I agree with ghostwalker and 32.

If you want some help then show me what you have done so far and I might be able to help you understand. Just giving you the answer won't help you in future.

Reply 7

otrivine

OP

14

Original post by msmith2512

I agree with ghostwalker and 32.

If you want some help then show me what you have done so far and I might be able to help you understand. Just giving you the answer won't help you in future.

ok i got for the first 3 terms 1,5,9
then i used the sum formula
sn=n/2(2+4n-4)=2000
sn=n(8n-4)=4000
8n squared -4n-4000
quadratic formula so i got 23

Reply 8

ghostwalker

17

Original post by otrivine

ok i got for the first 3 terms 1,5,9
then i used the sum formula
sn=n/2(2+4n-4)=2000

OK up to here.

sn=n(8n-4)=4000

At this point you've multiplied by 4 on the LHS (you got rid of the 2 and multiplied the bracket by 2), and by 2 on the RHS.

Reply 9

otrivine

OP

14

Original post by ghostwalker

OK up to here.

At this point you've multiplied by 4 on the LHS (you got rid of the 2 and multiplied the bracket by 2), and by 2 on the RHS.

Yes is that correct? :smile:

Reply 10

ghostwalker

17

Original post by otrivine

Yes is that correct? :smile:

No.

What ever you do to one side of an equation, you must do to the other to maintain equality.

So, multiplying one side by 4, and the other by 2 isn't a valid option.

I'd suggest just simplifying the LHS and dividing what you have in the brackets by the 2.

Reply 11

otrivine

OP

14

Original post by ghostwalker

No.

What ever you do to one side of an equation, you must do to the other to maintain equality.

So, multiplying one side by 4, and the other by 2 isn't a valid option.

I'd suggest just simplifying the LHS and dividing what you have in the brackets by the 2.

So u mean multiply both sides by 2 but I did right?

Reply 12

ghostwalker

17

Original post by otrivine

So u mean multiply both sides by 2 but I did right?

No, you multiplied one side by 4.
Looking at one of two of your other threads, it looks like you're struggling with basic algebraic maniplulation; and that's an issue you really need to address.

I'll post the working for this one using two different methods.

Starting from

S_n=\frac{n}{2}(2+4n-4)=2000

Let's simplify that to

\frac{n}{2}(4n-2)=2000

--------------(X)

A)

If we multiply through by 2 on each side, we get:

n(4n-2)=4000

Then expanding brackets:

4n^2-2n=4000

And we can divide through by 2 and rearrange to get

2n^2-n-2000=0

B) Alternatively, from (X), we could have

divided the 2 into the brackets to get

n(2n-1)=2000

Expanding the brackets now, and rearranging we have, as before.

2n^2-n-2000=0

sequence

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