Question 258681
Let A be your starting number, d be the difference for the arithmetic progression.
{{{A+(A+d)+(A+d+d)=21}}}
{{{3A+3d=21}}}
{{{A+d=7}}}
Then from the geometric progression information,
First:{{{A+1}}}
Second:{{{A+d+5}}}
Third:{{{A+d+d+15=A+2d+15}}}
Since they are in geometric progression, each successive pair equals the common ratio.
{{{(A+d+5)/(A+1)=(A+2d+15)/(A+d+5)}}}
From eq. 1, you know,
{{{A+d=7}}}
Substitute,
{{{(A+d+5)/(A+1)=(A+2d+15)/(A+d+5)}}}
{{{(7+5)/(A+1)=(7+d+15)/(7+5)}}}
{{{12/(A+1)=(22+d)/12}}}
{{{(A+1)(22+d)=144}}}
Again using eq. 1,
{{{A+d=7}}}
{{{d=7-A}}}
{{{(A+1)(22+d)=144}}}
{{{(A+1)(22+7-A)=144}}}
{{{(A+1)(29-A)=144}}}
{{{29A-A^2+29-A=144}}}
{{{-A^2+28A+29=144}}}
{{{-A^2+28A-115=0}}}
{{{A^2-28A+115=0}}}
You can factor, 
{{{(A-23)(A-5)=0}}}
Two solutions:
{{{A-23=0}}}
{{{A=23}}}
{{{d=7-A}}}
{{{d=7-23=-16}}}
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{{{A-5=0}}}
{{{A=5}}}
{{{d=7-A}}}
{{{d=7-5=2}}}
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I leave it to you to verify the solutions for both.