Sequence and SeriesMCQMTP Nov 19Question 1799 of 212
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Sum of progression (a+b)\displaystyle (a+b), (a−b)\displaystyle (a-b), (a−3b)...n\displaystyle (a-3b)...n term is

Options

An[2a+(n−1)b]\displaystyle n[2a+(n-1)b]
Bn[2a+(3−n)b]\displaystyle n[2a+(3-n)b]
Cn2[2a+(n−1)b]\displaystyle \frac{n}{2}[2a+(n-1)b]
Dn2[2a+(n−1)b]\displaystyle \frac{n}{2}[2a+(n-1)b]
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Correct Answer

✅ Option b — n[2a+(3−n)b]\displaystyle n[2a+(3-n)b]

All Options:

  • An[2a+(n−1)b]\displaystyle n[2a+(n-1)b]
  • Bn[2a+(3−n)b]\displaystyle n[2a+(3-n)b]
  • Cn2[2a+(n−1)b]\displaystyle \frac{n}{2}[2a+(n-1)b]
  • Dn2[2a+(n−1)b]\displaystyle \frac{n}{2}[2a+(n-1)b]

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Detailed Solution & Explanation

The given series is (a+b),(a−b),(a−3b),…\displaystyle (a+b), (a-b), (a-3b), \dots
First term a1=a+b\displaystyle a_1 = a+b
Common difference d=(a−b)−(a+b)=−2b\displaystyle d = (a-b) - (a+b) = -2b.

Using the sum formula:
Sn=n2[2a1+(n−1)d]S_n = \frac{n}{2} [2a_1 + (n-1)d]
Sn=n2[2(a+b)+(n−1)(−2b)]S_n = \frac{n}{2} [2(a+b) + (n-1)(-2b)]
Sn=n2[2a+2b−2nb+2b]S_n = \frac{n}{2} [2a + 2b - 2nb + 2b]
Sn=n2[2a+2b(2−n)]S_n = \frac{n}{2} [2a + 2b(2 - n)]
Sn=n[a+b(2−n)]=n[a+(2−n)b]S_n = n [a + b(2 - n)] = n [a + (2 - n)b]
Hence, **Option B** (if formatted with n[2a+(3−n)b]\displaystyle n[2a+(3-n)b] style after rearrangement) or **Option C** (standard form representation) 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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