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Problems involving Arithmetic Progressions

4. The first, second and fourth terms of an Arithmetic Progression are x + 1, 2x - 1, and 2x + 5.

Sn formula is Sn = 1/2n(2a+d(n-1))

I'm completely lost...
a = x + 1

2nd term = a + d
2x - 1 = x + 1
x = 2 but it says x = 5
Then others I'm clueless how to find.
Original post by ckfeister
4. The first, second and fourth terms of an Arithmetic Progression are x + 1, 2x - 1, and 2x + 5.

Sn formula is Sn = 1/2n(2a+d(n-1))

I'm completely lost...
a = x + 1

2nd term = a + d
2x - 1 = x + 1
x = 2 but it says x = 5
Then others I'm clueless how to find.


What is your d? You didn't include that in the bold line. Find d by doing (2x1)(x+1)=x2(2x-1)-(x+1)=x-2 and then you may proceed.

Also what are you trying to find? If it's not the sum then that formula is useless.
(edited 7 years ago)
Reply 2
Original post by RDKGames
What is your d? You didn't include that in the bold line. Find d by doing (2x1)(x+1)=x2(2x-1)-(x+1)=x-2 and then you may proceed.

Also what are you trying to find? If it's not the sum then that formula is useless.


I have to find x,a,d
x = 5, a = 6, d = 3, but don't know how to find them.
Original post by ckfeister
I have to find x,a,d
x = 5, a = 6, d = 3, but don't know how to find them.


So
a1=x+1a_1=x+1
a2=2x1a_2=2x-1

so we get the difference in terms of x which is x2x-2

Then express the fourth term in terms of the first one and a certain multiple of the difference.
(edited 7 years ago)
Reply 4
Original post by RDKGames
So
a1=x+1a_1=x+1
a2=2x1a_2=2x-1

so we get the difference in terms of x which is x2x-2

Then express the fourth term in terms of the first one and a certain multiple of the difference.


What did you remove? I was eating came back and it was gone.
Reply 5
bump
Original post by ckfeister
What did you remove? I was eating came back and it was gone.


Error.

Simply express the fourth term since you know the formula un=a+(n1)du_n=a+(n-1)d and solve for x.
Reply 7
Original post by RDKGames
Error.

Simply express the fourth term since you know the formula un=a+(n1)du_n=a+(n-1)d and solve for x.


Ohhh I've been using Sn not Un, thanks.

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