+1557 A geometric sequence has 400 terms. The first term is 1000 and the common ratio is -\frac{9}{10}. How many terms of this sequence are greater than 1?
+129881 We need to solve this
1000*(9/10)^(n-1) = 1
(.9)^(n -1) = 1/1000 take the log of both sides
log (.9)^(n-1) = log (1/1000)
And we can write
(n-1) log (.9) = log (1/1000)
n -1 = log (1/1000) / log (.9)
n -1 = 65.56
n = 66.56
The 66th term is 1000(.9)(66-1) = 1.061
The 67th term is 1000(.9)(67-1) = .955
So 66 terms are > 1
