Find the 6th term of the arithmetic sequence given an = (1/5)an-1, given a1 = 20
If I read this correctly as
an = (1/5) an-1 , where 1 = 20
then
1 = 20
2 = 4
3 = 4/5
etc
then this not an arithmetic but rather a geometric sequence
with a = 20 , r = 1/5
t6 = ar^5 = 20(1/5)^5 = 20/3125 = 4/625
Give the arithmetic sequence -7 ,-1, 5,11. Find the a.32term b.42 term.
To find the 6th term of the arithmetic sequence, we can use the formula for the nth term of an arithmetic sequence:
an = a1 + (n - 1)d
Where:
an is the nth term,
a1 is the first term,
n is the position of the term in the sequence,
d is the common difference between terms.
The given sequence is defined recursively as an = (1/5) * an-1, with a1 = 20. Since we know the first term a1, we can find the common difference d using the recursive formula.
Given:
a1 = 20
Using the recursive formula, we can find a2:
a2 = (1/5) * a1 = (1/5) * 20 = 4
Using the same logic, we can find a3, a4, a5:
a3 = (1/5) * a2 = (1/5) * 4 = 4/5
a4 = (1/5) * a3 = (1/5) * (4/5) = 4/25
a5 = (1/5) * a4 = (1/5) * (4/25) = 4/125
Now, we can find the common difference, d between any pair of consecutive terms:
d = a2 - a1 = 4 - 20 = -16
Now, we have the first term (a1 = 20) and the common difference (d = -16).
Using the formula for the nth term of an arithmetic sequence, we can find the 6th term (a6):
a6 = a1 + (6 - 1)d
= 20 + 5(-16)
= 20 - 80
= -60
Therefore, the 6th term of the given arithmetic sequence is -60.