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Q.
If $a, b, c, d$ are in G.P., then $\left(a^{3}+b^{3}\right)^{-1},\left(b^{3}+c^{3}\right)^{-1},\left(c^{3}+\right.$
$\left.d^{3}\right)^{-1}$ are in
Sequences and Series
Solution:
Let $b=a r, c=a r^{2}$ and $d=a r^{3}$
Then, $\quad \frac{1}{a^{3}+b^{3}}=\frac{1}{a^{3}\left(1+r^{3}\right)}, \frac{1}{b^{3}+c^{3}}=\frac{1}{a^{3} r^{3}\left(1+r^{3}\right)}$ 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 $1 / r^{3}$.