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

Step $1$ : Simplify Given data.

Given, $a, b, c$ and $d$ are in G.P.

So,

$∴ b^{2} = a c . . . (i)$

$c^{2} = b d$ $. . . (i i)$

$a d = b c$ $. . . (i i i)$

Step $2$ : Solve for prove.

Required to prove
$(a^{n} + b^{n}), (b^{n} + c^{n}), (c^{n} + a^{n})$ are in G.P. i.e.,

$(b^{n} + c^{n})^{2} = (a^{n} + b^{n}) (c^{n} + d^{n})$

Taking $L.H.S$

$(b^{n} + c^{n})^{2} = b^{2 n} + 2 b^{n} C^{n} + C^{2 n}$

$\Rightarrow L.H.S. = (b^{2})^{n} + 2 b^{2} c^{n} + (c^{2})^{n}$

$\Rightarrow L.H.S. = (a c)^{n} + 2 b^{n} c^{n} + (b d)^{n}$ [ Using $(i)$ and $(i i)$ ]

$\Rightarrow L.H.S. = a^{n} c^{n} + b^{n} c^{n} + b^{n} c^{n} + b^{n} d^{n}$

$= a^{n} c^{n} + b^{n} c^{n} + a^{n} d^{n} + b^{n} d^{n}$ [using $(i i i)$ ]

$= c^{n} (a^{n} + b^{n}) + d^{n} (a^{n} + b^{n})$

$= (a^{n} + b^{n}) (c^{n} + d^{n})$

$= R.H.S$

Therefore, $(a^{n} + b^{n}), (b^{n} + c^{n}),$ and $(c^{n} + d^{n})$ are in G.P.

Hence proved.

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