Question

Let $a_1=0$ and $a_1$, $a_2$, $a_3$, ..., $a_n$ be real numbers such that $\left|a_i\right|=|a_{i-1}+1|$ for all i then the AM of the numbers $a_1$, $a_2$, $a_3$, ..., $a_n$ has the value A where

• A. $A<-\frac{1}{2}$
• B. $A<-1$
• C. $A\ge -\frac{1}{2}$
• D. $A=-\frac{1}{2}$

Answer 1

The correct option is C $A\ge -\frac{1}{2}$

We know that $a_1=0$

Now

$|a_i|=|a_{i-1}+1|$

Considering

$a_i=a_{i-1}+1$

Hence

$a_2=1$

$a_3=2$

$a_4=3$ and so on

$a_n=n-1$

Hence

$\sum a_i$$=1+2+3+...(n-1)$
$=\frac{(n-1).n}{2}$.

Hence A.M

$=\frac{(n-1)n}{4}$ ...(i)

Considering $a_i=-(a_{i-1}+1)$, we get

$a_2=-1$

$a_3=0$

$a_4=-1$

$a_5=0$ and so on

$a_n=0$ if n is odd else $a_n=-1$

Hence if n is even

$\sum a_i$$=-\frac{n}{2}$

And if n is odd.

Then $n-1$ is even and hence

$\sum a_i$$=\frac{-(n-1)}{2}$.

Hence

$AM=\frac{-n}{4}$ if n is even or $\frac{-(n-1)}{4}$

Hence

$A=\frac{n(n-1)}{4}$ if $a_i=a_{i-1}+1$

or

If $a_i=-(a_{i-1}+1)$

And n is even then

$A=\frac{-n}{4}$

Else if n is odd

$A=\frac{-(n-1)}{4}$.

Now the minimum value of $n$ has to be 2.

Therefore the minimum value of $A$

$=\frac{-n}{4}$

$=\frac{-1}{2}$

$\because n=2$

Hence

$A\ge \frac{-1}{2}$.