Geometric Progression

Destined2REIGN SG

two convergent geometric progression each have a first term a. The sum to infinity of the first progression is 16 and the sum to infinity of the 2nd progression is 64. If the common ratio of the second progression is equal to the square of the common ratio of the first progression, find the value of a.

TenSaru SG

Wow I just did this not long ago... I'm also JC1 ^_^

a/(1-r) = 16

a/(1-r^2) = 64

a/(1-r)(1+r) = 64

16/(1+r) = 64 (sub a/(1-r)=16)

1/(1+r) = 4

1+r = 1/4

r= -3/4

a= 16[1-(-3/4)] = 28

Destined2REIGN SG

Thanks. Got 2 more.

The sum of the first n terms is 1-(3/4)^n. Obtain an expression for the rth term of the series. Prove this is a GP and state the values of the first term and common ratio.

AND

In a GP, the first term is 12 and 4th term is -3/2. Find the common ratio r, and Sn (sum of all n terms).
Find S (sum to infinity) and state the least value of n for which the magnitude of the difference between Sn and S is less than 0.001

wee_ws SG

Hi,

Given: S_n.

To find T_n, consider S_n - S_(n - 1).

To prove the series to be a GP, consider T_n / T_(n - 1) and show that it is a constant.

Given: T_1 = 12, T_4 = -3/2.

Consider forming 2 equations in order to find r and S_n.

For the last part, consider |S - S_n| < 0.001 and solve for n via algebra or GC.

Thanks!

Cheers,
Wen Shih

Destined2REIGN SG

given all terms of GP are positive, Find the first term if S6 = 1 and S12 = 65. I know I can do it via simulataneous. But is there a shortcut?

AND. in a GP, a = 12, U4 = -3/2. Find r ( common Ratio), and Sn (sum of all n terms).

Find S (sum to infinity) and least value of n for which the magnitude of difference between Sn and S is less than 0.001.

I found r = -1/2, Sn ={12[1-(-0.5^n)]}/{1-(-0.5)} and S =8.

But I cant do the part in bold.

wee_ws SG

Hi,

I believe you will obtain, after some simplification, this:

| 8 (-0.5)^n | < 0.001

This will lead us to conclude

8 (0.5)^n < 0.001,

of which n could be solved by taking logs.

For your first question, we have 2 unknowns (first term and common ratio) so we will need to form 2 equations.

Thanks!

Cheers,
Wen Shih

Destined2REIGN SG

I didnt get this | 8 (-0.5)^n | < 0.001. Could you show your working? anyway, how do you get the modulus sign?

anyway how do you show 1+ e^-x + e^-2x + e^-3x... has a sum to infinity (I know the ans must not have x inside. but how)?

wee_ws SG

Hi,

You have already obtained S and S_n yes?

Substitute both into |S - S_n| < 0.001, which is the requirement of the question, i.e. "the magnitude of the difference between Sn and S is less than 0.001".

We use the modulus because we are interested in the value of the difference regardless of the sign.

As for your next question, we will not be able to show that the sum has a sum to infinity. However, we will be able to find a condition for the sum to infinity to exist, which is |common ratio| < 1. From "1 + e^(-x) + e^(-2x) + e^(-3x) + ...", can you see the common ratio?

Thanks!

Cheers,
Wen Shih

Destined2REIGN SG

Is n > 12.9? (Smallest n is 13?)

and all i need to show is that common ratio is e^-x = 1/e^x which is smaller than 1?

wee_ws SG

Hi,

Check whether the condition |S - S_13| is indeed < 0.001?

Check what the question wants you to do. Thanks!

Cheers,
Wen Shih