QOTD-Quantitative Aptitude-Sum of the Series

2019-06-25 | Team PendulumEdu

What will be the sum of the series $3+\frac{5}{8}+\frac{7}{8^{2}}+\frac{9}{8^{3}}+......\infty =?$

   a. $\frac{13}{4}$

   b. $\frac{21}{8}$

   c. $\frac{75}{23}$

   d. $\frac{80}{21}$

Correct answer (d)

Solution:

Let us assume

$X=3+\frac{5}{8}+\frac{7}{8^{2}}+\frac{9}{8^{3}}+......\infty$ ....................(I)

Multiply both the sides of equation (I) by 1/8

$\frac{X}{8}=\frac{3}{8}+\frac{5}{8^{2}}+\frac{7}{8^{3}}+\frac{9}{8^{4}}+......\infty$………….. (II)

Subtract equation (I) from (II)

$X-\frac{X}{8}=3+\frac{5}{8}-\frac{3}{8}+\frac{7}{8^{2}}-\frac{5}{8^{2}}+\frac{9}{8^{3}}-\frac{7}{8^{3}}+......\infty$

$\frac{7X}{8}=3+\frac{2}{8}+\frac{2}{8^{2}}+\frac{2}{8^{3}}+......\infty$

Sum of any infinite geometric progression for -1 < r < 1 is given as $=\frac{a}{(1-r)}$

Where, a is the first term and r is common ratio

Here, the terms $(\frac{2}{8}+\frac{2}{8^{2}}+\frac{2}{8^{3}}+......\infty )$ constitute a geometric progression having first term (a) as $\frac{2}{8}$and common ratio (r) as $\frac{2}{8}$

Thus,

$\frac{7X}{8}=3+\frac{\frac{2}{8}}{1-\frac{2}{8}}$

$X=\frac{80}{21}$

Hence, (d) is the correct answer.

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