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Maths Series Question

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
Reply 1
badabing
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


Well, almost right, apart from I think on the second part, you should've reversed the inequality here when dividing by ln0.8:
(n-1)ln0.8<ln0.05
n-1>ln0.05/ln0.8
because ln0.8 is negative. So you end up with:
n>14.4251 so n=15
(I think)
Original post by Djeorge
Well, almost right, apart from I think on the second part, you should've reversed the inequality here when dividing by ln0.8:
(n-1)ln0.8<ln0.05
n-1>ln0.05/ln0.8
because ln0.8 is negative. So you end up with:
n>14.4251 so n=15
(I think)

I know this is an old thread but wouldn't it be n>14 due to rounding?
Reply 3
Original post by M.Johnson2111
I know this is an old thread but wouldn't it be n>14 due to rounding?

Very old thread and not read fully, but rounding is not appropriate.
When n=14, it's greater than one. When n=15, it's less than one. So it's the latter.
Original post by mqb2766
Very old thread and not read fully, but rounding is not appropriate.
When n=14, it's greater than one. When n=15, it's less than one. So it's the latter.

Got to be the oldest necro I've ever seen...
Reply 5
Original post by zetamcfc
Got to be the oldest necro I've ever seen...

n=14 years old ...
This is a very old thread but I have to say I agree that n = 15

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