Question

If a, b, c are in AP, then which of the following will not be in AP?

• A. $a^{2}bc,a{b}^{2}c,ab{c}^2$
• B. $\frac{1}{bc},\frac{1}{ac},\frac{1}{ab}$
• C. $\frac{a}{b},\frac{b}{c},\frac{c}{a}$
• D. $\sqrt{\frac{a}{bc}},\sqrt{\frac{b}{ac}},\sqrt{\frac{c}{ab}}$

Answer 1

The correct option is C $\frac{a}{b},\frac{b}{c},\frac{c}{a}$

Given: a, b, c are in AP
$\Rightarrow b–a=c–b$
$\Rightarrow 2b=a+c$…..(i)
Now, consider: $a^{2}bc+ab{c}^2$
$=abc(a+c)$
$=abc\times 2b$ [Using (i)]
$=2a{b}^{2}c$
Thus, $a^{2}bc,a{b}^{2}c,ab{c}^2$ are in AP.
Consider:
$\frac{1}{bc}+\frac{1}{ab}=\frac{1}{b}[\frac{1}{c}+\frac{1}{a}]$
$=\frac{1}{b}\left(\frac{a+c}{ac}\right)$
$=\frac{1}{b}\left(\frac{2b}{ac}\right)$ [Using (i)]
$=\frac{2}{ac}$
Thus, $\frac{1}{bc},\frac{1}{ac},\frac{1}{ab}$ are in AP
Consider: $\frac{a}{b}+\frac{c}{a}=\frac{a^2+bc}{ab}\ne \frac{2b}{c}$
Thus, $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are not in AP
Consider; $\sqrt{\frac{a}{bc}}+\sqrt{\frac{c}{ab}}=\sqrt{\frac{a}{bc}}+\sqrt{\frac{c}{ab}}$
$=\frac{(\sqrt{a}{)}^2+(\sqrt{c}{)}^2}{\sqrt{abc}}$
$=\frac{2b}{\sqrt{abc}}$
$=2\sqrt{\frac{b}{ac}}$

Thus, $\sqrt{\frac{a}{bc}},\sqrt{\frac{b}{ac}},\sqrt{\frac{c}{ab}}$ are in AP. Therefore, all the options except for (3) are in AP. Hence, the correct answer is option (3).