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Quantitative Aptitude Question on Sequence and Series

The sum of the infinite series 15(15−17)+(15)2(15−17)2−(17)2+(15)3(15−17)3+…\frac{1}{5} \left( \frac{1}{5} - \frac{1}{7} \right) + \left( \frac{1}{5} \right)^2 \left( \frac{1}{5} - \frac{1}{7} \right)^2 - \left( \frac{1}{7} \right)^2 + \left( \frac{1}{5} \right)^3 \left( \frac{1}{5} - \frac{1}{7} \right)^3 + \dots is equal to

A

7408\frac{7}{408}

B

5408\frac{5}{408}

C

7816\frac{7}{816}

D

5816\frac{5}{816}

Answer

5408\frac{5}{408}

Explanation

Solution

Let's denote the given series as S:

S=15(15−17)+(15)2[(15)2−(17)2]+(15)3[(15)3−(17)3]+…S = \frac{1}{5}\left(\frac{1}{5} - \frac{1}{7}\right) + \left(\frac{1}{5}\right)^2\left[\left(\frac{1}{5}\right)^2 - \left(\frac{1}{7}\right)^2\right] + \left(\frac{1}{5}\right)^3\left[\left(\frac{1}{5}\right)^3 - \left(\frac{1}{7}\right)^3\right] + \dots

We can rewrite this as:

S=(15)2−(15)(17)+(15)4−(15)2(17)2+(15)6−(15)3(17)3+…S = \left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)\left(\frac{1}{7}\right) + \left(\frac{1}{5}\right)^4 - \left(\frac{1}{5}\right)^2\left(\frac{1}{7}\right)^2 + \left(\frac{1}{5}\right)^6 - \left(\frac{1}{5}\right)^3\left(\frac{1}{7}\right)^3 + \dots

Now, let's group the terms:

S=[(15)2+(15)4+(15)6+… ]−[(15)(17)+(15)2(17)2+(15)3(17)3+… ]S = \left[\left(\frac{1}{5}\right)^2 + \left(\frac{1}{5}\right)^4 + \left(\frac{1}{5}\right)^6 + \dots\right] - \left[\left(\frac{1}{5}\right)\left(\frac{1}{7}\right) + \left(\frac{1}{5}\right)^2\left(\frac{1}{7}\right)^2 + \left(\frac{1}{5}\right)^3\left(\frac{1}{7}\right)^3 + \dots\right]

The first series is a geometric series with first term a=(15)2a = \left(\frac{1}{5}\right)^2 and common ratio r=(15)2r = \left(\frac{1}{5}\right)^2.

The second series is also a geometric series with first term a=(15)(17)a = \left(\frac{1}{5}\right)\left(\frac{1}{7}\right) and common ratio r=(15)(17)r = \left(\frac{1}{5}\right)\left(\frac{1}{7}\right).

Using the formula for the sum of an infinite geometric series (S=a1−rS = \frac{a}{1-r}), we can find the sum of both series and subtract them to get the value of S.

After calculations, we get: S=5408S = \frac{5}{408}

Therefore, the sum of the infinite series is 5/408.