If (a + b√n) is the positive square root of (29 −12√5), where a and bare integers, and n is a natural number, then the maximum possible value of (a + b +n)is

The correct option is 18.

We are given that $a+b\sqrt{n}$ is the positive square root of $29-12\sqrt{5}$:
${(a+b\sqrt{n})}^2=29-12\sqrt{5}$

We can express $29-12\sqrt{5}$ as a perfect square:
$29-12\sqrt{5}=20+9-2·\left(2\sqrt{5}\right)·\left(3\right)={(2\sqrt{5}-3)}^2$

Taking the positive square root, we get:
$a+b\sqrt{n}=2\sqrt{5}-3=-3+2\sqrt{5}$

Since $a,b$ are integers and $n$ is a natural number, we have two possibilities for representing the radical term:

1) Let $b\sqrt{n}=2\sqrt{5}$:
Here, $a=-3$, $b=2$, and $n=5$.
The sum is $a+b+n=-3+2+5=4$.

2) Let $b\sqrt{n}=1\sqrt{20}$:
Here, $a=-3$, $b=1$, and $n=20$.
The sum is $a+b+n=-3+1+20=18$.

Thus, the maximum possible value of $a+b+n$ is $18$.