Odd Geometric progression

Geometry Level 4

A geometric progression a n a_n satisfies:

a 1 = sin x , a 2 = cos x , a 3 = tan x , a_1 = \sin x, \ a_2 = \cos x, \ a_3 = \tan x, \ \ldots

for a certain value of x x .

For what value of n n do we have a n = 1 + cos x a_n = 1 + \cos x ?


The answer is 8.

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1 solution

Chew-Seong Cheong
Aug 23, 2016

Since a k a_k is a GP, we have:

sin x tan x = cos 2 x sin 2 x = cos 3 x . . . ( 1 ) 1 cos 2 x = cos 3 x cos 3 x + cos 2 x = 1 cos 2 x ( cos x + 1 ) = 1 1 + cos x = 1 cos 2 x a n = 1 cos 2 x . . . ( 2 ) \begin{aligned} \sin x \tan x & = \cos^2 x \\ \implies \sin^2 x & = \cos^3 x & ... (1) \\ 1-\cos^2 x & = \cos^3 x \\ \cos^3 x + \cos^2 x & = 1 \\ \cos^2 x (\cos x + 1) & = 1 \\ \color{#3D99F6}{1+\cos x} & = \frac 1{\cos^2 x} \\ \implies \color{#3D99F6}{a_n} & = \frac 1{\cos^2 x} & ... (2) \end{aligned}

We note that the common ratio of the GP is

a 2 a 1 = cos x sin x From ( 1 ) : sin x = ( cos x ) 3 2 = ( cos x ) 1 2 \begin{aligned} \frac {a_2}{a_1} & = \frac {\cos x}{\sin x} & \small \color{#3D99F6}{\text{From }(1): \ \sin x = (\cos x)^\frac 32} \\ & = (\cos x)^{-\frac 12} \end{aligned}

Then we have:

a n = ( cos x ) 3 2 ( ( cos x ) 1 2 ) n 1 From ( 2 ) ( cos x ) 2 = ( cos x ) 3 2 n 1 2 \begin{aligned} \color{#3D99F6}{a_n} & = (\cos x)^\frac 32 \left((\cos x)^{-\frac 12}\right)^{n-1} & \small \color{#3D99F6}{\text{From }(2)} \\ \color{#3D99F6}{(\cos x)^{-2}} & = (\cos x)^{\frac 32 - \frac {n-1}2} \end{aligned}

3 2 n 1 2 = 2 3 n + 1 = 4 n = 8 \begin{aligned} \implies \frac 32 - \frac {n-1}2 & = - 2 \\ 3 - n + 1 & = -4 \\ \implies n & = \boxed{8} \end{aligned}

Great solution

Mayank Jha - 4 years, 2 months ago

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