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In a geometric progression containing $6$ terms, the first term is $2$ and the sum of $6$ terms is $728$.

What is the common ratio?

Rócherz
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Achari S Ganesha
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  • Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Nov 30 '16 at 10:03

2 Answers2

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Hint. By definition of geometric progression, the sum of the six terms is given by $$2+2k+2k^2+2k^3+2k^4+2k^5=728\implies k(1+k+k^2+k^3+k^4)=\frac{728-2}{2}=363$$ where $k$ is the common ratio. Can you take it from here?

P.S. Note that the equation $k(1+k+k^2+k^3+k^4)=363$ has a unique positive solution (the polynomial on the left is strictly increasing for $k> 0$). Try with some integer value.

Robert Z
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The partial sum $$2\sum_{k=0}^{5}x^{k-1}$$ can be simplified to

$$2\cdot \frac{1-x^6}{1-x}=728$$

$$ \frac{1-x^6}{1-x}=364$$

This is a polynomial equation

$x^6-364x+363=0$

Now apply the Rational root theorem. For this purpose you have to factorize $363$. After you have found a solution you have to check it.

callculus42
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