Geometric Progression: sum of 1st 3 terms is 4.88, sum to infinity is 10

[Simonsky]

Having trouble with the following -doesn't seem to be enough information:

A geometric progression has first term a and common ration r. The sum of the first three terms is 4.88 and sum to infinity is 10.

1) Write down two equations involving a and r 9 I can do that bit: a=ar=ar^2 = 4.88 and a/(1-r) = 10

but then it asks: Show that 1-r^3 = 0.488 -can't seem to get that!

2) Then it asks: mk(Subscript) is the kth term calculate the sum to infinity of mk^2.

Thanks for any help here!

 

[Dr.Peterson]

A geometric progression has first term a and common ratio r. The sum of the first three terms is 4.88 and sum to infinity is 10.

1) Write down two equations involving a and r 9 I can do that bit: a=ar=ar^2 = 4.88 and a/(1-r) = 10

but then it asks: Show that 1-r^3 = 0.488 -can't seem to get that!

2) Then it asks: mk(Subscript) is the kth term calculate the sum to infinity of mk^2.

I think when you said "a=ar=ar^2 = 4.88", you meant "a+ar+ar^2 = 4.88". Am I right?

You are probably expected to use the formula you have learned for the sum of a finite number of terms of a geometric progression. What is that formula? What you wrote is valid, but makes it a little harder to get to the required conclusion.

We can look at the second part once you get the first. Note that you can write subscripts using the button marked x₂. The question is, as I read it,

If m_k is the kth term, calculate the sum to infinity of (m_k)².

Is that right? I would start by writing an expression for m_k and then (m_k)².

 

[Simonsky]

I think when you said "a=ar=ar^2 = 4.88", you meant "a+ar+ar^2 = 4.88". Am I right?

You are probably expected to use the formula you have learned for the sum of a finite number of terms of a geometric progression. What is that formula? What you wrote is valid, but makes it a little harder to get to the required conclusion.

We can look at the second part once you get the first. Note that you can write subscripts using the button marked x₂. The question is, as I read it,



Is that right? I would start by writing an expression for m_k and then (m_k)².

Dr. Peterson-thanks for response -you were right about my silly typos! The only formula I know for the sum of an infinite geometric progression is: a/1-r

 

[Dr.Peterson]

Dr. Peterson-thanks for response -you were right about my silly typos! The only formula I know for the sum of an infinite geometric progression is: a/(1-r)

You should have learned a formula for the sum of FINITELY MANY terms before the sum of the infinite series. If not, search for it: for example, here.

Alternatively, you may know that 1+r+r^2 is one factor of (1 - r)^3. Is the factoring of a difference of cubes familiar?

 

[Simonsky]

You should have learned a formula for the sum of FINITELY MANY terms before the sum of the infinite series. If not, search for it: for example, here.

Alternatively, you may know that 1+r+r^2 is one factor of (1 - r)^3. Is the factoring of a difference of cubes familiar?

O.K I've worked it out -you have to use both formulas (finite and infinite). So:-

a(1-r^3)/(1-r ) = 4.88 -------1-r^3 = +4.88(1-r)/a ---------then take infinite geo. series formula a/1-r = 10 -----10(1-r) = a then apply to finite formula:

1-r^3 = 4.88/10 =0.488 so then r^3 = 0.512 ---cubed root of r^3 = 0.8 so: a/(1-r) = a/0.2 and a/0.2 = 10 so a =2.

Thanks for prodding me!

Still stuck on second part of question:

kth term of progression is mk. Calculate Sigma k=1.....infinity (mk)^2

Any hints welcome.

 

[Dr.Peterson]

O.K I've worked it out -you have to use both formulas (finite and infinite). So:-

a(1-r^3)/(1-r ) = 4.88 -------1-r^3 = +4.88(1-r)/a ---------then take infinite geo. series formula a/1-r = 10 -----10(1-r) = a then apply to finite formula:

1-r^3 = 4.88/10 =0.488 so then r^3 = 0.512 ---cubed root of r^3 = 0.8 so: a/(1-r) = a/0.2 and a/0.2 = 10 so a =2.

Thanks for prodding me!

Still stuck on second part of question:

kth term of progression is mk. Calculate Sigma k=1.....infinity (mk)^2

Any hints welcome.

It's a little hard to follow what you wrote, but you seem to have gone beyond part (1), showing that 1-r^3 = 0.488. I would show it this way, which I think is basically what you did:

a*(1 - r^3)/(1 - r) = 4.88
a/(1 - r) = 10

Solving the second equation for a, a = 10(1 - r).

Putting this in place of a in the first equation, 10(1 - r)*(1 - r^3)/1 - r) = 4.88, so

10(1 - r^3) = 4.88

which leads directly to the answer.

You got this, and then solved for r = 0.8 and a = 2. Good.

Now for part (2).

What is the formula for the kth term of the geometric progression with given a and r? Put in the values you have, and show me the simplified formula for the given progression m_k.

Then what is the formula for (m_k)²? Just square the formula you just wrote, and simplify again. You will find that this is another geometric progression, so you can find its sum!

A general tip for solving unfamiliar problems: take it one step at a time. You may have no idea how to solve the whole thing, but if you do whatever you see that you CAN do, then you can often see something else you can do, until you realize the whole thing can be solved. Be a bold explorer!