Let a1=0 and a1,a2,a3,…,an be real numbers such that ai=ai−1+1 for all i, then the A.M. of the numbers a1,a2,a3,…,an has the value A where

A
$A < - \frac{1}{2}$
B
$A < - 1$
C
$A \geq - \frac{1}{2}$
D
$A = - \frac{1}{2}$

Solution:

We have,

$|a_{i}| = |a_{i - 1} + 1|$ for all $i = 2, 3, \dots n + 1$

$\begin{array}{l} \Rightarrow a_{i}^{2} = {(a_{i - 1} + 1)}^{2} \\ \Rightarrow a_{i}^{2} - a_{i - 1}^{2} = 2 a_{i - 1} + 1 \\ \Rightarrow \sum_{i = 2}^{n + 1} a_{i}^{2} - a_{i - 1}^{2} = \sum_{i = 2}^{n + 1} (2 a_{i - 1} + 1) \\ \Rightarrow a_{n + 1}^{2} - a_{1}^{2} = 2 (a_{1} + a_{2} + \dots + a_{n}) + n \\ \Rightarrow a_{n + 1}^{2} = 2 n A + n [∵ A = \frac{a_{1} + a_{2} + \dots + a_{n}}{n} and a_{1} = 0] \\ \Rightarrow A = \frac{1}{2 n} (a_{n + 1}^{2} - n) \Rightarrow A = \frac{a_{n + 1}^{2}}{2 n} - \frac{1}{2} \Rightarrow A \geq - \frac{1}{2} \end{array}$