An algebra problem by edil tizon

It is given that 2,4,x form a geometric progression.Therefore we get the value of x as 8.It is also given that 3,x,y form arithmetic progression.Since we already know the value of x we find that the value of y is 13.

By the mean property of the geometric progression, we have 2x=42, or x=8. By the mean property of the arithmetic progression, we have 3+y=2x=2.8=16, so y=16−3=13.

If 2, 4, and $x$ form a geometric series, then we can use the simple and common function: $f(x)=2^x$

If we plug in 3 (since $x$ is the 3^{\text{rd} term), we get $2^{3}=8$ . Therefore, $x=8$ .

We can use this to calculate $y$ , since we know $3, 8, y$ is an arithmetic series (linear):

$8-3=5$ $f(x)=3+5(x-1)$ $f(3)=3+5(3-1)=3+10=\boxed{13}$