# Arithmetic Progressions' Help Needed

• November 6th 2010, 12:05 AM
ashishgaurav
Arithmetic Progressions' Help Needed
I need help regarding the following question:

Four different integers are in A.P. One of these integers is equal to the squares of the rest integers. Find the numbers.

It is like this:
If a, b, c, d are in A.P. then d=a^2 +b^2 +c^2

I know the answer, but I simply can't figure out the explanation. The A.P. is -1,0,1,2

But I do not understand how it came.

Any help will be appreciated.
• November 6th 2010, 02:12 AM
earboth
Quote:

Originally Posted by ashishgaurav
I need help regarding the following question:

Four different integers are in A.P. One of these integers is equal to the squares of the rest integers. Find the numbers. <=== is this the complete text of the question?

It is like this:
If a, b, c, d are in A.P. then d=a^2 +b^2 +c^2

I know the answer, but I simply can't figure out the explanation. The A.P. is -1,0,1,2

But I do not understand how it came.

Any help will be appreciated.

Here is how I would do this problem:

$\begin{array}{lcl}a=a&~\implies &a^2 = a^2\\b=a+x&\implies & b^2=a^2+2ax+x^2\\c=a+2x &\implies & c^2 = a^2+4ax+4x^2\\d= a+3x & &\end{array}$

where x is the constant displacement between 2 consecutive numbers.

You have to solve the equation

$a+3x = a^2+(a^2+2ax+x^2)+(a^2+4ax+4x^2) \implies 5x^2+(6a-3)x+3a^2-a=0$

But in my opinion there is missing a condition which determines the relation between a and x. (Crying)
• November 6th 2010, 02:23 AM
Quote:

Originally Posted by ashishgaurav
I need help regarding the following question:

Four different integers are in A.P. One of these integers is equal to the squares of the rest integers. Find the numbers.

It is like this:
If a, b, c, d are in A.P. then d=a^2 +b^2 +c^2

I know the answer, but I simply can't figure out the explanation. The A.P. is -1,0,1,2

But I do not understand how it came.

Any help will be appreciated.

Did you mean sum of squares of the rest of the integers? Taking that to be what you meant, begin by writing relevant equations to the problem.

b-a=c-b
2b=a+c --- 1

d-c=c-b
2c=b+d
b=2c-d --- 2

Sub 2 into 1, 2(2c-d)=a+c
a=3c-2d -- 3

$a=b^2+c^2+d^2$ --- 4

Substitute 3 and 2 into 4

After simplifying, you are left with $5c^2+2d^2-3c+2d-4cd=0$

Then let c be 0(you could have picked d to be 0), then solve for d and the rest of the unknowns.