concept to reach the solution of the given problem.
Complete step by step solution:
Let the first three terms of the G.P. be \[a,ar,a{r^2}\].
Given that their sum is equal to 70 i.e.,
\[
\Rightarrow a + ar + a{r^2} = 70 \\
\therefore a\left( {1 + r + {r^2}} \right) = 70...............................\left( 1 \right) \\
\]
Also, given that the extreme terms are multiplied by 4 and the middle term is multiplied by 5. Then the terms are given by \[4a,5ar,4a{r^2}\]. And these terms are in A.P.
We know that the condition to be in A.P. for the three terms \[x,y,z\] is given by \[2y = x + z\].
Since, \[4a,5ar,4a{r^2}\] are in A.P. and by the condition to be in A.P. we have
\[
\Rightarrow 2\left( {5ar} \right) = 4a + 4a{r^2} \\
\Rightarrow 10ar = 4a + 4a{r^2} \\
\Rightarrow 2a\left( {5r} \right) = 2a\left( {2 + 2{r^2}} \right) \\
\]
Cancelling the common terms, we have
\[
\Rightarrow 5r = 2 + 2{r^2} \\
\Rightarrow 2{r^2} - 5r + 2 = 0 \\
\Rightarrow 2{r^2} - 4r - r + 2 = 0 \\
\Rightarrow 2r\left( {r - 2} \right) - 1\left( {r - 2} \right) = 0 \\
\Rightarrow \left( {2r - 1} \right)\left( {r - 2} \right) = 0 \\
\therefore r = \dfrac{1}{2},2 \\
\]
We know that for an infinite term of G.P. the common ratio must be less than one i.e., \[r < 1\].
So, \[r = \dfrac{1}{2}\].
Substituting \[r = \dfrac{1}{2}\] in equation \[\left( 1 \right)\], we have
\[
\Rightarrow a\left( {1 + \dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) = 70 \\
\Rightarrow a\left( {1 + \dfrac{1}{2} + \dfrac{1}{4}} \right) = 70 \\
\Rightarrow a\left( {\dfrac{{4 + 2 + 1}}{4}} \right) = 70 \\
\Rightarrow a\left( {\dfrac{7}{4}} \right) = 70 \\
\Rightarrow a = 70 \times \dfrac{4}{7} \\
\therefore a = 40 \\
\]
We know that the sum of infinite terms in G.P. with the first term \[a\] and common ratio \[r\] is given by \[{S_\infty } = \dfrac{a}{{1 - r}}\].
Hence, the sum of the infinite terms of G.P. is \[{S_\infty } = \dfrac{{40}}{{1 - \dfrac{1}{2}}} = \dfrac{{40}}{{\dfrac{1}{2}}} = 40 \times 2 = 80\]
Thus, the correct option is D. 80
Note: For an infinite term of G.P. the common ratio must be less than one i.e., \[r < 1\] and their sum is given by \[{S_\infty } = \dfrac{a}{{1 - r}}\]. The extreme terms in a sequence are the first and last terms of that sequence.
Complete step by step solution:
Let the first three terms of the G.P. be \[a,ar,a{r^2}\].
Given that their sum is equal to 70 i.e.,
\[
\Rightarrow a + ar + a{r^2} = 70 \\
\therefore a\left( {1 + r + {r^2}} \right) = 70...............................\left( 1 \right) \\
\]
Also, given that the extreme terms are multiplied by 4 and the middle term is multiplied by 5. Then the terms are given by \[4a,5ar,4a{r^2}\]. And these terms are in A.P.
We know that the condition to be in A.P. for the three terms \[x,y,z\] is given by \[2y = x + z\].
Since, \[4a,5ar,4a{r^2}\] are in A.P. and by the condition to be in A.P. we have
\[
\Rightarrow 2\left( {5ar} \right) = 4a + 4a{r^2} \\
\Rightarrow 10ar = 4a + 4a{r^2} \\
\Rightarrow 2a\left( {5r} \right) = 2a\left( {2 + 2{r^2}} \right) \\
\]
Cancelling the common terms, we have
\[
\Rightarrow 5r = 2 + 2{r^2} \\
\Rightarrow 2{r^2} - 5r + 2 = 0 \\
\Rightarrow 2{r^2} - 4r - r + 2 = 0 \\
\Rightarrow 2r\left( {r - 2} \right) - 1\left( {r - 2} \right) = 0 \\
\Rightarrow \left( {2r - 1} \right)\left( {r - 2} \right) = 0 \\
\therefore r = \dfrac{1}{2},2 \\
\]
We know that for an infinite term of G.P. the common ratio must be less than one i.e., \[r < 1\].
So, \[r = \dfrac{1}{2}\].
Substituting \[r = \dfrac{1}{2}\] in equation \[\left( 1 \right)\], we have
\[
\Rightarrow a\left( {1 + \dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) = 70 \\
\Rightarrow a\left( {1 + \dfrac{1}{2} + \dfrac{1}{4}} \right) = 70 \\
\Rightarrow a\left( {\dfrac{{4 + 2 + 1}}{4}} \right) = 70 \\
\Rightarrow a\left( {\dfrac{7}{4}} \right) = 70 \\
\Rightarrow a = 70 \times \dfrac{4}{7} \\
\therefore a = 40 \\
\]
We know that the sum of infinite terms in G.P. with the first term \[a\] and common ratio \[r\] is given by \[{S_\infty } = \dfrac{a}{{1 - r}}\].
Hence, the sum of the infinite terms of G.P. is \[{S_\infty } = \dfrac{{40}}{{1 - \dfrac{1}{2}}} = \dfrac{{40}}{{\dfrac{1}{2}}} = 40 \times 2 = 80\]
Thus, the correct option is D. 80
Note: For an infinite term of G.P. the common ratio must be less than one i.e., \[r < 1\] and their sum is given by \[{S_\infty } = \dfrac{a}{{1 - r}}\]. The extreme terms in a sequence are the first and last terms of that sequence.
Recently Updated Pages
What are the figures of speech in the poem Wind class 11 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Two tankers contain 850 litres and 680 litres of petrol class 10 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when eggshell is added to nitric acid class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why was Kamaraj called as Kingmaker class 10 social studies CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What makes elections in India democratic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
What is 1 divided by 0 class 8 maths 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)
Give 10 examples for herbs , shrubs , climbers , creepers
![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)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![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)