Question

If $p^{th}, q^{th}$ $r^{th}$ terms of an A.P. are equal to corresponding terms of a G.P. and these terms are respectively x, y, z, then $x^{y-z} \cdot y^{z-x} \cdot z^{x-y}$ equals

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

Answer: (B)

Let first term of an A.P. be a and c.d. be d and first term of a G.P. be A and c.r. be R,then
$a+\left(p-1\right)d=AR^{p-1}=x$
$\Rightarrow p-1=\left(x-a\right)/d\,...\left(1\right)$
$a+\left(q-1\right)d=Ar^{q-1}=y$
$\Rightarrow q-1=\left(y-a\right)/d\,...\left(2\right)$
$a+\left(r-1\right)d=AR^{r-1}=z$
$r-1=\left(z-a\right)/d\,...\left(3\right)$
$\therefore $ Given expression
$=\left(AR^{p-1}\right)^{y-z} \cdot \left(AR^{q-1}\right)^{z-x} \cdot \left(AR^{r-1}\right)^{x-y}$
$=A^{0}\,R^{\left(p-1\right)\left(y-z\right) + \left(q-1\right)\left(z-x\right)+\left(r-1\right)\left(x-y\right)}$
$=A^{0}R^{\left[\left(x-a\right)\left(y-z\right)+\left(y-a\right)\left(z-x\right)+\left(z-a\right)\left(x-y\right)\right]/d}\,\left[By \left(1\right), \left(2\right)\right]$
$=A^{0}R^{0}=1$