Sequence and SeriesMCQMTP Apr 21Question 1847 of 212
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The sum of the series 0.5+0.55+0.555+\displaystyle 0.5+0.55+0.555+\dots to n\displaystyle n terms is:

Options

A5n9+581(1(0.1)n)\displaystyle \frac{5n}{9} + \frac{5}{81}(1 - (0.1)^n)
B5n9581(1(0.1)n)\displaystyle \frac{5n}{9} - \frac{5}{81}(1 - (0.1)^n)
C5n9+581[1(0.1)n]\displaystyle \frac{5n}{9} + \frac{5}{81}[1 - (0.1)^n]
DNone
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Correct Answer

Option b5n9581(1(0.1)n)\displaystyle \frac{5n}{9} - \frac{5}{81}(1 - (0.1)^n)

All Options:

  • A5n9+581(1(0.1)n)\displaystyle \frac{5n}{9} + \frac{5}{81}(1 - (0.1)^n)
  • B5n9581(1(0.1)n)\displaystyle \frac{5n}{9} - \frac{5}{81}(1 - (0.1)^n)
  • C5n9+581[1(0.1)n]\displaystyle \frac{5n}{9} + \frac{5}{81}[1 - (0.1)^n]
  • DNone

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Detailed Solution & Explanation

Let the sum be Sn=0.5+0.55+0.555+\displaystyle S_n = 0.5 + 0.55 + 0.555 + \dots
Sn=5[0.1+0.11+0.111+]S_n = 5 [0.1 + 0.11 + 0.111 + \dots]
Sn=59[0.9+0.99+0.999+]S_n = \frac{5}{9} [0.9 + 0.99 + 0.999 + \dots]
Sn=59[(10.1)+(10.01)+(10.001)+]S_n = \frac{5}{9} [(1-0.1) + (1-0.01) + (1-0.001) + \dots]
Sn=59[n(0.1+0.01++(0.1)n)]S_n = \frac{5}{9} \left[n - \left(0.1 + 0.01 + \dots + (0.1)^n\right)\right]
Sn=59[n0.1(1(0.1)n)10.1]S_n = \frac{5}{9} \left[n - \frac{0.1(1-(0.1)^n)}{1-0.1}\right]
Sn=59[n19(1(0.1)n)]=5n9581(1(0.1)n)S_n = \frac{5}{9} \left[n - \frac{1}{9}(1-(0.1)^n)\right] = \frac{5n}{9} - \frac{5}{81}(1-(0.1)^n)
Hence, **Option B** is the correct answer.

About This Chapter: Sequence and Series

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Arithmetic & Geometric Progressions

This chapter covers Arithmetic Progressions (AP) and Geometric Progressions (GP). Students learn how to find the nth term, sum of n terms, arithmetic/geometric means, and sum to infinity of a GP.

View Official ICAI Syllabus

Exam Strategy Tip

For complex 'sum of series' questions, a great hack is to substitute n = 1 and n = 2 into the question and the options to see which one matches, completely bypassing the formula.

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