Question

If $a_1=0$ and $a_1,a_2,a_3,\dots ,a_n$ 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 $a_1,a_2,\dots ,a_n$ has value $x$ where

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

Answer 1

Answer: (B)

We have, $\left|a_i\right|=\left|a_{i-1}+1\right|$
$\Rightarrow a_i^2=a_{i-1}^2+2a_{i-1}+1$
Putting $i=1,2,3,\dots ,n+1$, we get
$a_1^2=0$
$a_2^2=a_1^2+2a_1+1$
$a_3^2=a_2^2+2a_2+1$
$⋮⋮⋮$
$a_n^2=a_{n-1}^2+2a_{n-1}+1$
$a_{n+1}^2=a_n^2+2a_n+1$
On adding, we get $\sum _{i=1}^{n+1}{a}_i^2=\sum _{i=1}^n{a}_i^2+2\sum _{i=1}^n{a}_i+n$
$\Rightarrow 2\sum _{i=1}^n{a}_i=-n+a_{n+1}^2\ge -n$
$\Rightarrow \frac{a_1+a_2+\dots +a_n}{n}\ge -\frac{1}{2}$
$\Rightarrow x\ge -\frac{1}{2}.$