Sequence and SeriesMCQMTP March 21Question 1794 of 212
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If the sum of n\displaystyle n terms of an A.P. is 3n2n\displaystyle 3n^2-n and its common difference is 6\displaystyle 6, then its third term is:

Options

A10
B12
C14
D16
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Correct Answer

Option c14

All Options:

  • A10
  • B12
  • C14
  • D16

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Detailed Solution & Explanation

Given the sum of n\displaystyle n terms Sn=3n2n\displaystyle S_n = 3n^2 - n.
The nth\displaystyle n^{\text{th}} term tn\displaystyle t_n is:
tn=SnSn1t_n = S_n - S_{n-1}
tn=3n2n[3(n1)2(n1)]t_n = 3n^2 - n - [3(n-1)^2 - (n-1)]
tn=3n2n[3n26n+3n+1]t_n = 3n^2 - n - [3n^2 - 6n + 3 - n + 1]
tn=6n4t_n = 6n - 4

To find the third term (t3\displaystyle t_3):
t3=6(3)4=184=14t_3 = 6(3) - 4 = 18 - 4 = 14
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.

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