Geometric series (1 Viewer)

[enigma_1]

How do I do a? I'm getting 11 but the answer's 10.

 

[enigma_1]

bump

 

[RealiseNothing]

The general term is :



We want to find the largest value of such that:











So is the largest value of such that the inequality holds, thus there are 10 terms where the series exceeds

 

[enigma_1]

Thanks! How would you do c? I keep getting one number over or under - why is this?

 

[integral95]

is the answer 132?

it's the some thing that realise did....

 

[QZP]

Thanks! How would you do c? I keep getting one number over or under - why is this?

Show me your working out. Your logic is flawed

is the answer 132?

it's the some thing that realise did....

That's correct

 

[enigma_1]

Can someone fix up my working? Where did I stuff up I don't even know. My school teacher doesn't even know how to do these questions - hence my flawed logic.

 

[rumbleroar]

You screwed up on the 6th line

 

[rumbleroar]

Try converting 10^-6 into a decimal and log LHS instead, I think you caused yourself problems when you tried to log both sides

 

[enigma_1]

I still aint getting it bro

 

[enigma_1]

No, its correct.

@Hyperbole, the problem with your working out is that the series is DECREASING. If you wanted to follow through, flip the inequality sign from the start and conclude that the 133 term is the one that is BELOW what is required. HENCE, the 132 term is correct answer.

ahhhhh no wonder!!!!!!!!!!!!!!!!!! i think its the wording that threw me off

 

[Drongoski]

don't know if this has been fixed.



because log 0.9 is NEGATIVE.

 

[enigma_1]

don't know if this has been fixed.



because log 0.9 is NEGATIVE.

but I don't get why log 0.9 is negative?

 

[enigma_1]

As in, how would you know without typing it into calculator?

 

[Drongoski]

because log x is negative if 0 < x < 1, whatever the base.

 

[enigma_1]

because log x is negative if 0 < x < 1, whatever the base.

THANK YOUUUUU, I finally get it now!!!!!!

 

[QZP]

don't know if this has been fixed.



because log 0.9 is NEGATIVE.

Dangit, didn't see this. Nice work