Sequences And Series Ques 83

The sum of first 20 terms of the sequence $0.7,0.77,0.777, \ldots$, is

(a) $\frac{7}{81}\left(179-10^{-20}\right)$

(b) $\frac{7}{9}\left(99-10^{-20}\right)$

(c) $\frac{7}{81}\left(179+10^{-20}\right)$

(d) $\frac{7}{9}\left(99+10^{-20}\right)$

(2013 Main)

Show Answer

Answer:

Correct Answer: 83.(c)

Solution:

Formula:

Geometric Progression:

  1. Let $S=0.7+0.77+0.777+\ldots$

$=\frac{7}{10}+\frac{77}{10^{2}}+\frac{777}{10^{3}}+\ldots$ upto 20 terms

$=7 [\frac{1}{10}+\frac{11}{10^{2}}+\frac{111}{10^{3}}+\ldots$ upto 20 terms]

$=\frac{7}{9} [\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots$ upto 20 terms]

$=\frac{7}{9} [(1-\frac{1}{10})+(1-\frac{1}{10^{2}})+(1-\frac{1}{10^{3}})$ +… +upto 20 terms]

$ =\frac{7}{9}[(1+1+\ldots+\text { upto } 20 \text { terms }) $

$ -(\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+\ldots+\text { upto } 20 \text { terms } )]$

$ =\frac{7}{9}[ 20-\frac{\frac{1}{10} \{1-(\frac{1}{10})^{20}\}}{1-\frac{1}{10}}] $

$ [ \sum _{i=1}^{20}=20 \text { and sum of } n \text { terms of GP ,} S_n=\frac{a\left(1-r^{n}\right)}{1-r} \text { when }(r<1) ] $

$ =\frac{7}{9} [\frac{179}{9}+\frac{1}{9} (\frac{1}{10})^{20} ]$

$=\frac{7}{81}[179+(10)^{-20}]$



Table of Contents

sathee Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on

Please select your preferred language
कृपया अपनी पसंदीदा भाषा चुनें