Question 1208641
<br>
Let the four terms of the AP be a, a+d, a+2d, and a+3d.<br>
Then the four terms of the GP are a+1, a+d+1, a+2d+3, and a+3d+9.<br>
In a GP, the ratio of consecutive terms is a constant.<br>
first, second, and third terms...<br>
{{{(a+d+1)/(a+1)=(a+2d+3)/(a+d+1)}}}<br>
{{{(a+d+1)(a+d+1)=(a+1)(a+2d+3)}}}
{{{a^2+ad+a+ad+d^2+d+a+d+1=a^2+2ad+3a+a+2d+3}}}
{{{d^2+2a+1=4a+3}}}
{{{d^2=2a+2}}} [1]
second, third, and fourth terms...<br>
{{{(a+3d+9)/(a+2d+3)=(a+2d+3)/(a+d+1)}}}
{{{a^2+ad+a+3ad+3d^2+3d+9a+9d+9=a^2+2ad+3a+2ad+4d^2+6d+3a+6d+9}}}
{{{4a=d^2}}} [2]<br>
Solve [1] and [2] simultaneously<br>
{{{2a+2=4a}}}
{{{2a=2}}}
{{{a=1}}}
{{{d^2=4a=4}}}
{{{d=2}}}<br>
The AP has first term 1 and common difference 2.<br>
ANSWER: 1, 3, 5, 7<br>
CHECK: The GP is 1+1=2, 3+1=4, 5+3=8, and 7+9=16<br>