Answer:
- an = 2·(4/5)^(n-1)
- 2, 8/5, 32/25, 128/125
Step-by-step explanation:
The sum of an infinite geometric series is ...
S = a1/(1 -r)
where r is the common ratio. The sum will only exist if |r| < 1.
The problem statement tells us S = 10 and a1 = 2, so we have ...
10 = 2/(1 -r)
r = 1 -2/10 = 4/5
So the n-th term of the series is ...
an = a1·r^(n-1)
an = 2·(4/5)^(n-1)
For values of n = 1 to 4, the terms are ...
2, 8/5, 32/25, 128/125
