Let `a_1=0` and `a_1,a_2,a_3` …. , `a_n` be real numbers such that `|a_i|=|a_(i-1) + 1|` for all I then the A.M. Of the number `a_1,a_2 ,a_3`…., `a_n` has the value A where : (a) `A lt -1/2` (b) `A lt -1` (c) `A ge -1/2` (d) A=-2

Options

A. `Alt-(1)/(2)`
B. `Alt-1`
C. `Age-(1)/(2)`
D. `A=-(1)/(2)`

Answer

C

Explanation

We have,
`|a_(i)|=|a_(i-1)+1|` for all i=2,3, . . . . n+1
`rArr" "a_(i)^(2)=(a_(i-1)+1)^(2)`
`rArr" "a_(i)^(2)-a_(i-1).^(2)=2_(i-1)+1`
`rArr" "underset(i=2)overset(n+1)suma_(i)^(2)-a_(i-1^(2))=underset(i=2)overset(n+1)sum(2a_(i-1)+1)`
`rArr" "a_(n+1).^(2)-a_(1)^(2)=2(a_(1)+a_(2)+ . . .. +a_(n))+n`
`rArr" "a_(n+1).^(2)=2nA+n" "{:[becauseA=(a_(1)+a_(2)+ . . . .+a_(n))/(n)anda_(1)=0]:}`
`rArr" "A=(1)/(2n)(a_(n+1).^(2)-n)rArrA=(a_(n+1))/(2n).^(2)-(1)/(2)rArrAge-(1)/(2)`