Question

If $a,b,c$ are in $A.P.$ and $a^2,b^2,c^2$ are in $H.P,$ then

• A. $a=b=c$
• B. $2b=3a+c$
• C. $b^2=\sqrt{\left(ac/8\right)}$
• D. None of these

Answer 1

The correct option is A $a=b=c$

$a,b,c\text{are in}AP$

$2b=a+c-\left(1\right)$

$a^2,b^2,c^2\text{are in}HP$

$\frac{1}{a^2},\frac{1}{b^2}\cdot \frac{1}{c^2}\text{are in}AP$

$\frac{2}{b^2}=\frac{1}{a^2}+\frac{1}{c^2}$

$\frac{2}{{\left(\frac{a+c}{2}\right)}^2}=\frac{1}{a^2}+\frac{1}{c^2}$

$\frac{8}{a^2+c^2+2ac}=\frac{a^2+c^2}{a^2{c}^2}$

$8a^2{c}^2={(a^2+c^2)}^2+2ac(a^2+c^2)$

$8a^2{c}^2=a^4+c^4+2a^2{c}^2+2a^{3}c+2c^{3}a$

$a^4+c^4-6a^2{c}^2+2a^{3}c+2a{c}^3=0$

${(a^2-c^2)}^2-4a^2{c}^2+2a^{3}c+2a{c}^3=0$

$(a-c{)}^2(a+c{)}^2+2ac(a^2+c^2-2ac)=0$

$(a-c{)}^2(a+c{)}^2+2ac(a-c{)}^2=0$

$(a-c{)}^2\left(\right(a+c{)}^2+2ac)=0$

$a-c=0$ or $(a+c{)}^2+2ac=0$

$a=c-\left(2\right)$

from equation $\left(1\right)$ and $\left(2\right)$

$\begin{array}{cc}2b& =c+c\\ b& =c\\ a& =b=c\end{array}$