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Given that the product of the first $n$ terms of a geometric sequence is $P$ ,

$ \bullet$ find $\displaystyle \prod_{k=1}^{\frac{n}{3}}a_{3k-1}$

in terms $P$ if the terms are $a_1,a_2,a_3...a_n$

$ \bullet$ hence prove that $S\ge n\sqrt{P}$ where S is the sum

$S=a_1+a_2+a_3+...+a_n$

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How is $S$ defined? And what are your thoughts on the question? –  Servaes Aug 28 '13 at 8:10
    
i just edited it,now $S$ is defined –  Jonas12 Aug 28 '13 at 10:53
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3 Answers

The inequality $S\ge n\sqrt P$ doesn't hold in general. A geometric sequence has the form $a_n=ar^n$ for some $a$ and $r$, so in general we have

$$P=a^nr^{1+2+\cdots+ n}= a^nr^{n(n+1)/2}$$

while

$$S=a(r+r^2+\cdots +r^n)$$

It's pretty clear by examination of these formulas that $S$ can be a lot smaller than $n\sqrt P$; for example, if $r=1$ and $a=4$, we have $S=4n$ while $n\sqrt P= n2^n$, so $S\lt n\sqrt P$ as soon as $n\gt2$.

Perhaps the OP means the $n$th root of $P$ rather than $n$ times the square root (or, more likely, $n$ times the $n$th root of $P$).

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$S\ge n\sqrt[n]{P}$ you are correct @Barry Cipra –  Jonas12 Aug 28 '13 at 19:19
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so far i considered

$\Large\prod_{k=1}^{\frac{n}{3}}a_{3k-1}=a_2a_5a_8a_{11}...a_{n-1}$

now we use geometric mean

$\large \frac{a_{n-1}}{a_{n-2}}=\frac{a_{n}}{a_{n-1}}$

$a_{n-1}^2=a_{n-2}a_{n}$ hence

$a_2^2=a_1a_3,a_5^2=a_4a_6...,a_{n-1}^2=a_{n-2}a_{n}$

so we have

$a_1a_2a_3a_4...a_n=a_2^2a_2a_3^2a_3...a_{3n}=a_2^3a_5^3a_{11}^3...a_{n-1}^3=(a_2a_5a_{11}...a_{n-1})^3=P$

$\Large\prod_{k=1}^{\frac{n}{3}}a_{3k-1}=\sqrt[3]{P}$

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Are you sure you have the correct question?

Take the case of a series with only one term where that term equals 1.

Then S=P=1 and therefore $S\ge 2\sqrt{P}$ is NOT true for this case.

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sorry i meant $S\ge n\sqrt{P}$ –  Jonas12 Aug 28 '13 at 18:18
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