Question 714016
If x is the value of the first term, and d is the difference between terms, then the first 6 terms of the arithmetic series are:
 
{{{x}}}, {{{x+d}}}, {{{x+2d}}}, {{{x+3d}}}, {{{x+4d}}}, {{{x+5d}}}
 
The sum of these 6 terms is 12, so we have:
 
{{{6x + 15d = 12}}}
 
The 4rd, 7th, and 16th terms, a geometric series, are 
 
{{{x+3d}}}, {{{x+6d}}}, and {{{x+15d}}}
 
So we know that
 
{{{(x+6d)/(x+3d) = (x+15d)/(x+6d)}}}
 
Multiplying both sides by {{{(x+3d)*(x+6d)}}}
 
{{{(x+6d)*(x+6d) = (x+15d)*(x+3d)}}}
 
{{{x^2 + 12xd + 36d^2 = x^2 + 18xd + 45d^2}}}
 
{{{18xd - 12xd = 36d^2 - 45d^2}}}
 
{{{6xd = -9d^2}}}
 
{{{x = (-3/2)*d}}} where {{{d<>0}}} (If {{{d=0}}} we have the trivial sequence {{{2}}}, {{{2}}}, {{{2}}}, {{{2}}}, {{{2}}}, ...)
 
Plugging {{{x = (-3/2)d}}} into {{{6x + 15d = 12}}}
 
{{{6*(-3/2)d + 15d = 12}}}
 
{{{-9d + 15d = 12}}}
 
{{{6d = 12}}}
 
{{{d = 2}}}
 
And plugging {{{d = 2}}} into {{{6x + 15d = 12}}}
 
{{{6x + 15(2) = 12}}}
 
{{{6x + 30 = 12}}}
 
{{{6x = -18}}}
 
{{{x = -3}}}
 
So our series is:
 
{{{-3}}}, {{{-1}}}, {{{1}}}, {{{3}}}, {{{5}}}, {{{7}}}, ...
 
The 4th, 7th, and 16th terms are {{{3}}}, {{{9}}}, and {{{27}}}
 
So the difference in the arithmetic sequence is 2 and the ratio in the geometric sequence is 3.