<s>[INST] Find 4 numbers which create a ratio Find four numbers which create ratio if its known that sum of first and last is equal to 14, sum of middle two is equal to 11 and sum of squares of all numbers is equal to 221
I got only that sum of product first with last and product of two middle is equal to 48
I have written this question by phone so please sorry if I did mistake
I hope I explain my point clear
 [/INST] If we interpret the phrase "four numbers which create ratio" as "four numbers that make a proportion", i.e. "the ratio of the first two numbers equals the ratio of the last two numbers," then we have four equations in the four variables $a,b,c,d$:
$$\frac ab=\frac cd$$
$$a+d=14$$
$$b+c=11$$
$$a^2+b^2+c^2+d^2=221$$
From the middle two we get $d=14-a$ and $c=11-b$. Substituting those into the first equation we get
$$\frac ab=\frac {11-b}{14-a}$$
$$a(14-a)=b(11-b)$$
$$a^2-14a=b^2-11b$$
Substituting the expressions for $c$ and $d$ into the last of the four original equations,
$$a^2+b^2+(11-b)^2+(14-a)^2=221$$
$$a^2+b^2+121-22b+b^2+196-28a+a^2=221$$
$$2(a^2-14a)+2(b^2-11b)+96=0$$
$$2(a^2-14a)+2(a^2-14a)+4\cdot 24=0$$
$$a^2-14a+24=0$$
$$(a-12)(a-2)=0$$
$$a=12 \quad\text{or}\quad a=2$$
If we take $a=12$ then we get
$$12^2-14\cdot 12=b^2-11b$$
$$b^2-11b+24=0$$
$$(b-8)(b-3)=0$$
$$b=8 \quad\text{or}\quad b=3$$
The first gives us the numbers $12,8,3,2$ in order; the second gives us $12,3,8,2$, the same four numbers in a different order.
Similarly, if we had taken $a=2$ we still would have gotten $b=8$ or $b=3$, giving us the answers $2,8,3,12$ or $2,3,8,12$, the very same four numbers in different orders.
Our final answer is the four numbers $12,8,3,2$, in four different possible orders.
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