- ✓${a^2}b,\,{c^2}a,\,{b^2}c$ are in $A.P.$
- B${a^2}b,\,{b^2}c,\,{c^2}a$ are in $H.P.$
- C${a^2}b,\,{b^2}c,\,{c^2}a$ are in $G.P.$
- DNone of these
==> $\frac{{2c}}{b} = \frac{b}{a} + \frac{a}{c}$
$ \Rightarrow \frac{{2c}}{b} = \frac{{bc + {a^2}}}{{ac}}$
==> $2a{c^2} = {b^2}c + b{a^2}$
$\therefore \,{a^2}b,\,{c^2}a$ and ${b^2}c$ are in $A.P.$
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$(A)$ $\frac{|\vec{c}|^2}{2}-|\vec{a}|=12$
$(B)$ $\frac{|\vec{c}|^2}{2}+|\vec{a}|=30$
$(C)$ $|\vec{a} \times \vec{b}+\vec{c} \times \vec{a}|=48 \sqrt{3}$
$(D)$ $\vec{a} \cdot \vec{b}=-72$
$\sin ^{-1}\left(\sum_{i=1}^{\infty} x^{i+1}-x \sum_{i=1}^{\infty}\left(\frac{x}{2}\right)^i\right)=\frac{\pi}{2}-\cos ^{-1}\left(\sum_{i=1}^{\infty}\left(-\frac{x}{2}\right)^i-\sum_{i=1}^{\infty}(-x)^i\right)$
lying in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is. . . . .
(Here, the inverse trigonometric functions $\sin ^{-1} x$ and $\cos ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $[0, \pi]$, respectively.)