Algebraic Sequences

A linear sequence has common difference of 8. Three consecutive terms in the sequence are added together to give a total of 126. Find the three terms.

So far I've tried this:
8n+8(n+1)+8(n+2)=126
8n+8n+8+8n+16=126
24n+24=126
24n=102
n = 4.25
This is not possible.
I know the answers are 34,42,50 but don't know how to get to them. If they said the expression of the sequence was 8n+x, I may not get this problem. am I wrong?

Reply 1

Zacken

22

Original post by bob12344321

A linear sequence has common difference of 8. Three consecutive terms in the sequence are added together to give a total of 126. Find the three terms.

So far I've tried this:
8n+8(n+1)+8(n+2)=126
8n+8n+8+8n+16=126
24n+24=126
24n=102
n = 4.25
This is not possible.
I know the answers are 34,42,50 but don't know how to get to them. If they said the expression of the sequence was 8n+x, I may not get this problem. am I wrong?

Let's call the 'first' term

a

, then the second term would be

a+8

and the third would be

a + 8 + 8 = a + 16

.

Thence their sum is

3a + 24 = 126

.

Also, this thread would be better put in the maths forum instead of the maths exam subforum.

Reply 2

Zacken

22

Original post by bob12344321

A linear sequence has common difference of 8. Three consecutive terms in the sequence are added together to give a total of 126. Find the three terms.

So far I've tried this:
8n+8(n+1)+8(n+2)=126
8n+8n+8+8n+16=126
24n+24=126
24n=102
n = 4.25
This is not possible.
I know the answers are 34,42,50 but don't know how to get to them. If they said the expression of the sequence was 8n+x, I may not get this problem. am I wrong?

Your method is wrong, but it's not totally invalid, you just called your first term

8n

so

n=4.35 \Rightarrow 8n = 34, \mathrm{ etc... }

Reply 3

bob12344321

OP

3

Original post by Zacken

Also, this thread would be better put in the maths forum instead of the maths exam subforum.

I didn't want to post the thread here but I couldn't find the maths subforum and because I was in a rush I just posted it here, anyways thank you for the answer

Reply 4

Zacken

22

Original post by bob12344321

I didn't want to post the thread here but I couldn't find the maths subforum and because I was in a rush I just posted it here, anyways thank you for the answer

Yeah, that's fine - you're welcome. I was just poinging it out for future reference.

Also, it's the other way around. You've posted in a sub-section of the maths forum.

Your post would be better placed here: http://www.thestudentroom.co.uk/forumdisplay.php?f=38

Reply 5

razzor

10

Original post by bob12344321

A linear sequence has common difference of 8. Three consecutive terms in the sequence are added together to give a total of 126. Find the three terms.

So far I've tried this:
8n+8(n+1)+8(n+2)=126
8n+8n+8+8n+16=126
24n+24=126
24n=102
n = 4.25
This is not possible.
I know the answers are 34,42,50 but don't know how to get to them. If they said the expression of the sequence was 8n+x, I may not get this problem. am I wrong?

Apologies if you have already sorted it, but I'll give you extra guidence.

First, what does 'common difference' (c.d.) actually mean?

In the sequence 1, 3, 5, 7..., the c.d. is 2 because we are adding 2 each time.
Similiarly, in 5, 9, 13, 17..., the c.d. is 4

In our mystery sequence, we know that the c.d. is 8 (I.e. we add 8 onto the previous term).
Let us call the first term needed n.
Any thoughts on how we might set up our equation?

(edited 8 years ago)

Algebraic Sequences

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