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Sum of Series 4/19 + 44/19^2 + 444/19^3 + ... | JEE

JEE Maths question with a full step-by-step solution.

Question

The sum of the series

\frac{4}{19}+\frac{44}{19^2}+\frac{444}{19^3}+\cdots

to infinity is

\frac{38}{81}

correct

\frac{4}{19}

\frac{36}{371}

\frac{18}{19}

Solution

Step 1: Let

S

be the sum and form

S-\dfrac{S}{19}

S-\frac{S}{19}=\frac{4}{19}+\frac{40}{19^2}+\frac{400}{19^3}+\cdots.

Step 2: The right side is a G.P. with first term

\dfrac{4}{19}

and ratio

\dfrac{10}{19}

\frac{18S}{19}=\frac{4}{19}\cdot\frac{1}{1-\frac{10}{19}}=\frac{4}{19}\cdot\frac{19}{9}=\frac49.

Step 3: So

S=\dfrac49\cdot\dfrac{19}{18}=\dfrac{38}{81}.

Correct answer: (1)

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