Question:
If $a, b, c$ are in GP, prove that $a^{3}, b^{3}, c^{3}$ are in GP
Solution:
To prove: $a^{3}, b^{3}, c^{3}$ are in GP
Given: $a, b, c$ are in GP
Proof: As a, b, c are in GP
$\Rightarrow b^{2}=a c$
Cubing both sides
$\Rightarrow\left(b^{2}\right)^{3}=(a c)^{3}$
$\Rightarrow b^{6}=a^{3} c^{3}$
$\Rightarrow \frac{b^{3}}{a^{3}}=\frac{c^{3}}{b^{3}}=$ common ratio $=r$
From the above equation, we can say that $a^{3}, b^{3}, c^{3}$ are in GP
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