1 Answers
Lokesh Verma
·2009-03-09 23:01:46
\text{this is same as }\\ \sum_{i=1}^{n-1}{(a_{i}x+a_{i-1})}^2\leq 0
\text{Which is true only when }\\ {(a_{i}x+a_{i-1})}^2 = 0 \text{ for all i} \\ a_{i}x+a_{i-1}=0 \\ so \frac{a_{i}}{a_{i-1}}=\frac{-1}{x}\\ \text{Hence it is in GP}