Ink on Sequences #1

$\boxed{1}.$

Let $b=a+d$ and $c=a+2d$ .

We see that $a^2+(a+d)^2=(a+2d)^2$ .

Solving the equation yields $a=3d$ .

$\therefore\boxed{\text{Possible}}$

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$\boxed{2}.$

Let $b=ar$ and $c=ar^2$ .

We see that $a^2+a^2r^2=a^2r^4$ and $r>1$ .

$r^4-r^2-1=0$

$r^2=\dfrac{1+\sqrt{5}}{2}>1$

$\therefore\boxed{\text{Possible}}$

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$\boxed{4}.$

From $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$ , we see that

$b=\dfrac{2ac}{a+c}$ .

Then $a^2+\left(\dfrac{2ac}{a+c}\right)^2=c^2$ .

$a^4+2a^3c+4a^2c^2-2ac^3-c^4=0$ . Divide both sides by $a^2c^2$ .

$\left(\dfrac{a^2}{c^2}-\dfrac{c^2}{a^2}\right)+2\left(\dfrac{a}{c}-\dfrac{c}{a}\right)+4=0$

Let $\dfrac{a}{c}-\dfrac{c}{a}=t$ and we get $t<0$ since $a-c<0$ .

$\dfrac{a^2}{c^2}+\dfrac{c^2}{a^2}=t^2+2$

$\dfrac{a}{c}+\dfrac{c}{a}=\sqrt{t^2+4}$ .

Therefore, $t\sqrt{t^2+4}+2t+4=0$ .

$t^2(t^2+4)=4t^2+16t+16$

$t^4-16t-16=0$ .

Let $f(t)=t^4-16t-16$ and we see that $f''(t)=12t^2\geq0$ .

So $y=f(t)$ is convex downward within all intervals.

Note that $f(0)=-16<0$ .

Therefore $f(t)=0$ has one negative root and one positive root.

Then we can see that $\dfrac{a}{c}-\dfrac{c}{a}=t$ has real roots.

$\therefore\boxed{\text{Possible}}$

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$\boxed{8}.$

We know that $c=a+b$ and

$a^2+b^2=(a+b)^2$ , which yields $ab=0$ .

This doesn't satisfy the condition of the problem where side lengths of $a$ , $b$ and $c$ are supposed to form a triangle.

$\therefore\boxed{\text{Impossible}}$

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Therefore, the answer is $1+2+4=\boxed{7}$ .