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.

0 LIKES

0

COMMENTS

Comments

COMMENT

Share Blog