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

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Solution

Given $a, b, c, d$ are in GP

Let $r$ be common ratio of GP

Therefore, $b = a r$
$c = a r^{2}$
$d = a r^{3}$

Now, $a^{n} + b^{n} = a^{n} + a^{n} r^{n} = a^{n} (1 + r^{n})$
Similarly $b^{n} + c^{n} = a^{n} r^{n} + a^{n} r^{2 n} = a^{n} r^{n} (1 + r^{n})$
And $c^{n} + d^{n} = a^{n} r^{2 n} + a^{n} r^{3 n} = a^{n} r^{2 n} (1 + r^{n})$

We can see the terms $a^{n} + b^{n}, b^{n} + c^{n}, c^{n} + d^{n}$ are in GP with common ratio $r^{n}$

Therefore the given terms are also in G.P.

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Similar questions

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.

(a^{n} + b^{n}), (b^{n} + c^{n}), (c^{n} + d^{n})

a, b, c

(a^{n} + b^{n}), (b^{n} + c^{n}), (c^{n} + d^{n})

a N = {a x : x \in N}

b n \cap c N = d N

b, c \in N

{a_{n}}, {b_{n}}, {c_{n}}

(i) a_{n} + b_{n} {+ c}_{n} = 2 n + 1

(i i) a_{n} b_{n} + b_{n} c_{n} + {+ c}_{n} a_{n} = 2 n - 1

(i i i) a_{n} b_{n} c_{n} = - 1

(i v) a_{n} < b_{n} < c_{n}

lim n \to \infty n a_{n}

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