Sequences

We have been given this problem and it has caused some argument between my group.
\sum _N^{2N}U_n=2750\:
Some of us have solved with 2N and got a 14 for N, and some have just used N and got 20. Anyone got any ideas?

You kinda need to define $U_n$ first lol...

We haven't been given U_N

Uhhhh can you present it in a clearer form pls?

Then you can't possibly answer this question.

What I did was use S_n = n/2 {2a+(n-1)d} and used n = 2N, I forgot to mention we are given a = 5 and d = 3 oops

N = 0
U_0 = 2750

Do I get a prize?

So then you DO have $U_n = 5+3(n-1)$ defined...

Well ok yeah you can use that formula for sure, but first you need to split

$\displaystyle \sum_{n=N}^{2N} U_n = \sum_{n=1}^{2N} U_n - \sum_{n=1}^{N-1} U_n$

and use it on both of them. Then simplify and solve for $N$.

unfortunately not

So my working of $S_n = 2N/2(2a+(n-1)d)$
$2750 = 2N/2(10+6N-3)[br]5500 = 2N(20+12N-6)[br]5500 = 40N+28N^2-12N[br]0 = 28N^2 + 20N - 5500$
Solve for N with the quadratic formula
$N = 14.56... or N = -15.86$
but $N$ cannot be -ve so $N = 14$
Im currently doing a level and i havent been taught the way you described

So my working of $S_n = 2N/2(2a+(2N-1)d)$
$2750 = 2N/2(10+6N-3)[br]5500 = 2N(20+12N-6)[br]5500 = 40N+24N^2-12N[br]0 = 24N^2 + 28N - 5500$
Solve for N with the quadratic formula
$N = 14.56... or N = -15.86$
but $N$ cannot be -ve so $N = 14$
Im currently doing a level and i havent been taught the way you described

What I did was use S_n = n/2 {2a+(n-1)d} and used n = 2N, I forgot to mention we are given a = 5 and d = 3 oops

That's wrong because $S_{2N}$ works when you are counting up from the 1st term to the (2N)th term.
But the sum presented to you starts from the nth term and goes to the (2n)th term.

Hence motivating the split I presented.

You could have mentioned you are talking about an arithmetic series too. Defining variables without context is bad practice. Using "a" as the first value (which I think you have misunderstood, more on this later) and "d" as the difference is not how everyone does it everywhere. I'm going to guess that you mean $U_n=5+3(n-1)$.

Write out the series:

(5+3(N-1))+(5+3(N))+(5+3(N+1))+...+(5+3(2N-3))+(5+3(2N-2))+(5+3(2N-1))

This is an arithmetic series where the first term is 5+3(N-1) and the common difference is 3. So you should be using d=3 and a=5+3(N-1) instead of a=5 in your formula. Splitting sums is optional, you can just deal directly with a single arithmetic series.

As I said, i am only 4 week into year 13 and so we havent been taught how to do that split method. I my teacher wouldnt set a problem without us having the knowledge to solve it, is there any other method to solving this?

So i could use $S_n = \frac{2N or N?}{2}(2(5+3(n-1))+3(N-1))$?

It's the simplest approach and really simple to understand.