Arithmetic and geometric sequence

Question

Which two numbers should be placed between -5 and 49 so that the first three numbers form an arithmetic sequence, whereas the last three numbers form a geometric sequence?

Answer 1

These numbers are 1 and 7 : using arithmetic and geometric properties we find: $$ -5,a,b,49$$ $$ \\ \\ \\ \begin{cases} a=\frac { b-5 }{ 2 } \\ { b }^{ 2 }=49a \end{cases}\Rightarrow \begin{cases} b=2a+5 \\ \left( 2a+5 \right) ^{ 2 }-49a=0\Rightarrow \end{cases}4{ a }^{ 2 }-29a+25=0\Rightarrow a=1,\Rightarrow b=7$$

Answer 2

Hint Let $x, y$ be the numbers.

Since $-5,x,y$ is an arithmetic progression we have $$2x=y-5$$ Since $x,y,49$ is a geometric progression, we have $$y^2=49x=\frac{49}{2}(y-5)$$

This is a quadratic equation.

Answer 3

the last three numbers form a geometric sequence

Since $49$ is the square of a prime, its only two other factors are $1$ and $7$. Now all that's left to do is checking to see if $-5,1,7$ are in arithmetic progression. Indeed, $1-(-5)=7-1$.