# Thread: Finding the sum of the series from n +1 to 49

1. ## Finding the sum of the series from n +1 to 49

Question:
The sum of the integers from 1 to n-1 is equal to the sum of integers from the n + 1 to 49. Find n

I am able to find 1 to n - 1 which is
$s_{n-1}= \frac{1}{2}(n(n-1))$

but for (n+1) + (n+2) + ....... + 49

I try this: $(n + 1) + (n+2) + (n+3) + ....... + 49$
a = n+1 common difference = 1
nth term = 2n
thus
49th term $= 2(49) = 98$

let n = 1
first term = 2(1) = 2

thus sum of n+1 to 49

$s_{49}= \frac{49}{2}(2(2)+(49-1)(1)) = 49(26)=1274$
thus
$\frac{1}{2}(n(n-1)) = 1274$

thus n = -49.9801....... or n = 50.980...
but n is an integer
please can some point me in the right direction
how do I found the first term, common difference and number of terms for the series from (n+1) to 49

2. ## Re: Finding the sum of the series from n +1 to 49

what on earth.... :O

$\displaystyle \sum_1^{n-1} ~k = \dfrac{n(n-1)}{2}$

$\displaystyle \sum_1^{49} ~k = \dfrac{(49)(50)}{2} = 1225$

$\displaystyle \sum_1^{n} ~k = \dfrac{n(n+1)}{2}$

$1225 = \dfrac{n(n+1)}{2} + \displaystyle \sum_{n+1}^{49} ~k$

if $\displaystyle \sum_1^{n-1} ~k = \displaystyle \sum_{n+1}^{49} ~k$

$1225 = \dfrac{n(n+1)}{2} + \dfrac{n(n-1)}{2} = \dfrac{n}{2}(n+1+n-1) = n^2$

$n = \sqrt{1225} = 35$

3. ## Re: Finding the sum of the series from n +1 to 49

[1+2+3+....+(n-1)] n [(n+1)+(n+2)+....+49]

(n-1)n/2 = 49(50)/2 - n(n+1)/2 : solve for n

aNUTter way to look at Romsek's...

4. ## Re: Finding the sum of the series from n +1 to 49

Originally Posted by bigmansouf
Question:
The sum of the integers from 1 to n-1 is equal to the sum of integers from the n + 1 to 49. Find n
$\sum\limits_{k = 1}^{49} k = \frac{{49 \cdot 50}}{2} = 49 \cdot 25$
If $1<J<49,$ then $\boxed{\sum\limits_{k = 1}^{J-1} k }+J+\boxed{\sum\limits_{k =J+ 1}^{49} k }$.
We are given that the two boxed sums are equals thus that gives:
$2\dfrac{(J-1)(J)}{2}+J=49 \cdot 25$ This gives the solution $N=35$