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# Arithmetic Sequences

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1. Doing solomon paper F C1 and I am stuck on question 8.

The third term of an arithmetic sequence is 5 1/2

The sum of the first four terms of the series is 22 3/4

The first equation I got as:

a + 2d = 5 1/2 - which is correct

The second I am having trouble with:

I thought that it would be

4/2 (a + (a+d) + (a+2d)) = 22 3/4

Which simplifies to:

4/2 (3a + 3d) = 22 3/4

but this isnt correct and according to the ms the second equation should be

4/2 (2a+3d) = 22 3/4

Not sure why there is only 2 a instead of 3....

2. (Original post by christinajane)
Doing solomon paper F C1 and I am stuck on question 8.

The third term of an arithmetic sequence is 5 1/2

The sum of the first four terms of the series is 22 3/4

The first equation I got as:

a + 2d = 5 1/2 - which is correct

The second I am having trouble with:

I thought that it would be

4/2 (a + (a+d) + (a+2d)) = 22 3/4
Not sure how you got that one.

Sum to n terms is "n/2 (first + last)" and the last one is a+3d as there are four terms.

Alternatively "first + second + third + fourth".
3. I would have thought the sum of the first four terms would be:

(a) + (a + d) + (a+2d) + (a+3d) = 4a + 6d.

4. (Original post by Zacken)
My turn
5. (Original post by christinajane)
..
XX
If you don't understand my workings say something

6. (Original post by Zacken)
I would have thought the sum of the first four terms would be:

(a) + (a + d) + (a+2d) + (a+3d) = 4a + 6d.

Actually zacken that was the first equation I got but that wasnt right according to the mark scheme

They say its

4/2 (2a+3d = 22 3/4

Which then simplifies too

4a + 6d = 22 3/4

What I had initially done was like you:

a + (a+d) + (a+2d) + (a +3d)

which simplified to

4/2( 4a + 6d)

Then getting rid of fraction

8a +12d

So not really sure...
7. (Original post by ghostwalker)
Not sure how you got that one.

Sum to n terms is "n/2 (first + last)" and the last one is a+3d as there are four terms.

Alternatively "first + second + third + fourth".
Yeah it makes sense if I use that formula like you said actually:

n/2 (first + last)

But doesnt work out if you use the one I used initally??

first + second + third + fourth".

a + (a+d) + (a +2d) + (a+3d)

How would you know which one to use because the second doesnt appear to work? Or am I missing something :-(
8. (Original post by Moazer)
Why D0 u have to be So difficult just use the S_n formula
Difficult?

Just trying to understand

9. (Original post by christinajane)
Actually zacken that was the first equation I got but that wasnt right according to the mark scheme

They say its

4/2 (2a+3d = 22 3/4

Which then simplifies too

4a + 6d = 22 3/4

What I had initially done was like you:

a + (a+d) + (a+2d) + (a +3d)

which simplified to

4/2( 4a + 6d)

Then getting rid of fraction

8a +12d

So not really sure...

Not sure where the 4/2 comes from.

We have "sum of first four terms" = 22 3/4.

The sum of the first four terms is 4a + 6d.

So 4a + 6d = 22 3/4. There is no 4/2 (4a + 6d) = 22 3/4.

It's just plain ol' 4a + 6d = 22 3/4.

Remember that a + a+d + a+2d + a+3d = (a+a+a+a) + (d+2d+3d) = 4a + 6d, again, no 4/2(4a + 6d) or anything.
10. (Original post by christinajane)
Difficult?

Just trying to understand

Please ignore that nasty person. We have trolls in the maths forum sometime.
11. (Original post by christinajane)
Difficult?

Just trying to understand

apologies baby
12. (Original post by Zacken)
Please ignore that nasty person. We have trolls in the maths forum sometime.
Speak for yourself. Remember when you told that kid he was going to have a **** life because he's attending Warwick?
13. (Original post by Zacken)
Not sure where the 4/2 comes from.

We have "sum of first four terms" = 22 3/4.

The sum of the first four terms is 4a + 6d.

So 4a + 6d = 22 3/4. There is no 4/2 (4a + 6d) = 22 3/4.

It's just plain ol' 4a + 6d = 22 3/4.

Remember that a + a+d + a+2d + a+3d = (a+a+a+a) + (d+2d+3d) = 4a + 6d, again, no 4/2(4a + 6d) or anything.

They have the 4/2 in the mark scheme -

Maybe I am just getting confused with the two s-n formula
14. (Original post by christinajane)
They have the 4/2 in the mark scheme -

Maybe I am just getting confused with the two s-n formula
There's a 4/2 if you want to use the Sn formula, there's no 4/2 if you want to manually add them up.

Remember that 4/2 = 2.

So if you want, you could re-write 4a + 6d = 2(2a + 3d) = (4/2)(2a + 3d).
15. To put it simply,

Like you said (a + 2d) = 5 1/2
So if you know that the formula for sum of a series is Sn = n/2 (2a + (n-1)d )
Then we just sub it in as n = 4 so S4 = 4/2 (2a + (4-1)d )
So it becomes S4 = 2(2a + 3d)
Again simplifying it to 4a + 6d = 22 3/4
Now its basically simultaneous equations
1) (a + 2d) = 5 1/2
2) (4a +6d) = 22 3/4
Multiply the first equation by 3 to get the d terms to be the same
So.....1) now equals (3a + 6d) = 16 1/2
Now subtract them .......SSS (same sign subtract)
(4a + 6d) = 22 3/4
-
(3a +6d) = 16 1/2
a= 22 3/4 - 16 1/2 = 6 1/4

Just sub you a in to find d into any one of the formula
Thus d = -3/8
16. (Original post by Zacken)
There's a 4/2 if you want to use the Sn formula, there's no 4/2 if you want to manually add them up.

Remember that 4/2 = 2.

So if you want, you could re-write 4a + 6d = 2(2a + 3d) = (4/2)(2a + 3d).
Ahhhh ok yeah totally get it now.

Thank you - was getting confused with my s-n formula and just plain old adding them up like you say. Thought with it being a sum of thing I had to include the n/2

Crystal clear :-)
17. (Original post by christinajane)
Ahhhh ok yeah totally get it now.

Thank you - was getting confused with my s-n formula and just plain old adding them up like you say. Thought with it being a sum of thing I had to include the n/2

Crystal clear :-)
Ahh, nice - glad it's all clicked!

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