The 20 leaves of a 40-page book are numbered 1 through 40. The first and last leaves are ripped out. What is the sum of the remaining page numbers?
So what is the sum of 3+4+5+...37+38 ?
It is an arithemetic sum, first term 3, d=1
http://en.wikipedia.org/wiki/Arithmetic_progression
n= 36 terms
Sum= n/2 (2a+(n-1)d)
I got 738
738
Using the formula of an Arithmetic Progression, we can solve the answer.
Since 1+2+3+4+...+37+38+39+40 = 41*20 = 820,
the answer is 820-40-39-2-1 = 738.
(If there are N pages in a book, there are N/2 leaves)
Hope I helped!
Milowen
To find the sum of the remaining page numbers, we first need to determine which pages are left after the first and last leaves are ripped out.
Since the book has 40 pages, the first leaf would be page 1 and the last leaf would be page 40.
When both the first and last leaves are ripped out, we are left with a book that has 38 pages. However, the pages are still numbered from 1 to 40. This means that pages 1 and 40 are no longer part of the book, but all the other pages remain.
The sum of the remaining page numbers can be calculated by adding up all the page numbers from 2 to 39 (since 1 and 40 are excluded).
To find the sum of consecutive numbers, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(a + l)
Where:
Sn is the sum of the series,
n is the number of terms in the series,
a is the first term, and
l is the last term.
In this case, the first term (a) is 2 and the last term (l) is 39. Substituting these values into the formula, we can calculate the sum:
Sn = (38/2)(2 + 39)
= 19 * 41
= 779
Therefore, the sum of the remaining page numbers is 779.