Can someone plz check if my solution is right?
Here's the question...
An infinite geometric sequence has common ratio r. The sum to infinity of the series is five times the first term of the series
a)i) show that r = 0.8
ii) given that the first term of the series is 20, find the least value of n such that the nth term of the sequence is less than one
ANSWERS
a)
i)
a/1-r = 5a
a=5a(1-r)
a=5a-5ar
0=4a-5ar
0=a(4-5r)
4-5r=0
4=5r
r=4/5
=> r=0.8
ii)
a=20
r=0.8
un=ar^n-1
20 x 0.8^n-1 < 1
0.8^n-1 < 1/20
ln0.8^n-1 < ln (1/20)
n-1(ln0.8) < ln (1/20)
nln0.8 - ln0.8 < ln(1/20)
nln0.8 < ln(1/20) + ln0.8
n < (ln(1/20) + ln0.8)/ln0.8
n < 14.4251
since n must be an integer, n is 14