The problem asks us to find the sum of the first $n$ terms ($S_n$) for each given sequence. (a) $1, 3, 5, 7, 9, 11$ (b) $4, 2, 0, -2, -4$

1. Problem Description

The problem asks us to find the sum of the first $n$ terms ($S_n$) for each given sequence.

(a) $1, 3, 5, 7, 9, 11$

(b) $4, 2, 0, -2, -4$

2. Solution Steps

(a) The sequence $1, 3, 5, 7, 9, 11$ is an arithmetic progression with first term $a = 1$ and common difference $d = 2$. The $n$th term is given by $a_n = a + (n-1)d = 1 + (n-1)2 = 2n - 1$.

The sum of the first $n$ terms of an arithmetic progression is given by:

$S_n = \frac{n}{2}(a_1 + a_n)$

Substituting $a_1 = 1$ and $a_n = 2n - 1$, we get:

$S_n = \frac{n}{2}(1 + 2n - 1) = \frac{n}{2}(2n) = n^2$

(b) The sequence $4, 2, 0, -2, -4$ is an arithmetic progression with first term $a = 4$ and common difference $d = -2$. The $n$th term is given by $a_n = a + (n-1)d = 4 + (n-1)(-2) = 4 - 2n + 2 = 6 - 2n$.

The sum of the first $n$ terms of an arithmetic progression is given by:

$S_n = \frac{n}{2}(a_1 + a_n)$

Substituting $a_1 = 4$ and $a_n = 6 - 2n$, we get:

$S_n = \frac{n}{2}(4 + 6 - 2n) = \frac{n}{2}(10 - 2n) = n(5 - n) = 5n - n^2$

3. Final Answer

(a) $S_n = n^2$

(b) $S_n = 5n - n^2$