If $a, b, c, d$ are in G.P., prove that they are in the same place as if they were in a different place.

Question

If $a, b, c, d$ are in G.P., prove that $\left(a^{2}+b^{2}\right),\left(b^{2}+c^{2}\right),\left(c^{2}+d^{2}\right)$ are in G.P.

And in general,

If $a, b, c, d$ are in G.P., prove that $$ \left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right),\left(c^{n}+d^{n}\right) \text { are in G.P. } $$

Answer 1

Let $r$ be the common ratio of the G.P. $a, b, c, d$. Then, $b=a r ; c=a r^{2}$ and $d=a r^{3}$.

$\therefore \quad\left(a^{2}+b^{2}\right)=\left(a^{2}+a^{2} r^{2}\right)=a^{2}\left(1+r^{2}\right) ;$

$\left(b^{2}+c^{2}\right)=\left(a^{2} r^{2}+a^{2} r^{4}\right)=a^{2} r^{2}\left(1+r^{2}\right) ;$

and $\left(c^{2}+d^{2}\right)=\left(a^{2} r^{4}+a^{2} r^{6}\right)=a^{2} r^{4}\left(1+r^{2}\right)$.

Thus, $\left(b^{2}+c^{2}\right)^{2}=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)$.

This shows that $\left(a^{2}+b^{2}\right),\left(b^{2}+c^{2}\right)$ and $\left(c^{2}+d^{2}\right)$ are in G.P.

Now, $\left(a^{n}+b^{n}\right)=\left(a^{n}+a^{n} r^{n}\right)=a^{n}\left(1+r^{n}\right) ;$

$\left(b^{n}+c^{n}\right)=\left(a^{n} r^{n}+a^{n} r^{2n}\right)=a^{n} r^{n}\left(1+r^{n}\right) ;$

and $\left(c^{n}+d^{n}\right)=\left(a^{n} r^{2n}+a^{n} r^{3n}\right)=a^{n} r^{2n}\left(1+r^{n}\right)$.

Thus, $\left(b^{n}+c^{n}\right)^2=\left(a^{n}+b^{n}\right)\left(c^{n}+d^{n}\right)$.

This shows that $\left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right)$ and $\left(c^{n}+d^{n}\right)$ are in G.P.

Answer 2

Since $ \ a \ , \ b \ , \ c \ , \ d \ $ form a geometric progression (GP) , we have $ \ \frac{b}{a} \ = \ \frac{c}{b} \ = \ \frac{d}{c} \ = \ r $ $ \Rightarrow \ ac \ = \ b^2 \ \ , \ \ bd \ = \ c^2 \ \ , \ \ ad \ = \ bc \ \ . $ Thus, $$ (b^2 \ + \ c^2)^2 \ \ = \ \ (b^2)^2 \ + \ 2·(bc)^2 \ + \ (c^2)^2 $$ $$ = \ \ (ac)^2 \ + \ (bc)^2 \ + \ (ad)^2 \ + \ (bd)^2 \ \ = \ \ (a^2 \ + \ b^2) \ · \ (c^2 \ + \ d^2)$$ $$ \Rightarrow \ \ \frac{b^2 \ + \ c^2}{a^2 \ + \ b^2} \ \ = \ \ \frac{c^2 \ + \ d^2}{b^2 \ + \ c^2} \ \ , $$ from which we conclude that $ \ (a^2 \ + \ b^2) \ \ , \ \ (b^2 \ + \ c^2) \ \ , \ \ (c^2 \ + \ d^2) \ $ also form a GP.

The general proof is analogous: $$ (b^n \ + \ c^n)^2 \ \ = \ \ (b^n)^2 \ + \ 2·(bc)^n \ + \ (c^n)^2 \ \ = \ \ (b^2)^n \ + \ 2·(bc)^n \ + \ (c^2)^n $$ $$ = \ \ (ac)^n \ + \ (bc)^n \ + \ (ad)^n \ + \ (bd)^n \ \ = \ \ (a^n \ + \ b^n) \ · \ (c^n \ + \ d^n)$$ $$ \Rightarrow \ \ \frac{b^n \ + \ c^n}{a^n \ + \ b^n} \ \ = \ \ \frac{c^n \ + \ d^n}{b^n \ + \ c^n} \ \ , $$ and the general statement follows.