Arithmetic series

**Coach** · April 6th 2010, 09:27 AM

Hello all!

Once again, I am stuck on a problem, and no matter how I twist it, I always end up in equations that have no solutions.

Question:
The sum of three consecutive members of an arithmetic progression is 2, and the sum of the squares of these members is $\frac{14}{9}$ .

My solution(done with the help of the book's hints)
Let $a$ be the "middle" number of the three numbers.
$a-d+a+d=4-(a-d)-(a+d)$
$2a=4-2a$
$4a=4$
$a=1$
Then
$(a-d)^2+a^2+(a+d)^2=a^2-2ad+d^2+a^2+a^2+2ad+d^2$
$3a^2+2d^2=\frac{14}{9}$

ans if we substitute $a=1$ , we get

$2d^2=\frac{14}{9}-3$
$2d^2=-\frac{7}{9}$

But then the equation has no solutions.

All help is appreciated. Anyway, why is it so that some equations do not end up having solutions?

**Deadstar** · April 6th 2010, 10:25 AM

Originally Posted by Coach

Hello all!

Once again, I am stuck on a problem, and no matter how I twist it, I always end up in equations that have no solutions.

Question:
The sum of three consecutive members of an arithmetic progression is 2, and the sum of the squares of these members is $\frac{14}{9}$ .

My solution(done with the help of the book's hints)
Let $a$ be the "middle" number of the three numbers.
$a-d+a+d=4-(a-d)-(a+d)$
$2a=4-2a$
$4a=4$
$a=1$

where did you get the 4 from?

If you have 3 successive terms with difference d it should go...

$(a-d) + a + (a+d) = 2$

=> $a + a + a - d + d = 2$

=> $3a = 2$

=> $a = \frac{2}{3}$

And then,

$(a-d)^2 + a^2 + (a+d)^2 = \frac{14}{9}$

=> $\left ( \frac{2}{3} - d \right )^2 + \frac{4}{9} + \left ( \frac{2}{3} + d \right )^2 = \frac{14}{9}$

=> $\frac{4}{3} + 2d^2 = \frac{14}{9}$

=> $d^2 = \frac{1}{9}$

=> $d = \frac{1}{3}$

So your sequence is $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{3}{3}$ .

Add them up, it comes to 2.

Add the squares up...

$\frac{1}{9}+\frac{4}{9}+\frac{9}{9} = \frac{14}{9}$

**Coach** · April 6th 2010, 11:24 AM

Thank you so much!

I thought that the average of the first and the second term equals the "middle term". And hence $\frac{(a-d)+(a+d)}{2}=2$ and $(a-d)+(a+d)=4$ ,
but now I realized where I made a mistake.
Thanks a lot!

Originally Posted by Deadstar

where did you get the 4 from?

If you have 3 successive terms with difference d it should go...

$(a-d) + a + (a+d) = 2$

=> $a + a + a - d + d = 2$

=> $3a = 2$

=> $a = \frac{2}{3}$

And then,

$(a-d)^2 + a^2 + (a+d)^2 = \frac{14}{9}$

=> $\left ( \frac{2}{3} - d \right )^2 + \frac{4}{9} + \left ( \frac{2}{3} + d \right )^2 = \frac{14}{9}$

=> $\frac{4}{3} + 2d^2 = \frac{14}{9}$

=> $d^2 = \frac{1}{9}$

=> $d = \frac{1}{3}$

So your sequence is $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{3}{3}$ .

Add them up, it comes to 2.

Add the squares up...

$\frac{1}{9}+\frac{4}{9}+\frac{9}{9} = \frac{14}{9}$

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