Thread: Geometric Progression - Finding variable to satisfy inequality

Hi everyone,

Firstly, I hope I'm posting this in the right forum (pre-calculus isn't a term I'm familiar with.)

I have a question about a geometric series that I'm stuck on. I've had a go at entering it here but I'm pretty hopeless at the LaTex package, so instead I've attached the question as an image. Please have a look and see if you might be able to help - I'd be really grateful.

So far I have u₁ = 108 and u₂ = 72, and the sum to infinity = 324.

It's the last bit I'm stuck on, finding the smallest value of k for which the sum of terms between k and infinity is less than 2.5. (the answer is 13)

I tried S_k < 2.5 , re-arranged and simplified, then took logs of both sides so that the variable k wasn't a power, and rearranged to get k on it's own.

(With S_k being the formula for the sum of "k" terms, and using 108 as the value for a/first term)

Originally Posted by Bluebell

Hi everyone,

Firstly, I hope I'm posting this in the right forum (pre-calculus isn't a term I'm familiar with.)

I have a question about a geometric series that I'm stuck on. I've had a go at entering it here but I'm pretty hopeless at the LaTex package, so instead I've attached the question as an image. Please have a look and see if you might be able to help - I'd be really grateful.

So far I have u₁ = 108 and u₂ = 72, and the sum to infinity = 324.

Those are correct.

It's the last bit I'm stuck on, finding the smallest value of k for which the sum of terms between k and infinity is less than 2.5. (the answer is 13)

I tried S_k < 2.5 , re-arranged and simplified, then took logs of both sides so that the variable k wasn't a power, and rearranged to get k on it's own.

(With S_k being the formula for the sum of "k" terms, and using 108 as the value for a/first term)

Without seeing your work it's hard to tell where your error is. You should be working with$$
\sum_{n=k}^\infty 162\left( \frac 2 3\right)^k = \frac{162\left( \frac 2 3\right)^k}{\frac 1 3} = 486\left(\frac 2 3\right)^k =2.5$$
Solving for $k$ gives $12.9972$ approximately so the answer is $13$.