An arithmetic sequence with first term 1 and common difference not equal to 0 has second, tenth, and thirty-fourth terms that form a geometric sequence.
What is the thirty-fourth term of the arithmetic sequence?
The fourth term in the geometric sequence above appears as the nth term in the arithmetic sequence. Find the value of n.
First term =1, Common difference =1/3, Number of terms = 34. Here are the terms:
(1, 1.333333333, 1.666666667, 2, 2.333333333, 2.666666667, 3, 3.333333333, 3.666666667, 4, 4.333333333, 4.666666667, 5, 5.333333333, 5.666666667, 6, 6.333333333, 6.666666667, 7, 7.333333333, 7.666666667, 8, 8.333333333, 8.666666667, 9, 9.333333333, 9.666666667, 10, 10.33333333, 10.66666667, 11, 11.33333333, 11.66666667, 12).
The 34th term = 12
The common ratio of the geometric sequence =3
Nth term=F + D*(N - 1), where F=First term, D=Common difference, N=Number of terms
The 4th term in the geometric sequence =3 * 12 =36
36=1 + 1/3*(N - 1), solve for N
N = 106th term.
So, the 4th term of the geometric sequence, or 36, appears as 106th term in the arithmetic sequence.
We know that
1 + d = a (1)
1 + 9d = ar (2)
1 + 33d = ar^2 (3)
Subbing (1) into (2) we have that
1 + 9d = (1 + d)r ⇒ r = (1 + 9d) / ( 1 + d) (4)
Sub (1) and (4) into (3)
1 + 33d = (1 + d) (1 + 9d)^2 / (1 + d)^2 simplify
1 + 33d = (1 + 9d)^2 / (1+ d)
(1 + d) (1 + 33d) = (1 + 9d)^2 simplify
33d^2 + 34d + 1 = 81d^2 + 18d + 1
48d^2 - 16d = 0
16d ( 3d - 1) = 0
Setting both factors to 0 and solving for d produces d = 0 (reject) or d = 1/3 (accept)
And 1 + d = a .....so.....1 + 1/3 = a = 4/3
And 1 + 9d = ar
1 + 9(1/3) = (4/3)r
1 + 3 = (4/3)r
4 = (4/3)r
r = 3
So....the 34th term of the arithmetic series = 1 + 33d = 1 + 33(1/3) = 1 + 11 = 12
The fourth term in the geometric sequence above appears as the nth term in the arithmetic sequence. Find the value of n.
The 4th term of the geometric series is = ar^3 = (4/3)(3)^3 = 36
So....to find which term this is in the aritmetic series, we have
36 = 1 + (1/3)(n - 1)
35 = (1/3)(n - 1)
105 = n - 1
106 = n