Sequence and SeriesMCQMTP June 22Question 1802 of 212
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If the common difference of an AP equals to the first term, then the ratio of its mth\displaystyle m^{th} term and nth\displaystyle n^{th} term is:

Options

An:m\displaystyle n:m
Bm:n\displaystyle m:n
Cm2:n2\displaystyle m^2:n^2
Dnone of these
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Correct Answer

Option bm:n\displaystyle m:n

All Options:

  • An:m\displaystyle n:m
  • Bm:n\displaystyle m:n
  • Cm2:n2\displaystyle m^2:n^2
  • Dnone of these

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Detailed Solution & Explanation

We are given that the common difference d\displaystyle d is equal to the first term a\displaystyle a (d=a\displaystyle d = a).
The mth\displaystyle m^{\text{th}} term tm\displaystyle t_m is:
tm=a+(m1)d=a+(m1)a=mat_m = a + (m-1)d = a + (m-1)a = ma

The nth\displaystyle n^{\text{th}} term tn\displaystyle t_n is:
tn=a+(n1)d=a+(n1)a=nat_n = a + (n-1)d = a + (n-1)a = na

The ratio of the mth\displaystyle m^{\text{th}} term and the nth\displaystyle n^{\text{th}} term is:
Ratio=tmtn=mana=mn\text{Ratio} = \frac{t_m}{t_n} = \frac{ma}{na} = \frac{m}{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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