14bearmanl
Badges: 7
Rep:
?
#1
Report Thread starter 1 month ago
#1
I've been staring at this GCSE question all day - can anyone help?

What is the nth term of

3, 3root5, 15, 15root5, 75
0
reply
mqb2766
Badges: 19
Rep:
?
#2
Report 1 month ago
#2
(Original post by 14bearmanl)
I've been staring at this GCSE question all day - can anyone help?

What is the nth term of

3, 3root5, 15, 15root5, 75
Youve been taught two types of sequences. Which is it?
0
reply
-Imperator-
Badges: 14
Rep:
?
#3
Report 1 month ago
#3
What number do you multiply each of those terms by to get the next?
0
reply
14bearmanl
Badges: 7
Rep:
?
#4
Report Thread starter 1 month ago
#4
(Original post by mqb2766)
Youve been taught two types of sequences. Which is it?
Because it's a root - it looks like a quadratic sequence - each time increasing by a multiple of root5. Not sure what to do next?

3 (times root5) = 3 root 5, or is it 3 to the power of n times root 5 - getting really confused. What do I do next?
0
reply
mqb2766
Badges: 19
Rep:
?
#5
Report 1 month ago
#5
(Original post by 14bearmanl)
Because it's a root - it looks like a quadratic sequence - each time increasing by a multiple of root5. Not sure what to do next?

3 (times root5) = 3 root 5, or is it 3 to the power of n times root 5 - getting really confused. What do I do next?
For a quadratic sequence, when you difference successive terms that forms a linear or arithmetic sequence. Does that happen here? If not, what else could it be?
0
reply
OJlongley
Badges: 9
Rep:
?
#6
Report 1 month ago
#6
(Original post by 14bearmanl)
I've been staring at this GCSE question all day - can anyone help?

What is the nth term of

3, 3root5, 15, 15root5, 75
The change between each number is root 5
3 * sqrt5 = 3sqrt5
3 sqrt5 * sqrt5 = 3 * 5 = 15
15 * sqrt5 = 15 sqrt5
15 * sqrt5 * sqrt5 = 15 * 5 = 75
Work from there
0
reply
14bearmanl
Badges: 7
Rep:
?
#7
Report Thread starter 1 month ago
#7
(Original post by -Imperator-)
What number do you multiply each of those terms by to get the next?
Multiple by root 5? I tried turning them all into roots - and can see they are multiples of 5 - just don't know what to do next?

1 = root 9
2 = root 45
3 = root 225
4 = root 1124
5 = root 5625
0
reply
OJlongley
Badges: 9
Rep:
?
#8
Report 1 month ago
#8
If you're still stuck then here's the answer

Spoiler:
Show
3 * sqrt5^(n-1)
Three times square root 3 to the power of n minus one
This is because each term is the previous term multiplied by square root 5
So,
3
3*(sqrt5)^1
3*(sqrt5)^2
3*(sqrt5)^3
Each term is 3*sqrt5 with the power increasing by one every time. The power is one below the term number
Therefore 3 * sqrt5^(n-1)
0
reply
14bearmanl
Badges: 7
Rep:
?
#9
Report Thread starter 1 month ago
#9
(Original post by mqb2766)
For a quadratic sequence, when you difference successive terms that forms a linear or arithmetic sequence. Does that happen here? If not, what else could it be?
Ah so it is a geometric progression sequence using surds, because it is increasing by root 5 each time. I'm still not sure how to work out the nth term on this?
0
reply
mqb2766
Badges: 19
Rep:
?
#10
Report 1 month ago
#10
(Original post by 14bearmanl)
Ah so it is a geometric progression sequence using surds, because it is increasing by root 5 each time. I'm still not sure how to work out the nth term on this?
What do you need to know for a geometric sequence nth term?
0
reply
14bearmanl
Badges: 7
Rep:
?
#11
Report Thread starter 1 month ago
#11
Do I need to apply the formula for geometric sequence - I'd forgotten about this one. Am guessing I need to put the root 5 into the formula to get the nth term. Need to find some practice questions now. Thanks so much.
0
reply
14bearmanl
Badges: 7
Rep:
?
#12
Report Thread starter 1 month ago
#12
(Original post by OJlongley)
If you're still stuck then here's the answer

Spoiler:
Show
3 * sqrt5^(n-1)
Three times square root 3 to the power of n minus one
This is because each term is the previous term multiplied by square root 5
So,
3
3*(sqrt5)^1
3*(sqrt5)^2
3*(sqrt5)^3
Each term is 3*sqrt5 with the power increasing by one every time. The power is one below the term number
Therefore 3 * sqrt5^(n-1)
Thanks - I def need to practice these more. Thanks so much.
0
reply
mqb2766
Badges: 19
Rep:
?
#13
Report 1 month ago
#13
(Original post by 14bearmanl)
Do I need to apply the formula for geometric sequence - I'd forgotten about this one. Am guessing I need to put the root 5 into the formula to get the nth term. Need to find some practice questions now. Thanks so much.
As a brief summary, you have arithmetic sequences (linear and quadratic) and geometric sequences. You can tell them apart by working out the difference and ratio of successive terms.
* If the difference is constant, its a linear arithmetic sequence
* If the difference increases linearly, its a quadratic sequence
* If the ratio is constant, its a geometric sequence.
When given a sequence, this is almost always the first thing to determine.

For each of these sequences, you need to know the initial term and the difference or ratio to get a formula for the nth term
https://www.mathsisfun.com/algebra/s...es-series.html
Last edited by mqb2766; 1 month ago
0
reply
14bearmanl
Badges: 7
Rep:
?
#14
Report Thread starter 1 month ago
#14
(Original post by mqb2766)
As a brief summary, you have arithmetic sequences (linear and quadratic) and geometric sequences. You can tell them apart by working out the difference and ratio of successive terms.
* If the difference is constant, its a linear arithmetic sequence
* If the difference increases linearly, its a quadratic sequence
* If the ratio is constant, its a geometric sequence.
When given a sequence, this is almost always the first thing to determine.

For each of these sequences, you need to know the initial term and the difference or ratio to get a formula for the nth term
https://www.mathsisfun.com/algebra/s...es-series.html
That's great - I'm going to take a look at this now. Just been doing so many quadratic sequences lately and looking at differences rather than ratios, I'd forgotten how to do it. Thanks so much.
0
reply
mqb2766
Badges: 19
Rep:
?
#15
Report 1 month ago
#15
(Original post by 14bearmanl)
That's great - I'm going to take a look at this now. Just been doing so many quadratic sequences lately and looking at differences rather than ratios, I'd forgotten how to do it. Thanks so much.
No problem., Its not easy to compute the differences here because of the sqrt(5), so you could just assume that
sqrt(5) ~ 2
and do approximate differences. However, its fairly clear that the differences increase faster than linear and successive ratios gives a constant r=sqrt(5).
Last edited by mqb2766; 1 month ago
0
reply
OJlongley
Badges: 9
Rep:
?
#16
Report 1 month ago
#16
(Original post by 14bearmanl)
Thanks - I def need to practice these more. Thanks so much.
No problem. I'm doing A-level and it still took me a minute to get the ^n-1
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should the school day be extended to help students catch up?

Yes (93)
28.44%
No (234)
71.56%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise