The sum of the series 5.05 + 1.212 + 0.29088 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given series is $5.05 + 1.212 + 0.29088 + ... \infty$.This is an infinite geometric progression $(G.P.)$ where the first term $a = 5.05$.The c

Vedclass · Accepted Answer

$6.64474$

Vedclass · Answer

(D) The given series is $5.05 + 1.212 + 0.29088 + ... \infty$.This is an infinite geometric progression $(G.P.)$ where the first term $a = 5.05$.The common ratio $r$ is calculated as $r = \frac{1.212}{5.05} = 0.24$.Since $|r| Substituting the values,we get $S_{\infty} = \frac{5.05}{1 - 0.24} = \frac{5.05}{0.76}$.Calculating the division,$S_{\infty} = 6.6447368... \approx 6.64474$.

The sum of the series $5.05 + 1.212 + 0.29088 + ... \infty$ is

Similar Questions

If $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$ are in $A.P.,$ then :-

The sum of $100$ terms of the series $0.9 + 0.09 + 0.009 + \dots$ will be

Write the seventh term of the following $GP$: $39366, 13122, 4374, 1458, \ldots$

If $x = \frac{4}{3} - \frac{4x}{9} + \frac{4x^2}{27} - \dots \infty$,then $x$ is equal to

$0.5737373...... = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module