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# 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

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.

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.

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