Question 691008
Middle numbers can be very useful in arithmetic progressions, even if the wording does not refer to an A.P., as in "three consecutive odd numbers".
You may have been told that if three numbers are in arithmetic progression, the middle one is the average of the other two (and the average of all three).
So, the middle number is the sum, divided by {{{3}}}: {{{24/3=8}}}.
{{{d}}} = common difference of the A.P.
The three numbers in the A.P., in order, are:
{{{8-d}}}, {{{8}}}, and {{{8+d}}}
If the first is decreased by 1 , the second is decreased by 2 and third is not changed, then the resulting numbers are:
{{{8-d-1=highlight(7-d)}}}, {{{8-2=highlight(6)}}}, and {{{highlight(8+d)}}}
If those numbers are in a geometric progression, the common ratio,
calculated from the first two, or from the last two is the same, so
{{{6/(7-d)=(8+d)/6}}}
From that equation, I can solve for {{{d}}}.
{{{6/(7-d)=(8+d)/6}}} --> {{{(7-d)(8+d)=6*6}}} (equating the cross-products or multiplying both sides times {{{6(7-d)}}}, whichever you prefer as an explanation)
{{{(7-d)(8+d)=6*6}}} --> {{{56-d-d^2=36}}} --> {{{56-d-d^2-36=36-36}}} --> {{{30-d-d^2=0}}}
Rearranging, and multiplying both sides times {{{(-1)}}}, I get the equation in a form I like better:
{{{30-d-d^2=0}}} --> {{{-d^2-d+30=0}}} --> {{{(-1)(-d^2-d+30)=(-1)(0)}}}  --> {{{highlight(d^2+d-30=0)}}}
Now, I can solve that quadratic equation by factoring:
{{{d^2+d-30=0}}} --> {{{(d+6)(d-5)=0}}} --> {{{d=-6}}} or {{{d=5)))
 
With {{{d=5}}}, the 3 numbers are 3, 8, and 13:
{{{8-d=8-5}}} --> {{{8-d=highlight(3)}}}, {{{highlight(8)}}}, and {{{8+d=8+5}}} --> {{{8+d=highlight(13)}}}
 
With {{{d=-6}}}, the 3 numbers are 14, 8, and 2:
{{{8-d=8-(-6)}}} --> {{{8-d=8+6}}} --> {{{8-d=highlight(14)}}}, {{{highlight(8)}}}, and {{{8+d=8+(-6)}}} --> {{{8+d=8-6}}} --> {{{8+d=highlight(2)}}}