Sequence and SeriesMCQPYQ May 18Question 1867 of 212
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The sum of m\displaystyle m terms of the series 1+11+111+\displaystyle 1+11+111+\dots up to m\displaystyle m terms, is equal to:

Options

A19(10m+19m10)\displaystyle \frac{1}{9}(10^{m+1} - 9m - 10)
B19(10m+19m10)\displaystyle \frac{1}{9}(10^{m+1} - 9m - 10)
C10m+19m10\displaystyle 10^{m+1} - 9m - 10
DNone of these
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Correct Answer

Option a19(10m+19m10)\displaystyle \frac{1}{9}(10^{m+1} - 9m - 10)

All Options:

  • A19(10m+19m10)\displaystyle \frac{1}{9}(10^{m+1} - 9m - 10)
  • B19(10m+19m10)\displaystyle \frac{1}{9}(10^{m+1} - 9m - 10)
  • C10m+19m10\displaystyle 10^{m+1} - 9m - 10
  • DNone of these

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Detailed Solution & Explanation

Let the sum be Sm=1+11+111+\displaystyle S_m = 1 + 11 + 111 + \dots
Sm=19[9+99+999+]S_m = \frac{1}{9} [9 + 99 + 999 + \dots]
Sm=19[(101)+(1021)+(1031)+]S_m = \frac{1}{9} [(10-1) + (10^2-1) + (10^3-1) + \dots]
Sm=19[(10+102++10m)m]S_m = \frac{1}{9} [(10 + 10^2 + \dots + 10^m) - m]
Sm=19[10(10m1)9m]S_m = \frac{1}{9} \left[\frac{10(10^m-1)}{9} - m\right]
Sm=181(10m+19m10)S_m = \frac{1}{81} (10^{m+1} - 9m - 10)
Hence, **Option A** 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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