Question

In $△$ ABC, if $si{n}^2\frac{A}{2},si{n}^2\frac{C}{2}$ be in H. P. then a, b, c will be in

• A. A. P.
• B. G. P.
• C. H. P.
• D. None of these

Answer 1

The correct option is C H. P.

$\frac{1}{si{n}^2\frac{C}{2}}-\frac{1}{si{n}^2\frac{B}{2}}=\frac{1}{si{n}^2\frac{C}{2}}areinA.P.$
$\Rightarrow \frac{1}{si{n}^2\frac{C}{2}}-\frac{1}{si{n}^2\frac{B}{2}}=\frac{1}{si{n}^2\frac{B}{2}}-\frac{1}{si{n}^2\frac{A}{2}}$
$\Rightarrow \frac{ab}{(s-a)(s-b)}-\frac{ac}{(s-a)(s-c)}$
$=\frac{ac}{(s-a)(s-c)}-\frac{bc}{(s-b)(s-c)}$
$\Rightarrow \left(\frac{a}{s-a}\right)\left(\frac{b(s-c)-c(s-b)}{(s-b)(s-c)}\right)=\left(\frac{c}{s-c}\right)\left(\frac{a(s-b)-b(s-a)}{(s-a)(s-b)}\right)$
$\Rightarrow$ abs - abc - acs + abc = acs - abc - bcs + abc
$\Rightarrow$ ab - ac = ac - bc $\Rightarrow$ ab + bc = 2ac
or, $\frac{1}{c}+\frac{1}{a}=\frac{2}{b},i.e.,a,b,careinH.P.$
Note : Students should remember this question as a fact.