There are four numbers, the first three numbers are equal proportion series, the last three numbers are equal difference series, the sum of the first and last two terms is 21, and the middle two terms If it is a / Q, a, AQ, a (2q-1), how to solve it,

There are four numbers, the first three numbers are equal proportion series, the last three numbers are equal difference series, the sum of the first and last two terms is 21, and the middle two terms If it is a / Q, a, AQ, a (2q-1), how to solve it,

There are four numbers. The first three numbers form an equal ratio sequence, and the last three numbers form an equal ratio difference sequence. The sum of the first and last two terms is 21, and the sum of the middle two terms is 18
Is that the topic
Let these four numbers be a / Q, a, AQ, a (2q-1), then
a/q+a(2q-1)=21
a+a（2q²-q）=21q
a（1+2q²-q）=21q
a=21q/（1+2q²-q） （1）
a+aq=18
a（1+q）=18 （2）
Substituting (1) into (2) leads to
21q/（1+2q²-q）*（1+q）=18
21q+21q²=18+36q²-18q
15q²-39q+18=0
5q²-13q+6=0
（5q-3）（q-2）=0
The solution is q = 3 / 5 or q = 2
Substituting (2) into the solution, we get a = 45 / 4 or a = 6
So these four numbers are 75 / 4 45 / 4 27 / 4 9 / 4, or 3 6 12 18