MCQ
If $a,\,b,\,c$ are in $A.P.$ and ${a^2},\,{b^2},{c^2}$ are in $H.P.$, then
- A$a \ne b \ne c$
- B${a^2} = {b^2} = \frac{{{c^2}}}{2}$
- C$a,\,b,\,c$ are in $G.P.$
- ✓$\frac{{ - a}}{2},b,c$are in $G.P$
$⇒$ $2b = a + c,b -a = c -b$
${a^2},{b^2},{c^2}$ are in $H.P.$
$\frac{1}{{{b^2}}} - \frac{1}{{{a^2}}} = \frac{1}{{{c^2}}} - \frac{1}{{{b^2}}}$ $ \Rightarrow \frac{{{a^2} - {b^2}}}{{{a^2}{b^2}}} = \frac{{{b^2} - {c^2}}}{{{b^2}{c^2}}}$
$⇒$ $(a - b)[{c^2}(a + b) - {a^2}(b + c)] = 0$,
$[\because \,(b - c) = (a - b)]$
$⇒$ $a = b$ or ${c^2}a + {c^2}b - {a^2}b - {a^2}c = 0$
$⇒$ ${c^2}a + {c^2}b - {a^2}b - {a^2}c = 0$
$⇒$ $ac\,(c - a) = b\,({a^2} - {c^2})$
$⇒$ $ac = - b\,(c + a)$
$⇒$ $ - ac = b.2b$
$⇒$ ${b^2} = ( - a/2)\,c$,
$\therefore - a/2,b,c$ are in $G.P.$
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Statement $I$ : For any two non-zero complex numbers $\mathrm{z}_1, \mathrm{z}_2$
$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ and
Statement $II$ : If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three distinct complex numbers and a, b, c are three positive real numbers such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$, then
$\frac{\mathrm{a}^2}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^2}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^2}{\mathrm{x}-\mathrm{y}}=1$
Between the above two statements,