# 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. $$\frac{80}{21}$$

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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