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 $\displaystyle \frac{14}{9}$.

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

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

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

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 $\displaystyle \frac{14}{9}$.

My solution(done with the help of the book's hints)
Let $\displaystyle a$ be the "middle" number of the three numbers.
$\displaystyle a-d+a+d=4-(a-d)-(a+d)$
$\displaystyle 2a=4-2a$
$\displaystyle 4a=4$
$\displaystyle a=1$
where did you get the 4 from?

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

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

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

=> $\displaystyle 3a = 2$

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

And then,

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

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

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

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

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

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

Add them up, it comes to 2.

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

3. Thank you so much!

I thought that the average of the first and the second term equals the "middle term". And hence $\displaystyle \frac{(a-d)+(a+d)}{2}=2$ and $\displaystyle (a-d)+(a+d)=4$,
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...

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

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

=> $\displaystyle 3a = 2$

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

And then,

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

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

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

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

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

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

Add them up, it comes to 2.

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