Croatian Mathematical Society Competitions

Let the first term of the sequence be $a$.

The second term is $a+d$. The fourth term is $a+3d$.

Their product is $(a+d)(a+3d)=-d^2$.

We are to find the product of the third and fifth terms:

Third term: $a+2d$ Fifth term: $a+4d$

Their product is $(a+2d)(a+4d)$.

Let us expand $(a+d)(a+3d)$:

$(a+d)(a+3d)=a^2+3ad+ad+3d^2=a^2+4ad+3d^2=-d^2$

So: $a^2+4ad+3d^2=-d^2$ $a^2+4ad+4d^2=0$

Now, $(a+2d)(a+4d)=a^2+4ad+2ad+8d^2=a^2+6ad+8d^2$

From above, $a^2+4ad+4d^2=0$, so $a^2+4ad=-4d^2$

Therefore: $(a+2d)(a+4d)=(a^2+4ad)+2ad+8d^2=(-4d^2)+2ad+8d^2=2ad+4d^2$

But $a^2+4ad+4d^2=0$

Let us solve for $a$: $a^2+4ad+4d^2=0$ $a^2+4ad=-4d^2$

So $a^2+4ad=-4d^2$

Now, $2ad+4d^2=2ad+4d^2$

But we can express $a$ in terms of $d$: $a^2+4ad+4d^2=0$ $(a+2d{)}^2=0$ So $a+2d=0$ Thus, $a=-2d$

Now, third term: $a+2d=-2d+2d=0$ Fifth term: $a+4d=-2d+4d=2d$

Their product: $0\times 2d=0$

Answer: $0$