A problem by Aman Bansal
Level
pending
A single paper is torn from a novel. The sum of the page numbers on the remaining pages of the novel is 15000. Find the sum of page numbers on the torn paper.
The answer is 51.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
The given problem is clearly a Arithmetic Progression. so using the formula for sum of an AP ie n/2[2a+(n-1)d]
where a is first term, d is common difference, n is no of terms in AP.
Now, a=1 (first page of novel.)
d=1 (common difference )
n=?
so putting it in equation n[2 1 + (n - 1) 1]/2=15000( given sum of remaining pages)
it gives a quadratic equation: n^2 + n -30000=0
solving using dharacharya's formula
the value of n comes out to be ~172.70
ie the actual value of N is 173 (bcoz the n has to be whole no. and we know that pages were torn therefore we have to take the nearest whole no. of greater value)
now again putting it in sum of the AP formula, the sum comes out to be 15051
therefore the sum of torn pages is 5 1