If 35 - 515-745-9135 - - - ab- then where a and b are coprime

Let S= nbsp;35 +515+745+9135+ … ⇒ S=15[31 +53+79+927+ …∞] Since, 31 +53+79+927+ …∞ is AGP with a=3, d=2, r=13 ∴ S=15[31 13 +2×13(1 13)^2] =15[92 +32] ⇒ S= 65 ...

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Byju's Answer

Standard XII

Mathematics

A.G.P

If 35 + 515+7...

Question

If $\frac{3}{5} + \frac{5}{15} + \frac{7}{45} + \frac{9}{135} + \dots = \frac{a}{b},$ then
(where $a$ and $b$ are coprime)

a = 6

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b = 5

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a = 5

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b = 6

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Solution

The correct option is B $b = 5$
Let $S = \frac{3}{5} + \frac{5}{15} + \frac{7}{45} + \frac{9}{135} + \dots$
$\Rightarrow S = \frac{1}{5} [\frac{3}{1} + \frac{5}{3} + \frac{7}{9} + \frac{9}{27} + \dots \infty]$
Since, $\frac{3}{1} + \frac{5}{3} + \frac{7}{9} + \frac{9}{27} + \dots \infty$ is AGP with $a = 3, d = 2, r = \frac{1}{3}$
$∴ S = \frac{1}{5} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{3}{1 - \frac{1}{3}} + \frac{2 \times \frac{1}{3}}{{(1 - \frac{1}{3})}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$= \frac{1}{5} [\frac{9}{2} + \frac{3}{2}]$
$\Rightarrow S = \frac{6}{5}$

Alternate Solution:
Let $S = \frac{3}{5} + \frac{5}{15} + \frac{7}{45} + \frac{9}{135} + \dots$
$\frac{1}{3} S = \frac{3}{5 \cdot 3} + \frac{5}{5 \cdot 3^{2}} + \frac{7}{5 \cdot 3^{3}} + \dots$
Now,
$S - \frac{1}{3} S = \frac{3}{5} + \frac{2}{5 \cdot 3} + \frac{2}{5 \cdot 3^{2}} + \frac{2}{5 \cdot 3^{3}} + \dots$
$\Rightarrow \frac{2}{3} S = \frac{3}{5} + \frac{2}{5 \cdot 3} (1 + \frac{1}{3} + \frac{1}{3^{2}} + \dots)$
$\Rightarrow \frac{2}{3} S = \frac{3}{5} + \frac{2}{5 \cdot 3} \times \frac{1}{1 - \frac{1}{3}}$
$\Rightarrow \frac{2}{3} S = \frac{3}{5} + \frac{2}{5 \cdot 3} \times \frac{3}{2}$
$\Rightarrow \frac{2}{3} S = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$
$\Rightarrow S = \frac{6}{5}$

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If 35+515+745+9135+…=ab, then (where a and b are coprime)

If $\frac{3}{5} + \frac{5}{15} + \frac{7}{45} + \frac{9}{135} + \dots = \frac{a}{b},$ then
(where $a$ and $b$ are coprime)