# Geometric sequences!!

How do you find the number of terms in the following geometric sequence,
729, -243, 81....-1/3

Pls explain,
thanks
Mohit_C
How do you find the number of terms in the following geometric sequence,
729, -243, 81....-1/3

Pls explain,
thanks

Let the first term be 'a', the common ratio be 'r', and the number of terms be 'n'.

a = 729
r = -1/3

The last term = arn - 1.

-1/3 = 729*(-1/3)n - 1

-1/2187 = (-1/3)n - 1

Solving this gives n = 8.

Therefore, there are 8 terms in the sequence ...

Hope this helps,

~~Simba
consider what is happening with each term;

first term to second term occurs through division by -3, the same generates the next term

=> the common ratio = -1/3

we need to find when arn = -1/3

Using a=729

=> rn = -1/2187

this is the same as (-1/3)7

=> sequence goes from ar0 -> ar7 , and so contains 8 terms
alternatively, type 729 in your calculator and press =
then type ANS/-3
729 is your first term. count how many times you press "=" until you get to -1/3.
Ok...how did you get n=8 from

-1/2187=-1/3^(n-1)?

Coz i tried using log but didnt work. Explain.
Mohit_C
Ok...how did you get n=8 from

-1/2187=-1/3^(n-1)?

Coz i tried using log but didnt work. Explain.

-1/2187 = -1/3n-1

=> 2187 = 3n-1
take logs of this and solve for n
Undry1
-1/2187 = -1/3n-1

=> 2187 = 3n-1
take logs of this and solve for n

Ok...looking at that i assume that u divided both sides by -1 rite? or else, how did you do that just get rid of the -1/...?
-1/2187 = (-1)n-1/3n-1
multiplying by -1 gives
1/2187 = (-1)n/3n-1
by inspection you can see that (-1)^n must be positive, giving
1/2187 = 1/3n-1
=> 2187 = 3n-1
Undry1
-1/2187 = (-1)n-1/3n-1
multiplying by -1 gives
1/2187 = (-1)n/3n-1
by inspection you can see that (-1)^n must be positive, giving
1/2187 = 1/3n-1
=> 2187 = 3n-1

I dont get this, how do you get rid of the - and get a positive number. Pls explain.
Mohit_C
I dont get this, how do you get rid of the - and get a positive number. Pls explain.

after multiplying by -1
-1 * -1/2187 = -1 * (-1)n-1/3n-1

the left hand side becomes positive, therefore the right hand side must also be positive:
-1 * (-1)n-1/3n-1 is positive

now 3n-1 is always positive no matter what value n takes
so -1 * (-1)n-1 , i.e. (-1)n must be positive. the only way this is possible is if n is even, giving (-1)n=1
Ok....finally got it. Thanks m8!
good stuff.
Ok i got another one here:

C2 Examination style:

A father promises his daughter an eternal gift on her birthday. On day 1 she receives &#8364;75 and each following day she receives 2/3 of the amount given to her the day before. The father promises that this will go on for ever. After k days the total amount of money that the daughter will have received exceeds &#8364;200. c) Find the smallest value of k.

Thanks.
a= 75, r=2/3
75(1-(2/3)^k) / (1/3) > 200
1-(2/3)^k > 200/225
(2/3)^k < 1-(200/225)
klog(2/3) < log(1-(200/225))
k > 5.4 (inequality switches around because log(2/3) is negative)
k=6
Undry1
a= 75, r=2/3
75(1-(2/3)^k) / (1/3) > 200
1-(2/3)^k > 200/225
(2/3)^k < 1-(200/225)
klog(2/3) < log(1-(200/225))
k > 5.4 (inequality switches around because log(2/3) is negative)
k=6

thanks