If S = (7)/(5) + 9 5 ^ 2 + 13 5 ^ 3 + 19 5 ^ 4 + , then 160S is equal to .

Correct answer: 0 Given, if S = (7)/(5) + 9 5 ^ 2 + 13 5 ^ 3 + 19 5 ^ 4 + ∞ . . . (i) Dividing by 5 on both sides of series (S)/(5) = 7 5 ^ 2 + 9 5 ^ 3 + 13 5 ^ 4 + 19 5 ^ 5 + ∞ . . . (ii) Subtracting equations (i) and (ii) S - (S)/(5) = [(7)/(5) + 9 5 ^ 2 + 13 5 ^ 3 + 19 5 ^ 4 + ∞] - [ 7 5 ^ 2 + 9 5 ^ 3 + 13 5 ^ 4 + 19 5 ^ 5 + ∞] (4S)/(5) = [(7)/(5) + ( 9 5 ^ 2 - 7 5 ^ 2 ) + ( 13 5 ^ 3 - 9 5 ^ 3 ) + ( 19 5 ^ 4 - 13 5 ^ 4 ) + ∞] (4S)/(5) = (7)/(5) + 2 5 ^ 2 + 4 5 ^ 3 + 6 5 ^ 4 + 8 5 ^ 5 + ∞ ((4S)/(5) - (7)/(5)) = k = 2 5 ^ 2 + 4 5 ^ 3 + 6 5 ^ 4 + 8 5 ^ 5 + . . ∞ . . . (iii) Let ((4S)/(5) - (7)/(5)) = k (k)/(5) = 2 5 ^ 3 + 4 5 ^ 4 + 6 5 ^ 5 + 8 5 ^ 6 + . . ∞ . . . (iv) Subtracting equation (iii) and (iv) (4k)/(5) = 2 5 ^ 2 + 2 5 ^ 3 + 2 5 ^ 4 + 2 5 ^ 5 ∞ Common ratio r = 2 5 ^ 3 2 5 ^ 2 = (1)/(5), r = 2 5 ^ 4 2 5 ^ 3 = (1)/(5) (4k)/(5) = (2)/(25)\ (1)/(1 - (1)/(5))\ = (1)/(10) k = (1)/(8) Now (4S)/(5) - (7)/(5) = (1)/(8) (4S)/(5) = (7)/(5) + (1)/(8) S = (61)/(32) 160S = 160 × (61)/(32) = 305

JEE Main 202131 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 31 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2021 · 31 August · Evening Shift · Mathematics · Sequences and Series

If S = (7)/(5) + 9 5 ^ 2 + 13 5 ^ 3 + 19 5 ^ 4 + , then 160S is...

S = \dfrac{7}{5} + \dfrac{9}{{5}^{2}} + \dfrac{13}{{5}^{3}} + \dfrac{19}{{5}^{4}} + \ldots ,

then

160S

is equal to .

Correct answer

Step-by-step solution

Given, if $S = \dfrac{7}{5} + \dfrac{9}{{5}^{2}} + \dfrac{13}{{5}^{3}} + \dfrac{19}{{5}^{4}} + \ldots \ldots \infty . . . \left(i\right)$

Dividing by $5$ on both sides of series

$\dfrac{S}{5} = \dfrac{7}{{5}^{2}} + \dfrac{9}{{5}^{3}} + \dfrac{13}{{5}^{4}} + \dfrac{19}{{5}^{5}} + \ldots \ldots \infty . . . \left(ii\right)$

Subtracting equations \left(i\right) \text{ and } \left(ii\right) $$ S - \dfrac{S}{5} = \left[\dfrac{7}{5} + \dfrac{9}{{5}^{2}} + \dfrac{13}{{5}^{3}} + \dfrac{19}{{5}^{4}} + \ldots \ldots \infty\right] - \left[\dfrac{7}{{5}^{2}} + \dfrac{9}{{5}^{3}} + \dfrac{13}{{5}^{4}} + \dfrac{19}{{5}^{5}} + \ldots \ldots \infty\right] $$ \dfrac{4S}{5} = \left[\dfrac{7}{5} + \left(\dfrac{9}{{5}^{2}} - \dfrac{7}{{5}^{2}}\right) + \left(\dfrac{13}{{5}^{3}} - \dfrac{9}{{5}^{3}}\right) + \left(\dfrac{19}{{5}^{4}} - \dfrac{13}{{5}^{4}}\right) + \ldots \ldots \infty\right] $$ \dfrac{4S}{5} = \dfrac{7}{5} + \dfrac{2}{{5}^{2}} + \dfrac{4}{{5}^{3}} + \dfrac{6}{{5}^{4}} + \dfrac{8}{{5}^{5}} + \ldots \ldots \infty $\left(\dfrac{4S}{5} - \dfrac{7}{5}\right) = k = \dfrac{2}{{5}^{2}} + \dfrac{4}{{5}^{3}} + \dfrac{6}{{5}^{4}} + \dfrac{8}{{5}^{5}} + \ldots . . \infty . . . \left(iii\right)$

Let \left(\dfrac{4S}{5} - \dfrac{7}{5}\right) = k $$ \dfrac{k}{5} = \dfrac{2}{{5}^{3}} + \dfrac{4}{{5}^{4}} + \dfrac{6}{{5}^{5}} + \dfrac{8}{{5}^{6}} + \ldots . . \infty . . . \left(iv\right) $Subtracting equation \left(iii\right) \text{ and } \left(iv\right)$

$\dfrac{4k}{5} = \dfrac{2}{{5}^{2}} + \dfrac{2}{{5}^{3}} + \dfrac{2}{{5}^{4}} + \dfrac{2}{{5}^{5}} \ldots \ldots \infty$

Common ratio

$r = \dfrac{\dfrac{2}{{5}^{3}}}{\dfrac{2}{{5}^{2}}} = \dfrac{1}{5}, r = \dfrac{\dfrac{2}{{5}^{4}}}{\dfrac{2}{{5}^{3}}} = \dfrac{1}{5}$

$\dfrac{4k}{5} = \dfrac{2}{25}\left\{\dfrac{1}{1 - \dfrac{1}{5}}\right\} = \dfrac{1}{10}$

$k = \dfrac{1}{8}$

Now $\dfrac{4S}{5} - \dfrac{7}{5} = \dfrac{1}{8}$

$\dfrac{4S}{5} = \dfrac{7}{5} + \dfrac{1}{8}$

$S = \dfrac{61}{32}$

$160S = 160 \times \dfrac{61}{32}$

$= 305$

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