What is the sum of the first 10 terms of the | vedclass

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

Question

(B) Let the sum be $S = 0.7 + 0.77 + 0.777 + \dots$ up to $10$ terms.We can write this as $S = 7(0.1 + 0.11 + 0.111 + \dots)$.Multiply and divide by $

Vedclass · Accepted Answer

$\frac{7}{81} \left( 89 + \frac{1}{10^{10}} \right)$

Vedclass · Answer

(B) Let the sum be $S = 0.7 + 0.77 + 0.777 + \dots$ up to $10$ terms.We can write this as $S = 7(0.1 + 0.11 + 0.111 + \dots)$.Multiply and divide by $9$: $S = \frac{7}{9} (0.9 + 0.99 + 0.999 + \dots)$.This can be rewritten as $S = \frac{7}{9} [(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \dots]$.For $10$ terms,$S = \frac{7}{9} [10 - (0.1 + 0.01 + \dots + 0.1^{10})]$.The sum inside the bracket is a geometric progression with $a = 0.1$,$r = 0.1$,and $n = 10$.Sum $= \frac{a(1 - r^n)}{1 - r} = \frac{0.1(1 - 0.1^{10})}{1 - 0.1} = \frac{0.1(1 - 10^{-10})}{0.9} = \frac{1}{9} (1 - 10^{-10})$.Substituting this back: $S = \frac{7}{9} [10 - \frac{1}{9} (1 - 10^{-10})] = \frac{7}{9} [\frac{90 - 1 + 10^{-10}}{9}] = \frac{7}{81} (89 + 10^{-10})$.Thus,the sum is $\frac{7}{81} (89 + \frac{1}{10^{10}})$.

What is the sum of the first $10$ terms of the series $0.7 + 0.77 + 0.777 + \dots$?

Similar Questions

Find the $20^{\text{th}}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$ up to $n$ terms.

The value of the sum $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \ldots$ up to $n$ terms is equal to

Let $a, b, c, d$ be real numbers such that $\sum_{k=1}^n (a k^3+b k^2+c k+d)=n^4$,for every natural number $n$. Then,$|a|+|b|+|c|+|d|$ is equal to

If the sum of the first ten terms of the series ${\left( {1\frac{3}{5}} \right)^2} + {\left( {2\frac{2}{5}} \right)^2} + {\left( {3\frac{1}{5}} \right)^2} + {4^2} + \dots$ is $\frac{16}{5}m$,then $m$ is equal to:

The $20^{th}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module