Question

$\text{Let a, b, c are non-zero real numbers such that}\frac{1}{a},\frac{1}{b},\frac{1}{c}\text{are in arithmetic progression}\text{and a, b, -2c are in geometric progression, then which of the following statement(s) must be true?}$

• A. $a^2,b^2,4c^2\text{are in geometric progression}$
• B. $-2a,b,-2c\text{are in arithmetic progression}$
• C. $a^3+b^3+c^3-3abc=0$
• D. $a^2,b^2,c^2\text{are in harmonic progression}$

Answer 1

Answer: (A), (B), (C)

$a^3+b^3+c^3-3abc=0$

$\because \frac{1}{a},\frac{1}{b},\frac{1}{c}\text{are in arithmetic progression}\therefore b=\frac{2ac}{a+c}....\left[1\right]\text{and a, b, -2c are in geometric progression,}\therefore b^2=-2ac....\left[2\right]\text{From [1] and [2], we get}b=\frac{-b^2}{a+c}\Rightarrow a+b+c=0\therefore a^3+b^3+c^3-3abc=0\text{Also,}(a+c{)}^2=(-b{)}^2\Rightarrow a^2+c^2-b^2=b^2\Rightarrow a^2+c^2=2b^2\therefore a^2,b^2,c^2\text{are not in harmonic progression}\text{Also,}{b}^4=4a^2{c}^2\therefore a^2,b^2,4c^2\text{are in geometric progression}$