Question

If three distinct numbers $a,b,c$ are in $H.P.$ and their squares are in $A.P.,$ then $a:b:c$ can be

• A. $1-\sqrt{3}:-2:1+\sqrt{3}$
• B. $1-\sqrt{3}:2:1+\sqrt{3}$
• C. $2+\sqrt{3}:-2:2-\sqrt{3}$
• D. $1+\sqrt{3}:-2:2-\sqrt{3}$

Answer 1

The correct option is A $1-\sqrt{3}:-2:1+\sqrt{3}$

Given, $b=\frac{2ac}{a+c}$
$\Rightarrow \frac{b}{2}=\frac{ac}{a+c}=k\text{(say)}\Rightarrow b=2k,ac=k(a+c)$

Also, $a^2,b^2,c^2$ are in A.P., so
$2b^2=a^2+c^2=(a+c{)}^2-2ac\Rightarrow (a+c{)}^2-2k(a+c)-8k^2=0\Rightarrow k=\frac{2k\pm \sqrt{36k^2}}{2}\Rightarrow a+c=4k,-2k$

When $a+c=4k$, we get
$ac=4k^2\Rightarrow a(4k-a)=4k^2\Rightarrow a^2-4k\cdot a+4k^2=0\Rightarrow a=2k,c=2k$
Which is not possible as numbers are distinct.

When $a+c=-2k$, we get
$ac=-2k^2\Rightarrow a(-2k-a)=-2k^2\Rightarrow a^2+2k\cdot a-2k^2=0\Rightarrow (a+k{)}^2=3k^2\Rightarrow a=-k(1\pm \sqrt{3})$
Now,
$a=-k(1+\sqrt{3})\Rightarrow c=-k(1-\sqrt{3})a=-k(1-\sqrt{3})\Rightarrow c=-k(1+\sqrt{3})$

So, $a:b:c$ is
$-k(1+\sqrt{3}):2k:-k(1-\sqrt{3})\text{or}-k(1-\sqrt{3}):2k:-k(1+\sqrt{3})\therefore a:b:c=1+\sqrt{3}:-2:1-\sqrt{3}\text{or}1-\sqrt{3}:-2:1+\sqrt{3}$