# algebra

• June 26th 2009, 09:29 AM
abhishek arora
algebra
two consecutive no's outof 1,2,3,...,n-1,n are removed. the average of remaining no's is 105/4.find n and the removed numbers
• June 26th 2009, 09:52 AM
pankaj
Quote:

Originally Posted by abhishek arora
two consecutive no's outof 1,2,3,...,n-1,n are removed. the average of remaining no's is 105/4.find n and the removed numbers

Let the removed numbers be $p$ and $p+1$

$\frac{(1+2+...+n)-(p+p+1)}{n-2}=\frac{105}{4}$

$\frac{\frac{n(n+1)}{2}-(2p+1)}{n-2}=\frac{105}{4}$

$8p=2n^2-103n+206$

Now $1\leq p\leq n-1$

$8\leq 8p\leq 8n-8$

$8\leq 2n^2-103n+206\leq 8n-8$

On solving we have two inequalities

$2n^2-103n+198\geq 0;(n-2)(2n-99)\geq 0;n\geq \frac{99}{2};n\geq 50$

Since $n\geq 2$ and $n$ must be an integer.

Similarly solving the inequality, $2n^2-111n+214\leq 0; (n-2)(2n-107)\leq 0$ and thus $n\leq 53$

Therefore, $50\leq n\leq 53$

$n=50,51,52,53$

It is only for $n=50$ that p is an integer.Thus $p=7$

Therefore the number of pages in the book are $50$ and the removed pages are $7$ and $8$
• June 26th 2009, 10:01 AM
abhishek arora
thanks
thank you pankaj sir u are like an angel for me