We are given that $u_3 = 11$ and $u_{12} = 56$. The problem asks to find the value of $u_{45}$. We assume the sequence is arithmetic.

1. Problem Description

We are given that $u_3 = 11$ and $u_{12} = 56$. The problem asks to find the value of $u_{45}$. We assume the sequence is arithmetic.

2. Solution Steps

Let $u_n$ be an arithmetic sequence.

Then $u_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

We are given $u_3 = 11$ and $u_{12} = 56$. So

$u_3 = a + (3-1)d = a + 2d = 11$ (1)

$u_{12} = a + (12-1)d = a + 11d = 56$ (2)

Subtracting equation (1) from equation (2) gives:

$(a + 11d) - (a + 2d) = 56 - 11$

$9d = 45$

$d = \frac{45}{9} = 5$

Substituting $d = 5$ into equation (1):

$a + 2(5) = 11$

$a + 10 = 11$

$a = 11 - 10 = 1$

Thus, the arithmetic sequence is $u_n = 1 + (n-1)5$.

Now we want to find $u_{45}$.

$u_{45} = 1 + (45-1)5 = 1 + (44)5 = 1 + 220 = 221$.

3. Final Answer

221