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?

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.

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

m-7

2time the sumof9 and f
m divided by 5

To solve this problem, we have two equations:

Equation 1: 3a1² + 6a1d + 6d² = 126
Equation 2: a1² + 2a1d - 3d² = 0

We need to solve these two equations simultaneously to find the values of a1 and d.

To do this, we can try solving one equation for one variable and substituting it into the other equation. Let's solve Equation 2 for a1 in terms of d:

a1² + 2a1d - 3d² = 0

Factoring the left side:
(a1 + 3d)(a1 - d) = 0

This gives us two possibilities:
a1 + 3d = 0 or a1 - d = 0

Let's consider each case separately:

Case 1: a1 + 3d = 0

If a1 + 3d = 0, then a1 = -3d. Substituting this into Equation 1:

3(-3d)² + 6(-3d)d + 6d² = 126
27d² - 18d² + 6d² = 126
15d² = 126
d² = 126/15
d² = 8.4

Taking the square root of both sides:
d = ±√8.4

Since the terms of the arithmetic progression are usually integers, we can discard the negative value and take d = √8.4.

Now, substituting d = √8.4 into a1 + 3d = 0:
a1 + 3√8.4 = 0
a1 = -3√8.4

So in this case, the values of the terms would be a1 = -3√8.4 and d = √8.4.

Case 2: a1 - d = 0

If a1 - d = 0, then a1 = d. Substituting this into Equation 1:

3(d)² + 6(d)d + 6d² = 126
3d² + 6d² + 6d² = 126
15d² = 126
d² = 126/15
d² = 8.4

Taking the square root of both sides:
d = ±√8.4

Similarly, discarding the negative value, we take d = √8.4.

Now, substituting d = √8.4 into a1 - d = 0:
a1 - √8.4 = 0
a1 = √8.4

So in this case, the values of the terms would be a1 = √8.4 and d = √8.4.

Therefore, the possible values of the terms in the arithmetic progression are:
a1 = -3√8.4, a2 = -3√8.4 + √8.4, a3 = -3√8.4 + 2√8.4

or

a1 = √8.4, a2 = √8.4 + √8.4, a3 = √8.4 + 2√8.4

Both cases satisfy the conditions given in the problem statement.