Sequence and SeriesMCQMTP June 2023 Series IQuestion 1808 of 212
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Sum of n\displaystyle n terms of two AP's are in the ratio (3n+5):(5n+3)\displaystyle (3n+5):(5n+3) then ratio of their 10th\displaystyle 10^{th} term

Options

A31:49\displaystyle 31:49
B30:49\displaystyle 30:49
C28:49\displaystyle 28:49
DNone of these
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Correct Answer

Option a31:49\displaystyle 31:49

All Options:

  • A31:49\displaystyle 31:49
  • B30:49\displaystyle 30:49
  • C28:49\displaystyle 28:49
  • DNone of these

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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)=2a1+(n1)d12a2+(n1)d2=3n+55n+3\frac{S_n^{(1)}}{S_n^{(2)}} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n+5}{5n+3}

We want to find the ratio of their 10th\displaystyle 10^{\text{th}} terms:
t10(1)t10(2)=a1+9d1a2+9d2=2a1+18d12a2+18d2\frac{t_{10}^{(1)}}{t_{10}^{(2)}} = \frac{a_1 + 9d_1}{a_2 + 9d_2} = \frac{2a_1 + 18d_1}{2a_2 + 18d_2}
By comparing with the sum ratio, we substitute n1=18    n=19\displaystyle n-1 = 18 \implies n = 19:
t10(1)t10(2)=3(19)+55(19)+3=57+595+3=6298=3149\frac{t_{10}^{(1)}}{t_{10}^{(2)}} = \frac{3(19) + 5}{5(19) + 3} = \frac{57 + 5}{95 + 3} = \frac{62}{98} = \frac{31}{49}
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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