Arithmetic Sequences

I would have thought the sum of the first four terms would be:

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

Edit: ninjad.

Reply 3

ghostwalker

17

Original post by Zacken

Edit: ninjad.

My turn

Reply 4

Moazer

Original post by christinajane

..
XX

If you don't understand my workings say something

U_3 = a + 2d

\frac{11}{2} = a + 2d

S_4 = 2[2a + 3d]

\frac{91}{4} = 4a + 6d

\frac{11}{2} = a + 2d

\frac{91}{4} = 4a + 6d

a = \frac{11}{2} - 2d

\frac{91}{4} = 4[\frac{11}{2} - 2d]+ 6d

\frac{91}{4} = 22 - 8d + 6d

\frac{3}{4} = -2d

d = -\frac{3}{8}

(edited 7 years ago)

Reply 5

OP

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.

Edit: ninjad.

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...

Reply 6

OP

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 :-(

(edited 7 years ago)

Reply 7

OP

Original post by Moazer

Why D0 u have to be So difficult just use the S_n formula

Difficult?

Just trying to understand

Excuse me for asking questions

Reply 8

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.

Reply 9

Original post by christinajane

Difficult?

Just trying to understand

Excuse me for asking questions

Please ignore that nasty person. We have trolls in the maths forum sometime.

Reply 10

Moazer

Original post by christinajane

Difficult?

Just trying to understand

Excuse me for asking questions

apologies baby

Reply 11

Moazer

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?

Reply 12

OP

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

Reply 13

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).

Reply 14

wavesofmetanoia

9

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

Reply 15

OP

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 :-)

Reply 16