Question

If one $GM,g$ and two $A{M}^{\prime }s,p$ and $q$ are inserted between two numbers $a$ and $b$, then $(2p-q)(p-2q)$ is equal to

• A. $g^2$
• B. $-g^2$
• C. $2g$
• D. $3g^2$

Answer 1

Answer: (B)

$-g^2$

$GM,g=\sqrt{ab}$

common difference of the A.P $=\frac{b-a}{3}$

$\therefore p=a+\frac{b-a}{3}$

$=\frac{2a+b}{3}$

$q=a+2\left(\frac{b-a}{3}\right)$

$=\frac{3a+2b-2a}{3}=\frac{2b+a}{3}$

$\therefore (2p-q)(p-2q)$

$=[2\left(\frac{2a+b}{3}\right)-\frac{2b+a}{3}][\frac{2a+b}{3}....2\left(\frac{2b+a}{3}\right)]$

$=\left[\frac{4a+2b-2b-a}{3}\right]\left[\frac{2a+b-4b-2a}{3}\right]$

$=\left[a\right][-b]$

$=-ab$

$=-(\sqrt{ab}{)}^2$

$=-g^2$