a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11
You need this formula for the nth term:
an = a1 + (n-1)d
[Some books and teachers use "t" for "term" instead of "a".
I'll use "a".]
The 11th term of an arithmetic sequence is 57
a11 = a1 + (11-1)d
57 = a1 + 10d
and the sum of the first and the fourth terms is 29.
a1 + a4 = 29
Substitute n=4 in the formula also
a4 = a1 + (4-1)d
a4 = a1 + 3d
Now substitute for a4 in
a1 + a4 = 29
a1 + (a1 + 3d) = 29
a1 + a1 + 3d = 29
2a1 + 3d = 29
So you have this system of equations:
57 = a1 + 10d
2a1 + 3d = 29
Solve the first for a1
57-10d = a1
Substitute in
2a1 + 3d = 29
2(57-10d) + 3d = 29
114 - 20d + 3d = 29
114 - 17d = 29
-17d = -85
d =
d = 5
Substitute d=5 in
57-10d = a1
57-10(5) = a1
57-50 = a1
7 = a1
So the sequence starts with a1=7 and adds d=5 each time.
To check, let's write them all out and see if the llth terms comes out
to be 57.
7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57
The last one came out 57. Count them. There are 11. So it's correct.
Determine the first 3 terms of the sequence.
7, 12, 17
Edwin