Progressions 1

Algebra Level 4

Three distinct numbers $x , y , z$ form a geometric progression in that order, and
the numbers $x + y , y + z , z + x$ form an arithmetic progression in that order.

Given that the common ratio of the geometric progression has two distinct possible values $A$ and $B$ , find $A+B$ .

1 solution

Hung Woei Neoh
Jun 1, 2016

Given a geometric progression $x,y,z$

We let the first term be $a$ and the common ratio be $r$

$x=a,\;y=ar,z=ar^2$

Now, given that $x+y,y+z,z+x$ form an arithmetic progression. The common difference,

$d= (y+z) - (x+y) = (z+x) - (y+z)\\ \implies z-x = x-y$

Substitute $x,y,z$ in terms of $a$ and $r$ :

$ar^2 - a = a-ar$

Geometric progressions cannot have a term $0$ , therefore $a \neq 0$ . We divide the equation by $a$ :

$r^2 - 1 = 1 - r\\ r^2 + r - 2 = 0\\ (r+2)(r-1) = 0\\ r=-2,\;1$

Therefore, $A=-2,\; B = 1,\; A+B = -2+1 = \boxed{-1}$

Very nice solution , thanks for solving it(+1)!

Rishabh Tiwari - 5 years ago