sequences and sseries (1 Viewer)

[shkspeare]

ok heres a question which i dont see what is wrong with my working out

find the value of n for the first term of hte series 5000 + 1000 + 200 + ... is less than 1

ok so heres my working

Tn < 1
5000 x (1/5)^n-1 < 1
n-1 log(1/5) < log (1/5000)
n -1 < 5.29 ...
n < 6.29

therefore n is 6

but the answer is 7 ... arghh some1 tewll me wot is wrong with ma working out!! >.< thxthx

 

[ND]

Originally posted by shkspeare
n-1 log(1/5) < log (1/5000)
n -1 < 5.29 ...

Dividing through by a negative number...

 

[shkspeare]

which part is negative?

maybe i should have put (n-1) = log(1/5000) / log (1/5) to make it clearer

 

[ND]

The log of any number <1 is negative.

 

[shkspeare]

ohh yeah forgot ! *slaps forhead*

thanks =)

 

[shkspeare]

how about this question... totally stuck

the limiting sum of a geometric series is 5 and the second term is 6/5.

Find the first term and the common ratio of the series

 

[ezzy85]

a/1-r = 5
ar^n-1 = 6/5

you know taht n = 2 since its the second term
sub that in and solve simultanously for a and r. the only thing is i cant remember if the limiting sum is a/1-r or a/r-1, but thats the basic method to do it.

 

[shkspeare]

ah ok thx

and thx for the book/flowchart and the past papers from yesterday~~!! =)

 

[ezzy85]

no probs