Sequence and SeriesMCQMTP Nov 18Question 1787 of 212
All Questions

If 6th\displaystyle 6^{th} and 13th\displaystyle 13^{th} term of an A.P are 15\displaystyle 15 and 36\displaystyle 36 respectively the A.P is

Options

A2,5,8,11\displaystyle 2,5,8,11
B1,4,6,8\displaystyle 1,4,6,8
C4,1,2,5\displaystyle -4,-1,2,5
D0,3,6,9\displaystyle 0,3,6,9
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Correct Answer

Option d0,3,6,9\displaystyle 0,3,6,9

All Options:

  • A2,5,8,11\displaystyle 2,5,8,11
  • B1,4,6,8\displaystyle 1,4,6,8
  • C4,1,2,5\displaystyle -4,-1,2,5
  • D0,3,6,9\displaystyle 0,3,6,9

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Detailed Solution & Explanation

Let the first term be a\displaystyle a and the common difference be d\displaystyle d.
Given:
t6=a+5d=15t_6 = a + 5d = 15
t13=a+12d=36t_{13} = a + 12d = 36
Subtracting the first equation from the second:
7d=21    d=37d = 21 \implies d = 3
Substitute d=3\displaystyle d = 3 back into the first equation:
a+5(3)=15    a+15=15    a=0a + 5(3) = 15 \implies a + 15 = 15 \implies a = 0

So the A.P. is:
0,3,6,9,12,15,0, 3, 6, 9, 12, 15, \dots
Hence, **Option D** 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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