Alicia 28 Nov 2015, 06:03
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?
Trial and error, and reading over your examples.
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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.
Murray 29 Nov 2015, 08:06
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
Alicia 30 Nov 2015, 10:31
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?
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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?
Murray 01 Dec 2015, 06:48
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.
Alicia 01 Dec 2015, 23:45
`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
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`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
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You're welcome, Alicia
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