How many terms of the AP 20, 18, 16, 14, ... must be taken so that their sum is 110?

$a=20,d=-2$, so $S_n=\frac{n}{2}$$[40-2(n-1)]=n(21-n)$. Setting $n(21-n)=110$ gives $n^2-21n+110=0$, so $(n-10)(n-11)=0$ and $n=10$ or $n=11$. Both roots are positive whole numbers, so neither can be rejected. The reason both work is that $a_{11}=20+10(-2)=0$: adding the 11th term adds nothing, so $S_{10}=S_{11}=110$. This is the case where the usual "reject the impossible root" step does not apply.