Question

In $\Delta ABC$ with usual notations, if $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)},$ then $a^{-2},b^{-2},c^{-2}$ are in

• A. $GP$
• B. $AGP$
• C. $AP$
• D. $HP$

Answer 1

The correct option is D $HP$

In a $\Delta ABC$

$\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow \frac{sinA}{sinC}=\frac{sin(A-B)}{sin(B-C)}$

$\Rightarrow \frac{sin({180}^{\circ }-(B+C\left)\right)}{sin({180}^{\circ }-(A+B\left)\right)}=\frac{sin(A-B)}{sin(B-C)}$

$\Rightarrow \frac{sin(B+C)}{sin(A+B)}=\frac{sin(A-B)}{sin(B-C)}$ $\because sin(180-\theta )=sin\theta$

$\Rightarrow sin(B+C)sin(B-C)=sin(A-B)sin(A+B)$

using formula

$\Rightarrow si{n}^{2}B-si{n}^{2}C=si{n}^{2}A-si{n}^{2}B$

$\Rightarrow 2si{n}^{2}B=si{n}^{2}A+si{n}^{2}C$

From sin rule.

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k$

$\Rightarrow 2k^2{b}^2=k^2{a}^2+k^2{c}^2$

$\Rightarrow 2b^2=a^2+c^2\to a^2,b^2,c^2$ are in A.P.

$\Rightarrow \frac{2}{b^{-2}}=\frac{1}{a^{-2}}+\frac{1}{c^{-2}}\to a^{-2},b^{-2},c^{-2}$ are in H.P