Question

Let $a,b\in R$ be such that $a,a+2b,2a+b$ are in A.P and $(b+1{)}^2,ab+5,(a+1{)}^2$ are in G.P then $(a+b)$ equals

Answer 1

$a,a+2b,2a+b$ are in A.P
$\therefore 2(a+2b)=a+2a+b\Rightarrow a=3b...\left(i\right)$
and $(b+1{)}^2,ab+5,(a+1{)}^2$ are in G.P
$\therefore (ab+5{)}^2=(a+1{)}^2\cdot (b+1{)}^2$
$\Rightarrow (3b^2+5)=\pm (3b+1)(b+1)$
$\therefore 3b^2+5=3b^2+4b+1\Rightarrow b=1$
and $3b^2+5=-3b^2-4b-1\Rightarrow 3b^2+2b+3=0D<0$
So, $b=1,a=3$
$\therefore a+b=4$