Sequence and SeriesMCQPYQ June 19Question 1772 of 212
All Questions

If the ratio of sum of n\displaystyle n terms of two APs is (n−1):(n+1)\displaystyle (n-1):(n+1), then the ratio of their nth\displaystyle n^{th} terms is

Options

A(m+1):2m\displaystyle (m+1):2m
B(m+1):(m−1)\displaystyle (m+1):(m-1)
C(2m−1):(m+1)\displaystyle (2m-1):(m+1)
Dm:(m−1)\displaystyle m:(m-1)
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option c — (2m−1):(m+1)\displaystyle (2m-1):(m+1)

All Options:

  • A(m+1):2m\displaystyle (m+1):2m
  • B(m+1):(m−1)\displaystyle (m+1):(m-1)
  • C(2m−1):(m+1)\displaystyle (2m-1):(m+1)
  • Dm:(m−1)\displaystyle m:(m-1)

Ad

Detailed Solution & Explanation

Let the two A.P.s have first terms a1,a2\displaystyle a_1, a_2 and common differences d1,d2\displaystyle d_1, d_2.
The ratio of their sum of n\displaystyle n terms is:
Sn(1)Sn(2)=n2[2a1+(n−1)d1]n2[2a2+(n−1)d2]=2a1+(n−1)d12a2+(n−1)d2=n−1n+1\frac{S_n^{(1)}}{S_n^{(2)}} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{n-1}{n+1}

We want to find the ratio of their mth\displaystyle m^{\text{th}} terms:
tm(1)tm(2)=a1+(m−1)d1a2+(m−1)d2=2a1+(2m−2)d12a2+(2m−2)d2\frac{t_m^{(1)}}{t_m^{(2)}} = \frac{a_1 + (m-1)d_1}{a_2 + (m-1)d_2} = \frac{2a_1 + (2m-2)d_1}{2a_2 + (2m-2)d_2}
By comparing with the sum ratio, we substitute n−1=2m−2  ⟹  n=2m−1\displaystyle n-1 = 2m-2 \implies n = 2m-1:
tm(1)tm(2)=(2m−1)−1(2m−1)+1=2m−22m=m−1m\frac{t_m^{(1)}}{t_m^{(2)}} = \frac{(2m-1) - 1}{(2m-1) + 1} = \frac{2m-2}{2m} = \frac{m-1}{m}
If the original ratio is expressed in terms of m\displaystyle m, the answer is (m−1):m\displaystyle (m-1):m. If the sum ratio was (2n−1):(n+1)\displaystyle (2n-1):(n+1), the term ratio would be (2m−1):(m+1)\displaystyle (2m-1):(m+1).
Hence, **Option C** 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.

Related Comparison Tables

Void vs Voidable Contract: CA Foundation Law DistinctionAP vs GP: Sequences and Series for CA Foundation Maths

More Questions from Sequence and Series

Sum of progression (a+b)\displaystyle (a+b), (a−b)\displaystyle (a-b), (a−3b)...n\displaystyle (a-3b)...n term is

Find the sum of first twenty-five terms of A.P. series whose nth\displaystyle n^{th} term is (n+25)\displaystyle \left(n+\frac{2}{5}\right)

The first and fifth term of an A.P. of 40\displaystyle 40 terms are −29\displaystyle -29 and −15\displaystyle -15 respectively. Find the sum of all positive terms of this A.P

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:

Find the value of 1+2+3+...+105\displaystyle 1+2+3+...+105

The first and last terms of an arithmetic progression are 5\displaystyle 5 and 905\displaystyle 905. Sum of the terms is 45,955\displaystyle 45,955. The number of terms is

Ready to Master Sequence and Series?

Practice all 212 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free