Sequence and SeriesMCQPYQ Jan. 21Question 1829 of 212
All Questions

In a geometric progression, that 3rd\displaystyle 3^{rd} and 6th\displaystyle 6^{th} terms are respectively 1 and −1/8\displaystyle -1/8. The term (a)\displaystyle (a) and common ratio are respectively.

Options

A4 and 12\displaystyle \frac{1}{2}
B4 and −12\displaystyle -\frac{1}{2}
C4 and −12\displaystyle -\frac{1}{2}
D4 and 12\displaystyle \frac{1}{2}
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Correct Answer

✅ Option b — 4 and −12\displaystyle -\frac{1}{2}

All Options:

  • A4 and 12\displaystyle \frac{1}{2}
  • B4 and −12\displaystyle -\frac{1}{2}
  • C4 and −12\displaystyle -\frac{1}{2}
  • D4 and 12\displaystyle \frac{1}{2}

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Detailed Solution & Explanation

Let the first term be a\displaystyle a and the common ratio be r\displaystyle r.
Given:
t3=ar2=1— (1)t_3 = ar^2 = 1 \quad \text{--- (1)}
t6=ar5=−18— (2)t_6 = ar^5 = -\frac{1}{8} \quad \text{--- (2)}

Divide equation (2) by equation (1):
ar5ar2=−1/81  ⟹  r3=−18  ⟹  r=−12\frac{ar^5}{ar^2} = \frac{-1/8}{1} \implies r^3 = -\frac{1}{8} \implies r = -\frac{1}{2}

Substitute r=−12\displaystyle r = -\frac{1}{2} back into equation (1):
a(−12)2=1  ⟹  a(14)=1  ⟹  a=4a\left(-\frac{1}{2}\right)^2 = 1 \implies a\left(\frac{1}{4}\right) = 1 \implies a = 4
So the first term is 4\displaystyle 4 and the common ratio is −12\displaystyle -\frac{1}{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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