The problem asks to find $u_{50}$ and $S_{50}$ for an arithmetic sequence given that $u_2 = 6$ and $u_{10} = 22$. Here, $u_n$ represents the nth term and $S_n$ represents the sum of the first n terms of the arithmetic sequence.

1. Problem Description

The problem asks to find $u_{50}$ and $S_{50}$ for an arithmetic sequence given that $u_2 = 6$ and $u_{10} = 22$. Here, $u_n$ represents the nth term and $S_n$ represents the sum of the first n terms of the arithmetic sequence.

2. Solution Steps

First, recall the formula for the nth term of an arithmetic sequence:

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

We are given that $u_2 = 6$ and $u_{10} = 22$. Therefore:

$u_2 = a + (2-1)d = a + d = 6$ (1)

$u_{10} = a + (10-1)d = a + 9d = 22$ (2)

Subtract equation (1) from equation (2):

$(a + 9d) - (a + d) = 22 - 6$

$8d = 16$

$d = 2$

Substitute $d=2$ into equation (1):

$a + 2 = 6$

$a = 4$

Now that we know $a=4$ and $d=2$, we can find $u_{50}$:

$u_{50} = a + (50-1)d = 4 + (49)(2) = 4 + 98 = 102$

Next, we need to find the sum of the first 50 terms, $S_{50}$. Recall the formula for the sum of the first n terms of an arithmetic sequence:

$S_n = \frac{n}{2}(a + u_n)$

Therefore,

$S_{50} = \frac{50}{2}(a + u_{50}) = \frac{50}{2}(4 + 102) = 25(106) = 2650$

3. Final Answer

$u_{50} = 102$

$S_{50} = 2650$