stuck on series qu! (1 Viewer)

[bro_sky]

Anyone know how to answer this question? ...I keep getting ideas.. thinking "yeah that's smart"... but hey don't work....

Find the value for (n) for whcih the sum of the series 20 + 4 + (4/5)...... is greater 24.85

answer is (n) = 4

 

[crazylilmonkee]

isnt that a limiting sum question?

a=20
r=1/5

s(infinity)=20/91-1/50
=25

 

[PoLaRbEaR]

use sum of GP:
S = a(1 - r^n)/(1-r) a=20 r=1/5 S=24.85

20[1 - (1/5)^n]/(1-1/5) > 24.85
20[1 - (1/5)^n] > 24.85x4/5
1 - (1/5)^n > 19.88/20
(1/5)^n > 1 - 0.994
nln(1/5) > ln0.006
n > 3.1787...
n is a whole number so:
n = 4

EDIT: the answer should be greater than or equal to 4...

 

[crazylilmonkee]

oooooooooh u learn something new everyday

 

[bro_sky]

yes... this works well except that when the base is less than one, ie (1/5) when passing from an index inequation to a log inequation should you not reverse the inequality sign?

ie.

nIn(1/5) < In (0.006)

which doesn't work as (n) < 3.1787....

what do you reckon???

 

[PoLaRbEaR]

ln(0.006) is also less than 1...
so both sides end up with a negative sign, which you can just cancel out...

 

[bro_sky]

gotcha

smmmmmart!!!!!



ain't it great when it works out?

thax a million.