Sequence and SeriesMCQMTP Nov 19Question 1791 of 213
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The sum of n\displaystyle n terms of an A.P. is $3n^2+n$; then its $p^{th}$ term is

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option b — $6p-2$

All Options:

  • A$6p+2$
  • B$6p-2$
  • C$6p-1$
  • Dnone of these

Detailed Solution & Explanation

Given the sum of n\displaystyle n terms Sn=3n2+n\displaystyle S_n = 3n^2 + n.
The nth\displaystyle n^{th\displaystyle \text{th}} term tn\displaystyle t_n is:
tn=Sn−Sn−1t_n = S_n - S_{n-1}
tn=(3n2+n)−[3(n−1)2+(n−1)]t_n = (3n^2 + n) - [3(n-1)^2 + (n-1)]
tn=3n2+n−[3(n2−2n+1)+n−1]t_n = 3n^2 + n - [3(n^2 - 2n + 1) + n - 1]
tn=3n2+n−[3n2−6n+3+n−1]t_n = 3n^2 + n - [3n^2 - 6n + 3 + n - 1]
tn=3n2+n−3n2+5n−2=6n−2t_n = 3n^2 + n - 3n^2 + 5n - 2 = 6n - 2

Substituting p\displaystyle p for n\displaystyle n, the pth\displaystyle p^{th\displaystyle \text{th}} term is $6p - 2$.
Hence, **Option B** 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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