If $a, b, c$ are in AP and $x, y, z$ are in GP, then the value of $x^{(b-c)} y^{(c-a)} z^{(a-b)}$ is

The correct answer is Option a: $1$. Since $a, b, c$ are in A.P., we have the common difference $d$: $$b-a = c-b = d \implies a-b = -d \text{ and } b-c = -d \text{ and } c-a = 2d$$ Since $x, y, z$ are in G.P., we can write: $$y = xr, \ z = xr^2$$ Substitute these into the expression: $$E = x^{(b-c)} y^{(c-a)} z^{(a-b)}$$ $$E = x^{-d} \cdot (xr)^{2d} \cdot (xr^2)^{-d}$$ $$E = x^{-d + 2d - d} \cdot r^{2d - 2d}$$ $$E = x^0 \cdot r^0 = 1$$ Hence, **Option A** is the correct answer.

Sequence and SeriesMCQMTP Nov 20Question 1881 of 212

All Questions

If $\displaystyle a, b, c$ are in AP and $\displaystyle x, y, z$ are in GP, then the value of $\displaystyle x^{(b-c)} y^{(c-a)} z^{(a-b)}$ is

Options

\displaystyle 1

\displaystyle 0

\displaystyle b - c

DNone of these

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Correct Answer

✅ Option a — $\displaystyle 1$

All Options:

A $\displaystyle 1$
B $\displaystyle 0$
C $\displaystyle b - c$
DNone of these

Detailed Solution & Explanation

Since

\displaystyle a, b, c

are in A.P., we have the common difference

\displaystyle d

b-a = c-b = d \implies a-b = -d \text{ and } b-c = -d \text{ and } c-a = 2d

Since

\displaystyle x, y, z

are in G.P., we can write:

y = xr, \ z = xr^2

Substitute these into the expression:

E = x^{(b-c)} y^{(c-a)} z^{(a-b)}

E = x^{-d} \cdot (xr)^{2d} \cdot (xr^2)^{-d}

E = x^{-d + 2d - d} \cdot r^{2d - 2d}

E = x^0 \cdot r^0 = 1

Hence, **Option A** 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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If a,b,c\displaystyle a, b, ca,b,c are in AP and x,y,z\displaystyle x, y, zx,y,z are in GP, then the value of x(b−c)y(c−a)z(a−b)\displaystyle x^{(b-c)} y^{(c-a)} z^{(a-b)}x(b−c)y(c−a)z(a−b) is