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Let the three numbers in G.P be $\frac ar,a,ar$

\begin{align} &\frac ar+a+ar=31 \\ and\ &\frac {a^2}{r^2}+a^2+a^2r^2=651\\\ &\implies a(r+\frac1r+1)=31\\ and\ & a^2(r^2+\frac{1}{r^2}+1)=651 \end{align} How to solve further please help...

3 Answers 3

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Let's suppose that your numbers are $ a, ar$ and $ ar^2$. Then we have by hyphotesis $$ a+ar+ar^2=31, $$ $$ a^2+a^2r^2+a^2r^4=651. $$

Then $ a(1+r+r^2)=31$ and $ a^2(1+r^2+r^4)=651$. So $$ a^2(1+r^2+r^4)=a^2(1+r+r^2)(r^2-r+1)=(a(1+r+r^2))(a(r^2-r+1))=$$ $$=31a(r^2-r+1)=651.$$

Thus we deduce that $ a(r^2-r+1)=21$. This toguether with $ a(r^2+r+1)=31$ lead us to $2ar=10$, i.e, $ ar=5$ and $2(ar^2+a)=52$, i.e., $ a+ar^2=a+5r=26$. Thus if we replace $ r=5/a $ we obtain the quadratic equation in $ a $: $ a^2-26a+25=0$. An easy factorization shows us that this equation has $ a=1$ and $ a=25$ as solutions.

If $ a=1$, then $ r=5$ and hence the numbers are $1, 5$ and $25$.

If $ a=25$, then $ r=1/5$ and hence the numbers are $25, 5, 1$. In any case we have the same numbers, namely: $1, 5$ and $25$.

Remark: In the solution of the problem I used the identity $$x^4+x^2+1=(x^2+x+1)(x^2-x+1).$$

This identity is known, at least in the latin american mathematical community, as Argand's identity.

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    Can you please explain me the fifth line how $(r^2-r+1)$ how -r. – 2017-01-15
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    @bappy I don't understand what's your doubt. Anyways, I edited my answer. Do you find it clearer? – 2017-01-15
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Hint

You didn't specify if $a,r$ must be natural integers. I will suppose so.

The relation $a(r+\frac{1}{r}+1)=31$ can be writen :

$$a(r^2+r+1)=31r$$

So $(r^2+r+1)\mid31r$. Since $gcd(r^2+r+1,r)=1$, Gauss theorem shows that $r^2+r+1\mid31$.

Now $31$ is a prime number; so $r^2+r+1$ must be equal to $1$ or $31$. Clearly, the former can't hold. So $r^2+r+1=31$. This last equation has got only one positive solution : $5$

You should now use the hypothesis $a^2(r^2+\frac{1}{r^2}+1)=651$ ...

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    Will you please show me how to move to the answer by solving this problem. The answer in my book is given 1,5,25 are the three numbers. But i need to know how we get these numbers. – 2017-01-15
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    Since $r=5$, the equation $a^2(r^2+\frac{1}{r^2}+1)=651$ becomes $a^2=(651\times 25)/(625+25+1)$, that is $a^2=25$. Finally, since $a\ge0$, we get $a=5$. As a conclusion $(a,r)=(5,5)$ and so $(a/r,a,ar)=(1,5,25)$. – 2017-01-15
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    Why $ a $ and $ r $ have to be natural integers? – 2017-01-15
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    They don't have to : it is just a simplifying hypothesis which enables the use of Gauss theorem. – 2017-01-15
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Given

$a+ar+ar^2=31\\ a^2+a^2r^2+a^2r^4=651$

square the first equation
$a^2+a^2r^2+a^2r^4+2(a^2r+a^2r^3+a^2r^2)=961\\ a^2+a^2r^2+a^2r^4+2r(a^2+a^2r^2+a^2r)=961\\ 651+2r(a^2+a^2r^2+a^2r)=961\\ 2r(a^2+a^2r^2+a^2r)=310\\ 2ar(a+ar^2+ar)=310\\ 2ar(31)=310\\ ar=5\\ a=\dfrac{5}{r}$

From first equation
$a+ar+ar^2=31\\ a(1+r+r^2)=31\\ \dfrac{5}{r}(1+r+r^2)=31\\ \dfrac{1}{r}(1+r+r^2)=\dfrac{31}{5}\\ \dfrac{1}{r}+1+r=\dfrac{31}{5}\\ 1+r+r^2=\dfrac{31r}{5}\\ 5+5r+5r^2=31r\\ 5r^2-26r+5=0$

Solving this quadratic equation gives $r=5$

$a=\dfrac{5}{r}=1$

So numbers are

$1,5,25$

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    how do u get 2 in the line after square the first equation. – 2017-01-15
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    @bappy $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$ – 2017-01-15