Progression

Algebra Level 2

The 1st, 4th and 8th terms of an arithmetic progression , are the first three terms of a geometric progression . Find the common ratio of the geometric progression.

3 3 5 5 4 3 \dfrac{4}{3} 3 7 \dfrac{3}{7}

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1 solution

let a 1 a_1 be the first term of the arithmetic progression and d d be the common difference, then

a 1 = x , a 4 = x + 3 d a_1=x, a_4=x+3d and a 8 = x + 7 d a_8=x+7d

Since the above is the first three terms of a geometric progression, then

a 4 a 1 = a 8 a 4 \dfrac{a_4}{a_1}=\dfrac{a_8}{a_4} \color{#D61F06}\large \implies x + 3 d x = x + 7 d x + 3 d \dfrac{x+3d}{x}=\dfrac{x+7d}{x+3d} \color{#D61F06}\large \implies x 2 + 6 x d + 9 d 2 = x 2 + 7 x d x^2+6xd+9d^2=x^2+7xd \color{#D61F06}\large \implies 9 d 2 = x d 9d^2=xd \color{#D61F06}\large \implies x = 9 d x=9d

Therefore, the common ratio of the geometric progression is,

r = a 4 a 1 = x + 3 d x = 9 d + 3 d 9 d 12 d 9 d = r=\dfrac{a_4}{a_1}=\dfrac{x+3d}{x}=\dfrac{9d+3d}{9d}\dfrac{12d}{9d}= 4 3 \boxed{\dfrac{4}{3}}

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