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 1 progression, 2 sums, 3 numbers (Posted on 2017-03-09)
Is it possible to have three real numbers in geometric progression
having sum 50, and the sum of their reciprocals equal to 4.5?
If so, find all such sets of three numbers.
If not, prove that there are none.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Solution Comment 1 of 1
Let the median number be m and let the ratio be r.  Then the members of the progression are m/r, m, m*r.
Their sum is mr + m + m*r = m*(1/r + 1 + r).

The reciprocals of the progression are 1/(m*r), 1/m, r/m.
Their sum is 1/(m*r) + 1/m + r/m = (1/m)*(1/r + 1 + r).

From the given values m*(1/r + 1 + r)=50 and (1/m)*(1/r + 1 + r)=4.5.

Create a ratios equation be dividing the first equation by the second:
[m*(1/r + 1 + r)]/[(1/m)*(1/r + 1 + r) = 50/4.5
This simplifies to m^2 = 100/9.

Then m = 10/3 or -10/3.  If m=10/3 then (10/3)*(1/r + 1 + r) = 50, which simplifies to 1/r + r = 14.  This implies r = 7+4*sqrt(3) or 7-4*sqrt(3).  These two answers are actually the same geometric sequence in ascending and descending order.

If m=-10/3 then then (-10/3)*(1/r + 1 + r) = 50, which simplifies to 1/r + r = -16.  This implies r = -8+3*sqrt(7) or -8-3*sqrt(7).  Like the previous case these two answers describe the same sequence.

Then the two distinct geometric progressions are
{(70-40*sqrt(3))/3, 10/3, (70+40*sqrt(3))/3} and
{(80-30*sqrt(7))/3, -10/3, (80+30*sqrt(7))/3}.

 Posted by Brian Smith on 2017-03-09 10:44:30

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