Certainly! Let's consider an arithmetic sequence. In an arithmetic sequence, the difference between consecutive terms is constant. Let’s denote:
- The first term of the sequence by \( a \)
- The common difference by \( d \)
The four consecutive terms in the arithmetic sequence can be represented as follows:
1. The first term: \( a \)
2. The second term: \( a + d \)
3. The third term: \( a + 2d \)
4. The fourth term: \( a + 3d \)
We need to prove that the sum of the two terms on the ends (the first and fourth terms) is equal to the sum of the two terms in the middle (the second and third terms).
Step-by-Step Solution:
1. First Term and Fourth Term:
- The first term is \( a \).
- The fourth term is \( a + 3d \).
- Summing these two terms gives:
[tex]\[
a + (a + 3d) = a + a + 3d = 2a + 3d
\][/tex]
2. Second Term and Third Term:
- The second term is \( a + d \).
- The third term is \( a + 2d \).
- Summing these two terms gives:
[tex]\[
(a + d) + (a + 2d) = a + d + a + 2d = 2a + 3d
\][/tex]
3. Comparison:
- The sum of the first and fourth terms is \( 2a + 3d \).
- The sum of the second and third terms is also \( 2a + 3d \).
Since both sums are identical, we have shown that:
[tex]\[
a + (a + 3d) = (a + d) + (a + 2d)
\][/tex]
Thus, for any four consecutive terms of an arithmetic sequence, the sum of the two terms on the ends is equal to the sum of the two terms in the middle.
