The middle number of a finite sequence of consecutive integers is 2043. The sum of the numbers before 2043 is twice the sum of those after 2043. How many integers are in the sequence? Thank you.
+14913 How many integers are in the sequence?
Hello Guest!
\(\dfrac{a+z}{2}=2043\\ z=4086-a\\ 2042-a+1=2(z-2044+1)\\ 2042-a+1=2(4086-a-2044+1)\\ 2042-a+1=8172-2a-4088+2\\ \color{red}a=2043\\ \color{red}z=2043 \)
There are the same number of integers before and after the middle number.
The task cannot be solved.
The "finite sequence" consists of only one integer.
!
asinus: Don't understand the solution below yours? proceed as follows:
Simplify equation 1 as follows:
(n + 2042)/2 (2042 - n + 1) = (2042 - n + 1) (2042 - n + 2)
-1/2 (n - 2043) (n + 2042) = (n - 2044) (n - 2043)
-(3 n^2)/2 + (8175 n)/2 - 2089989 = 0
-3/2 (n - 2043) (n - 682) = 0
Divide both sides by - 3/2
(n - 2043) (n - 682) ==0
n ==2043, or n == 682 - this is the first term of the first half of the sequence
Do the same with equation 2 and should get:
m ==3,404 - this is the last term of the 2nd half of the sequence.
+14913 Hello Guest from Answer #2 #3!
The two parts of your sequence
682...2043...3404
are the same length.
However, the sum of the first part of the sequence should be twice as long as the second part of the sequence.
!
Hi asinus: The question is NOT accurately stated !! It has been reversed. You are absolutely right that there is no solution the way the question is transposed! The question should read:
The middle number of a finite sequence of consecutive integers is 2043. The sum of the numbers AFTER 2043 is twice the sum of those BEFORE 2043. How many integers are in the sequence? Thank you.
This is the question I solved !!! I automatically assumed that the question was reversed!.
