Question

If $a,b,c,d$ are in G.P., then ${\left(a^3+b^3\right)}^{-1},{\left(b^3+c^3\right)}^{-1},(c^3+$ ${d^3)}^{-1}$ are in

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

Answer 1

Answer: (B)

Let $b=ar,c=a{r}^2$ and $d=a{r}^3$
Then, $\frac{1}{a^3+b^3}=\frac{1}{a^3\left(1+r^3\right)},\frac{1}{b^3+c^3}=\frac{1}{a^3{r}^3\left(1+r^3\right)}$ and, $\frac{1}{c^3+d^3}=\frac{1}{a^3{r}^6\left(1+r^3\right)}$
Clearly, ${\left(a^3+b^3\right)}^{-1},{\left(b^3+c^3\right)}^{-1}$ and ${\left(c^3+d^3\right)}^{-1}$ are in $G.P$. with common ratio $1/r^3$.