## Task B-4.7.

Let $a, b$ and $c$ be consecutive terms of a geometric sequence. If $a+b+c=\frac{7}{3}$ and $a^{2}+b^{2}+c^{2}=$ $\frac{91}{9}$, determine $a, b$ and $c$.
## First Solution.

If $a, b, c$ are consecutive terms of a geometric sequence, then $b^{2}=a c .(*)$

The equation $a+b+c=\frac{7}{3}$ can be written as $a+c=\frac{7}{3}-b$ and squared. Then we have

$(a+c)^{2}=\left(\frac{7}{3}-b\right)^{2}$, which is

$a^{2}+c^{2}+2 a c=\frac{49}{9}-\frac{14}{3} b+b^{2}$.

From the given equation $a^{2}+b^{2}+c^{2}=\frac{91}{9}$, express $a^{2}+c^{2}=\frac{91}{9}-b^{2}$

and substitute it into the last equation. Also, apply the equation $(*)$.

This gives:

$\frac{91}{9}-b^{2}+2 b^{2}=\frac{49}{9}-\frac{14}{3} b+b^{2}$, which simplifies to

$\frac{14}{3} b=\frac{49}{9}-\frac{91}{9}=-\frac{42}{9}$, so

$b=-1$.

It remains to solve the system

$a+c=\frac{10}{3}$

$a c=1$

The system reduces to the quadratic equation $3 a^{2}-10 a+3=0$ (or $3 c^{2}-10 c+3=0$) \quad 1 point whose solutions are $a=3$ and $a=\frac{1}{3}$, and thus the solutions to the system are $\left(3, \frac{1}{3}\right),\left(\frac{1}{3}, 3\right)$.

1 point

Finally, $a=3, b=-1, c=\frac{1}{3}$ or $a=\frac{1}{3}, b=-1, c=3$.

1 point