Sequence and SeriesMCQMTP Sep 24 Series IIQuestion 1822 of 212
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The nth\displaystyle n^{th} term of the series whose sum to n\displaystyle n terms is 3n2+2n\displaystyle 3n^2+2n is

Options

A3n1\displaystyle 3n-1
B8n2\displaystyle 8n-2
C6n1\displaystyle 6n-1
D6n1\displaystyle 6n-1
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Correct Answer

Option d6n1\displaystyle 6n-1

All Options:

  • A3n1\displaystyle 3n-1
  • B8n2\displaystyle 8n-2
  • C6n1\displaystyle 6n-1
  • D6n1\displaystyle 6n-1

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Detailed Solution & Explanation

Given the sum of n\displaystyle n terms Sn=3n2+2n\displaystyle S_n = 3n^2 + 2n.
The nth\displaystyle n^{\text{th}} term tn\displaystyle t_n is:
tn=SnSn1t_n = S_n - S_{n-1}
tn=3n2+2n[3(n1)2+2(n1)]t_n = 3n^2 + 2n - [3(n-1)^2 + 2(n-1)]
tn=3n2+2n[3(n22n+1)+2n2]t_n = 3n^2 + 2n - [3(n^2 - 2n + 1) + 2n - 2]
tn=3n2+2n[3n26n+3+2n2]t_n = 3n^2 + 2n - [3n^2 - 6n + 3 + 2n - 2]
tn=3n2+2n3n2+4n1=6n1t_n = 3n^2 + 2n - 3n^2 + 4n - 1 = 6n - 1
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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