1. ## Arithmetic series

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?

2. 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.

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

3. 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!
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.

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