Question

If $a_{1}=0$ and $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are real numbers such that $\left|a_{i}\right|=\left|a_{i-1}+1\right|$ for all $i$ then the A.M. of the numbers $a_{1}, a_{2}, \ldots, a_{n}$ has value $x$ where

• A. $x \leq-\frac{1}{2}$
• B. $x \geq-\frac{1}{2}$
• C. $x<-\frac{1}{2}$
• D. None of these

Answer 1

Answer: (B)

We have, $\left|a_{i}\right|=\left|a_{i-1}+1\right|$
$\Rightarrow a_{i}^{2}=a_{i-1}^{2}+2 a_{i-1}+1$
Putting $i=1,2,3, \ldots, n+1$, we get
$a_{1}^{2}=0$
$a_{2}^{2}=a_{1}^{2}+2 a_{1}+1$
$a_{3}^{2}=a_{2}^{2}+2 a_{2}+1$
$\vdots \quad \vdots \quad \vdots$
$a_{n}^{2}=a_{n-1}^{2}+2 a_{n-1}+1$
$a_{n+1}^{2}=a_{n}^{2}+2 a_{n}+1$
On adding, we get $\displaystyle\sum_{i=1}^{n+1} a_{i}^{2}=\displaystyle\sum_{i=1}^{n} a_{i}^{2}+2 \displaystyle\sum_{i=1}^{n} a_{i}+n$
$\Rightarrow 2 \displaystyle\sum_{i=1}^{n} a_{i}=-n+a_{n+1}^{2} \geq-n$
$\Rightarrow \frac{a_{1}+a_{2}+\ldots+a_{n}}{n} \geq-\frac{1}{2}$
$ \Rightarrow x \geq-\frac{1}{2} .$