The 11th term of an arithmetic sequence is 57 and the sum of the first and fourth terms is 29.

Answer 1

Let the first term of the arithmetic sequence be denoted by a and the common difference be denoted by d.

Using the formula for the nth term of an arithmetic sequence:

a11 = a + 10d = 57

Given that the sum of the first and fourth terms is 29:

a + a + 3d = 29
2a + 3d = 29

Now we have two equations:

a + 10d = 57
2a + 3d = 29

Solving these two equations simultaneously, we find:

a = 11 and d = 4

Therefore, the arithmetic sequence is: 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 57

So, the 11th term of the arithmetic sequence is 57.