Question 1134409
<br>
The four numbers are<br>
a, ar, ar^2, ar^3<br>
Their sum is {{{a(1+r+r^2+r^3)}}}; their product is {{{a^4r^6}}}.  So<br>
{{{a(1+r+r^2+r^3) = 85}}} and {{{a^4r^6 = 4096}}}<br>
Since both the sum and product are whole numbers, the four numbers are almost certainly whole numbers.  So solve the problem by logical trial and error.<br>
One way is to start with the given product and find the numbers with that product whose sum is 85.<br>
{{{a^4r^6 = 4096 = 2^12}}}<br>
A bit of playing around with that finds a=1 and r=2^2=4 is a solution; that gives the four numbers as 1, 4, 16, and 64; and their sum is 85.<br>
Or another way would be to start with the sum of 85 and find four numbers with that sum whose product is 4096.  This one takes a bit more problem solving skill....<br>
{{{a(1+r+r^2+r^3) = 85}}}
{{{a(1+r)+ar^2(1+r) = 85}}}
{{{a(1+r)(1+r^2) = 85 = 5*17}}}<br>
Again a=1 and r=4 satisfies this equation, because 1+r = 5 and 1+r^2 = 17.<br>
And here is another way to start with the sum of 85 and find the solution by logical reasoning; this path is probably easier to see than the one described above.<br>
{{{a(1+r+r^2+r^3) = 85 = 5*17}}}
Since the numbers are probably whole numbers, this means a has to be either 1 or 5.  And a bit of playing with those possibilities quickly shows it can't be 5.  So a = 1, and it takes only a bit more trial and error to find that r=4.  Then verifying the solution by showing that the product is 4096 completes the problem.