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

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(D) The given series is $5.05 + 1.212 + 0.29088 + \dots \infty$.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 $S_{\infty} = \frac{5.05}{1 - 0.24} = \frac{5.05}{0.76} = 6.6447368... \approx 6.64474$.

Class 11 Mathematics Sequences and Series Geometric progression

Easy

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

A
$6.93378$
B
$6.87342$
C
$6.74384$
D
$6.64474$

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Class 11 Questions

Class 11

Mathematics Questions

Chapter

Sequences and Series

Subtopic

Geometric progression

Which term of the sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$ is $\frac{1}{19683}$?

Easy

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If $x > 1, y > 1, z > 1$ are in $G.P.$,then $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$ are in

EasyIIT 1998

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Find the geometric mean of the first $n$ natural numbers.

Medium

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If $a, b$ and $c$ are three distinct numbers in $G.P.$ and $a + b + c = xb$,then $x$ cannot be:

DifficultJEE MAIN 2019

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The first term of a $G.P.$ is $7$,the last term is $448$ and the sum of all terms is $889$. Then the common ratio is:

Easy

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

Similar Questions

Which term of the sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$ is $\frac{1}{19683}$?

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