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

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Solution

$∵$ a, b, c are in G.P.

Let $\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = k$

$∴ \frac{b}{a} = k \Rightarrow b = a k$

Also, $\frac{c}{b} = k \Rightarrow c = b k = (a k) . k = a k^{2}$

and $\frac{d}{c} = k \Rightarrow d = c k = (a k^{2}) . k = a k^{3}$

Now, to prove that $(a^{n} + b^{n}), (b^{n} + c^{n}), (c^{n} + d^{n})$ in G.P., we have:

$\frac{b^{n} + c^{n}}{a^{n} + b^{n}} = \frac{c^{n} + d^{n}}{b^{n} + c^{n}}$

Now, $\frac{b^{n} + c^{n}}{a^{n} + b^{n}} = \frac{(a k)^{n} + (a k^{2})^{n}}{a^{n} + (a k)^{n}}$

$= \frac{a^{n} k^{n} + a^{n} k^{2 n}}{a^{n} + a^{n} k^{n}}$

$= \frac{a^{n} k^{n} (1 + k^{n})}{a^{n} (1 + k^{n})}$

$= \frac{a^{n} k^{n}}{a^{n}} = k^{n}$

Also, $\frac{c^{n} + d^{n}}{b^{n} + c^{n}} = \frac{(a k^{2})^{n} + (a k^{3})^{n}}{(a k)^{n} + (a k^{2})^{n}}$

$= \frac{a^{n} k^{2 n} + a^{n} k^{3 n}}{a^{n} k^{n} + a^{n} k^{2 n}}$

$= \frac{a^{n} k^{n} + a^{n} k^{2 n}}{a^{n} + a^{n} k^{n}}$

$= \frac{a^{n} k^{n} (1 + k^{n})}{a^{n} (1 + k^{n})}$

$= \frac{a^{n} k^{n}}{a^{n}} = k^{n}$

Also, $\frac{c^{n} + d^{n}}{b^{n} + c^{n}} = \frac{(a k^{2})^{n} + (a k^{3})^{n}}{(a k)^{n} + (a k^{2})^{n}}$

$= \frac{a^{n} k^{2 n} + a^{n} k^{3 n}}{a^{n} k^{n} + a^{n} k^{2 n}}$

$= \frac{a^{n} k^{2 n} (1 + k^{n})}{a^{n} k^{n} (1 + k^{n})}$

$= \frac{a^{n} k^{2 n}}{a^{n} k^{n}} = k^{n}$

$∴ \frac{b^{n} + c^{n}}{a^{n} + b^{n}} = \frac{c^{n} + d^{n}}{b^{n} + c^{n}}$

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

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