IntMath Home » Forum home » Series and the Binomial Theorem » find the first term and common ratio

IntMath forum | Series and the Binomial Theorem

find the first term and common ratio [Solved!]

My question

How to find the first term and common ratio if given that sum of first and third term is 20 and sum of fourth and sixth term is 540?

Relevant page

2. Geometric Progressions

What I've done so far

Trial and error, and reading over your examples.

X

How to find the first term and common ratio if given that sum of first and third term is 20 and sum of fourth and sixth term is 540?
Relevant page

<a href="/series-binomial-theorem/2-geometric-progressions.php">2. Geometric Progressions</a>

What I've done so far

Trial and error, and reading over your examples.

Re: find the first term and common ratio

Hello Alicia

I'll get you started and hope you can go from there.

As you know, the terms in an arithmetic progression go:

`a, ar, ar^2, ar^3, ...`

From the question, you know that

`a + ar^2 = 20`

Next, factorise the LHS.

Now write the sum of the 4th and 6th terms and factorise. You should be able to recognise something.

Then divide the second answer by the first. Out will pop your answer.

Good luck

X

Hello Alicia

I'll get you started and hope you can go from there.

As you know, the terms in an arithmetic progression go:

`a, ar, ar^2, ar^3, ...`

From the question, you know that

`a + ar^2 = 20`

Next, factorise the LHS.

Now write the sum of the 4th and 6th terms and factorise. You should be able to recognise something.

Then divide the second answer by the first. Out will pop your answer.

Good luck

Re: find the first term and common ratio

Do you mean `a(1+r^2) = 20`?

The 4th terms is `ar^3` and the 6th is `ar^5`, so

`ar^3 + ar^5 = 540`

?? Where to go from there?

X

Do you mean `a(1+r^2) = 20`?

The 4th terms is `ar^3` and the 6th is `ar^5`, so

`ar^3 + ar^5 = 540`

?? Where to go from there?

Re: find the first term and common ratio

You need to factor your last expression, then divide the second line by the first.

X

You need to factor your last expression, then divide the second line by the first.

Re: find the first term and common ratio

`ar^3(1+r^2) = 540`

Do it mean this?

`(ar^3(1+r^2))/(a(1+r^2)) = 540/20`

Then

`r^3 = 27`

`r = 3`

I got it!

Putting that back in the first line gives me:

`a(1+3^2) = 10a = 20`

So `a=2`.

So first term is `2` and common ratio is `3`.

Plugging in to check:

The terms will be:

`2, 6, 18, 54, 162, 486, ...`

Sum of first and 3rd is `2 + 18 = 20` (OK)

Sum of 4th and 6th terms: `54 + 486 = 540`

Thank you so much

Regards
Alicia

X

`ar^3(1+r^2) = 540`

Do it mean this?

`(ar^3(1+r^2))/(a(1+r^2)) = 540/20`

Then

`r^3 = 27`

`r = 3`

I got it!

Putting that back in the first line gives me:

`a(1+3^2) = 10a = 20`

So `a=2`.

So first term is `2` and common ratio is `3`.

Plugging in to check:

The terms will be:

`2, 6, 18, 54, 162, 486, ...`

Sum of first and 3rd is `2 + 18 = 20` (OK)

Sum of 4th and 6th terms: `54 + 486 = 540`

Thank you so much

Regards
Alicia

Re: find the first term and common ratio

You're welcome, Alicia

X

You're welcome, Alicia

Reply

You need to be logged in to reply.

Search IntMath, blog and Forum