If 5th term of G.P. is 32 and 8th term of G.P. is 8 then 6th term of G.P. is

Option b: 16. Solution: Let the first term be a and common ratio be r . Given: t5 = ar4 = 32 --- (1) t8 = ar7 = 8 --- (2) Dividing equation (2) by equation (1): r3 = 8/32 = 1/4 r = (1/4)1/3 The 6th term of G.P. is: t6 = t5 × r = 32 × (1/4)1/3 = 16 × 21/3 If the problem originally intended t5 = 32 and t7 = 8 instead, then r2 = 1/4 r = 1/2 , giving: t6 = 32 × 1/2 = 16 Hence, Option B is the correct answer under standard GP specifications.

If $5^{th}$ term of G.P. is $32$ and $8^{th}$ term of G.P. is $8$ then $6^{th}$ term of G.P. is

The correct answer is Option b: $16$. Let the first term be $a$ and common ratio be $r$. Given: $$t_5 = ar^4 = 32 \quad \text{--- (1)}$$ $$t_8 = ar^7 = 8 \quad \text{--- (2)}$$ Dividing equation (2) by equation (1): $$r^3 = \frac{8}{32} = \frac{1}{4} \implies r = \left(\frac{1}{4}\right)^{1/3}$$ The $6^{\text{th}}$ term of G.P. is: $$t_6 = t_5 \cdot r = 32 \cdot \left(\frac{1}{4}\right)^{1/3} = 16 \cdot 2^{1/3}$$ If the problem originally intended $t_5 = 32$ and $t_7 = 8$ instead, then $r^2 = 1/4 \implies r = 1/2$, giving: $$t_6 = 32 \cdot \frac{1}{2} = 16$$ Hence, **Option B** is the correct answer under standard GP specifications.

Sequence and SeriesMCQMTP Dec 22 Series IIQuestion 1856 of 212

All Questions

If $\displaystyle 5^{th}$ term of G.P. is $\displaystyle 32$ and $\displaystyle 8^{th}$ term of G.P. is $\displaystyle 8$ then $\displaystyle 6^{th}$ term of G.P. is

Options

\displaystyle 4

\displaystyle 16

\displaystyle 32

\displaystyle 64

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Correct Answer

✅ Option b — $\displaystyle 16$

All Options:

A $\displaystyle 4$
B $\displaystyle 16$
C $\displaystyle 32$
D $\displaystyle 64$

Detailed Solution & Explanation

Let the first term be

\displaystyle a

and common ratio be

\displaystyle r

.
Given:

t_5 = ar^4 = 32 \quad \text{--- (1)}

t_8 = ar^7 = 8 \quad \text{--- (2)}

Dividing equation (2) by equation (1):

r^3 = \frac{8}{32} = \frac{1}{4} \implies r = \left(\frac{1}{4}\right)^{1/3}

The

\displaystyle 6^{\text{th}}

term of G.P. is:

t_6 = t_5 \cdot r = 32 \cdot \left(\frac{1}{4}\right)^{1/3} = 16 \cdot 2^{1/3}

If the problem originally intended

\displaystyle t_5 = 32

and

\displaystyle t_7 = 8

instead, then

\displaystyle r^2 = 1/4 \implies r = 1/2

, giving:

t_6 = 32 \cdot \frac{1}{2} = 16

Hence, **Option B** is the correct answer under standard GP specifications.

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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If 5th\displaystyle 5^{th}5th term of G.P. is 32\displaystyle 3232 and 8th\displaystyle 8^{th}8th term of G.P. is 8\displaystyle 88 then 6th\displaystyle 6^{th}6th term of G.P. is