JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Sequences and Series

Let $a,b=ar,c=a{r}^2$ be in G.P., where $r$ is an integer. The arithmetic mean condition gives $\frac{a+ar+a{r}^2}{3}=ar+2$, so $a+a{r}^2=2ar+6$, giving $a(r^2-2r+1)=6$, i.e., $a(r-1{)}^2=6$. For $r$ to be an integer, we need $(r-1{)}^2=\frac{6}{a}$. Since $(r-1{)}^2$ is a perfect square, $a$ must divide 6 and $\frac{6}{a}$ must be a perfect square. Trying $a=6$ gives $(r-1{)}^2=1$, so $r=2$ (taking the positive value). Thus $a=6,b=12,c=24$. Now $\frac{a^2+a-14}{a+1}=\frac{36+6-14}{7}=\frac{28}{7}=4$.