JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 19)
The sum of first four terms of a geometric progression (G. P.) is $${{65} \over {12}}$$ and the sum of their respective reciprocals is $${{65} \over {18}}$$. If the product of first three terms of the G.P. is 1, and the third term is $$\alpha$$, then 2$$\alpha$$ is _________.
Answer
3
Explanation
Let the terms are $$a,ar,a{r^2},a{r^3}$$
$$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$$ ..........(1)
$${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$$
$${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$$ ...............(2)
Doing $${{(1)} \over {(2)}},$$
$${a^2}{r^3} = {{18} \over {12}} = {3 \over 2}$$
Also given, $${a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}$$
$${4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}$$
$$\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}$$
$$2\alpha = 3$$
$$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$$ ..........(1)
$${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$$
$${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$$ ...............(2)
Doing $${{(1)} \over {(2)}},$$
$${a^2}{r^3} = {{18} \over {12}} = {3 \over 2}$$
Also given, $${a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}$$
$${4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}$$
$$\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}$$
$$2\alpha = 3$$
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