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QJEEM21Ma7fbf: JEE Main 2021 Mathematics - Class 11 JEEM21Ma7fbf Solution: The sum of first four terms of a geometric progression (G. P.)…

The sum of first four terms of a geometric progression (G. P.) is 6512{{65} \over {12}} and the sum of their respective reciprocals is 6518{{65} \over {18}} . If the product of first three terms of the G.P. is 1, and the third term is α\alpha , then 2 α\alpha is _________.

Step-by-Step Verified Solution

Let the terms are a,ar,ar2,ar3a,ar,a{r^2},a{r^3}

a+ar+ar2+ar3=6512a + ar + a{r^2} + a{r^3} = {{65} \over {12}} ..........(1)

1a+1ar+1ar2+1ar3=6518{1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}

1a(r3+r2+r+1r3)=6518{1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}} ...............(2)

Doing (1)(2),{{(1)} \over {(2)}},

a2r3=1812=32{a^2}{r^3} = {{18} \over {12}} = {3 \over 2}

Also given, a3r3=1⇒a(32)=1⇒a=23{a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}

49r3=32⇒r3=3323⇒r=32{4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}

α=ar2=23.(32)2=32\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}

2α=32\alpha = 3

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