The 11th tern of an arithmetic sequence is 57 and the sum of the first and fourth term is 29.Determine the first three term of the sequence
Let's denote the first term of the arithmetic sequence as 'a' and the common difference as 'd'.
We're given that the 11th term is 57, so we can use the formula for the nth term of an arithmetic sequence to write the equation:
a + 10d = 57 (1)
We're also given that the sum of the first and fourth term is 29.
The first term is 'a' and the fourth term can be written as 'a + 3d'. So, we can write the equation:
a + (a + 3d) = 29 (2)
Simplifying equation (2), we get:
2a + 3d = 29 (3)
Now, we have a system of equations with equation (1) and equation (3).
To solve this system, we can multiply equation (1) by 2 and subtract equation (3) from it to eliminate 'a':
2(a + 10d) - (2a + 3d) = 2(57) - 29
2a + 20d - 2a - 3d = 114 - 29
17d = 85
d = 85/17
d = 5
Substituting the value of 'd' back into equation (1), we can solve for 'a':
a + 10(5) = 57
a + 50 = 57
a = 7
Therefore, the first term of the arithmetic sequence is 7, the second term is 7+5 = 12, and the third term is 7+2(5) = 17.
So, the first three terms of the sequence are 7, 12, and 17.