# Arithemtic and geometric progession

1. May 28, 2013

### Government\$

1. The problem statement, all variables and given/known data
Numbers a,b,c are consecutive members of increasing arithmetic progression, and numbers a,b,c+1 are consecutive members of geometric progression. If a+b+c=18 then a^2 +b^2 + c^2=?

3. The attempt at a solution
$a + b + c= 18$

$a + a +d +a + 2d = 18$

$3a + 3d = 18$

$3(a+d)= 18$

$a+d=6=b$

$a + b + c + 1 = 19$

$a + aq + aq^{2} = 19$

$aq=b=6 \rightarrow q=\frac{6}{a}$

$a + aq + aq^{2} = 19 \rightarrow a +6 + \frac{36}{a} = 19$

$a +6 + \frac{36}{a} = 19 /*a$

$a^{2} +6a + 36= 19a \rightarrow a^{2} -13a + 36= 0$

$\frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm 5}{2}$

$a_{1}=9$ and $a_{2}=4$

If $a_{1}=9$ then $d=-3$ but i have increasing arithemtic progression so i must take $a_{2}=4$.

So $a_{2}=4$ and $d=2$ since $a+d=6$.

Then $a=4$, $b=6$, $c=8$

$4^{2} + 6^{2} + 8^{2} = 16 + 36 + 64 = 116$ but my textbooks says that $133$ is solution. What's the problem here?

Last edited: May 28, 2013
2. May 28, 2013

### haruspex

Clearly your solution satisfies all the conditions. If you've quoted the problem correctly then it's an error in the book.

3. May 28, 2013

### butan1ol

the author might have accidentally plug in ( c + 1 ) as c in a^2+b^2+c^2...
4^2 + 6^2 + 9^2 yields 133