If $\left(3-x,\:x,\:\sqrt{9-x}\right)$ is an arithmetic sequence, find its sixth term.

Question

To me it seems that this isn't an arithmetic sequence, because of the third term. I found that X is 5, but if that's the case the difference between the terms is not constant. How can I solve this?

Answer 1

Observe that if $(3-x,x,\sqrt{9-x})$ is an arithmetic progression then

$$d=x-(3-x)=\sqrt{9-x}-x$$

Solve this equation to obtain $x=\frac{17}{9}$.

Finally, observe that $d=\frac{7}{9}$ and so $a_6=a_1+5d=3-x+5d=\frac{10}{9}+5\cdot\frac{7}{9}=5$

Edit:

From a comment below, if we allow the value of the square root to be the negative value we can obtain $x=0$ and $d=-3$ so $a_6=3+5(-3)=-12$

Answer 2

To be an arithmetic sequence, we need $x-(3-x) = \sqrt{9-x}-x $ or $3x-3 =\sqrt{9-x} $ or $9x^2-18x+9 = 9-x $ or $9x^2-17x = 0$.

This gives $x = 0$ or $x = 17/9$.

For $x=0$, this gives $3, 0, \sqrt{9}$, so we have to take the negative square root.

For $x=17/9$, this gives $3-17/9 = 10/9, 17/9, \sqrt{9-17/9} =\sqrt{64/9} =8/3 =24/9$ which works.

Now get the sixth term.

Answer 3

To be an arithmetic sequence the differences in terms must be constant.

So $x - (3-x) = 2x -3$ must equal $\sqrt{9-x} - x$

So $\sqrt{9-x} - x$ must equal $2x -3$.

So... just solve that.... It's a straight forward quadratic.

"I found that $X$ is $5$". I'm utterly perplexed as to how you found that.

"but if that's the case the difference between the terms is not constant" But keeping the difference constant is the ONLY criterion we know of for the value of $x$. If $x =5$ doesn't keep the difference constant I utterly fail to see how on earth you could have derived $x = 5$ in the first place.

$\sqrt{9-x} - x = 2x - 3; x \le 9$

$\sqrt{9-x} = 3x -3 \ge 0; x\le 9$

$9-x = 9(x^2 - 2x +1); 1 \le x \le 9$.

$9x^2 - 17x = 0; 1\le x \le 9$

$9x =17; 1\le x \le 9$

$x = \frac {17}9$.

And that is a an arithmethic series.

$3 - x = \frac {10}9$ and $x = \frac {17}9$ and $\sqrt {9 -\frac{17}9}= \sqrt{\frac {94}9} = \frac 83$.

And the difference is constant:

$\frac {17}9 - \frac {10}9 = \frac 79$. And $\frac 83 - \frac {17}9 = \frac 79$.

Not a problem.

But $x$ must be $\frac {17}9$.