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

We have, |ai|=|ai−1+1| ⇒   ai2=a1−12+2ai−1+1i=1,2,3,......, n+1,  we geta12=0a22=a12+2a1+1a32=a22+2a2+1……………..an2=an−12+2an−1+1an+12=an2+2an+1On adding, we get∑i=1n+1ai2=∑i=1nai2+2∑i=1nai+n⇒    2∑i=1nai=−n+an+12≥−n⇒     a1+a2+a3+........+ann≥−12, from the question  x=a1+a2+a3+........+ann⇒x≥−12