Compose two progressions: an arithmetic and a geometric one, each consisting of four terms; such that if

you add the corresponding terms of both progressions, the resulting numbers should be: 27, 27, 39, 87.
Let $a, a+d, a+2 d, a+3 d$ be the desired arithmetic progression, and $b, b q, b q^{2}, b q^{3}$ be the desired geometric progression. According to the problem,

$a+b=27, a+d+b q=27, a+2 d+b q^{2}=39, a+3 d+b q^{3}=87$

Subtract the first equation from the second, the second from the third, and the third from the fourth: $d+b(q-1)=0, d+b q(q-1)=$ $12, d+b q^{2}(q-1)=48$

From the first equation, we get $b(q-1)=-d$; substitute this expression into the second and third equations: $d-d q$ $=12, d-d q^{2}=48$

Dividing the last equation by the second-to-last equation, we get $1+q=4$, so $q=3$. Therefore, $d=-6, b$ $=3$ and $a=24$.

## Answer

$24,18,12,6$ and $3,9,27,81$.