I would need help with example: The three numbers are consecutive terms of arithmetic progression and the sum of their second powers is 126. The first number 3 times smaller, the second number with no change and the third number 4 times greater, are the 3 consequtive terms of geometric progression. What are the values of these terms?

Question

I can not move with it. I would be grateful for any help.
Thank you.

from arithmetic:
a1
a2=a1+d
a3=a1+2d
a1²+(a1+d)²+(a1+2d)²=126
3a1²+6a1d+6d²=126

from geometric:
a1/3
a1+d
4a1+8d

4a1+8d/a1+d=3a1+3d/a1
a1²+2a1d-3d²=0

How should I continue? I can not divide them so what is the next step?
Thank you.

Answer 1

first of all, let's make your definitions less confusing.
Since a is the standard for first term
let your AS terms be
a , a+d, and a+2d

then a^2 + (a+d)^2 + (a+2d)^2 = 126
a^2 + a^2 + 2ad + d^2 + a^2 + 4ad + 4d^2 = 126
3a^2 + 6ad + 5d^2 = 126 notice you had 6d^2 , #1

new terms:
a---> a/3
a+d ---> a+d
a+2d --> 4a + 8d

then (a+d)/(a/3) = (4a+8d)/(a+d)
etc.
a^2 + 2ad - 3d^2 = 0 , #2 , you had that also, great!

I agree that division will not work because of the zero, also it would be very difficult to factor out either the a or the d

how about elimination, (we are going to be lucky here)

#1 as is : 3a^2 + 6ad + 5d^2 = 126
#2 times 3: 3a^2 + 6ad - 9d^2 = 0
subtract them:
14d^2 = 126
d^2 = 9
d = ± 3

if d = 3 , in #2
a^2 + 6a - 27 = 0
(a+9)(a-3) = 0
a = -9 or a = 3

if d = -3 ........

I know you can carry on.

NOTE: if our a^2 and ad's had not canceled, we would have a very very nasty system of equations.
I will make one little change in our equations in Wolfram

ours:
http://www.wolframalpha.com/input/?i=solve+x%5E2+%2B+2xy+-+3y%5E2+%3D+0+%2C+3x%5E2+%2B+6xy+%2B+5y%5E2+%3D+126

small change: 6xy to 5xy
http://www.wolframalpha.com/input/?i=solve+x%5E2+%2B+2xy+-+3y%5E2+%3D+0+%2C+3x%5E2+%2B+5xy+%2B+5y%5E2+%3D+126

Answer 2

m-7
2time the sumof9 and f
m divided by 5