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.

$\displaystyle a-d+a+d=4-(a-d)-(a+d)$

$\displaystyle 2a=4-2a$

$\displaystyle 4a=4$

$\displaystyle a=1$