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

Open in App

Solution

It is given that $a, b, c$ and $d$ are in G.P.

$∴ b^{2} = a c . . . (1), . c^{2} = b d . . . (2), a d = b c . . . . (3)$
It has to be proved that $(a^{n} + b^{n}), (b^{n} + c^{n}), (c^{n} + d^{n})$ are in G.P.
i.e., $(b^{n} + c^{n})^{2} = (a^{n} + b^{n}) (c^{n} + d^{n})$
Consider
$L.H.S. = (b^{n} + c^{n})^{2} = b^{2 n} + {2 b}^{n} c^{n} + c^{2 n}$

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

$= {(a c)}^{2} + {2 b}^{n} c^{n} + {(b d)}^{n}, [Using (1) and (2)]$

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

$= a^{n} c^{n} + b^{n} c^{n} + a^{n} d^{n} + b^{n} d^{n}, [Using (3)]$

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

$= (a^{n} + b^{n}) (c^{n} + d^{n}) = R.H.S.$

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

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

Suggest Corrections

Similar questions

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

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, b, c, d

(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}

Join BYJU'S Learning Program