Let <a_n> be an A.P. of natural numbers with common difference l such that  a₁ + a₂ + a₃ + a₄ = 18 and a₁a₂a₃a₄ + 4 = 361 . Then max {a₁, a₂, a₃, a₄} is equal to

The correct answer is 6.

Let the four terms of the Arithmetic Progression (A.P.) of natural numbers be:
$a_1$, $a_2=a_1+l$, $a_3=a_1+2l$, and $a_4=a_1+3l$
where $l$ is the common difference, and both $a_1$ and $l$ are natural numbers ($a_1,l\in \mathbb{N}$).

We are given the sum of the four terms:
$a_1+a_2+a_3+a_4=18$

Substituting the terms in terms of $a_1$ and $l$:
$a_1+(a_1+l)+(a_1+2l)+(a_1+3l)=18$
$4a_1+6l=18$
Dividing the entire equation by 2:
$2a_1+3l=9$

Since $a_1$ and $l$ must be natural numbers (positive integers starting from 1):
If $l=1$:
$2a_1+3\left(1\right)=9\Rightarrow 2a_1=6\Rightarrow a_1=3$ (which is a natural number).
If $l=2$:
$2a_1+3\left(2\right)=9\Rightarrow 2a_1=3$ (no natural number solution for $a_1$).
If $l\ge 3$:
$2a_1=9-3l\le 0$ (no positive integer solution for $a_1$).

Therefore, the unique solution is $a_1=3$ and $l=1$.

Thus, the four terms of the A.P. are:
$a_1=3$
$a_2=4$
$a_3=5$
$a_4=6$

Let us verify the second condition:
Note that the standard mathematical relation for four terms in A.P. is:
$a_1{a}_2{a}_3{a}_4+l^4=k^2$ (a perfect square)
With $l=1$, the equation becomes $a_1{a}_2{a}_3{a}_4+1^4=361$ (since $l^4$ was typographically represented as 4 in the question text):
$3\times 4\times 5\times 6+1=360+1=361$
This perfectly satisfies the condition.

The maximum of the terms is:
$max\{a_1,a_2,a_3,a_4\}=a_4=6$