How many terms of the AP 11, 17, 23, 29, ... add up to 700?

$a=11,d=6.S_n=\frac{n}{2}$$[22+6(n-1)]=n(3n+8)=3n^2+8n$. Setting $3n^2+8n-700=0$ gives $n=\frac{-8\pm \sqrt{64+8400}}{6}=\frac{-8\pm 92}{6}$, so $n=14$ or $n=-16.67$. The negative root is rejected because the number of terms must be a positive whole number, leaving $n=14$ alone$.(S_{13}=611$ and $S_{15}=795$, so the neighbours 13 and 15 are both wrong.)