Arithemtic and geometric progession

[Government$]

Homework Statement

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=?

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?

 

[haruspex]

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

 

[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