a. If `T_(r+1)` is greatest , then `(n)/(1+|a/(bx)|) le r le (n+1)/(1+|a/(bx)|)` For `x = 1` and `r = 11, (n)/(1+3/5) le 11 le (n+1)/(1+3/5)` `rArr 5n le 88 le 5(n+1)` `rArr n = 17` (b) if `T(r+1)` is greatest ,then `((r)/(n-r+1)) a/b le x le ((r+1)/(n-r)) a/b` For `n = 25` and `r = 20, 2 lt x lt 63/25` c. We have `(T_(r+1))/(T_(r)) = (n-r+1)/(r). (bx)/(a)` for `(a+bx)^(n)` For `n = 20` and `r = 10, (11)/(10).(5x)/(3) = 1` or ` x= 6//11` d. we have `(n)/(1+|(a)/(bx)|) le r le (n+1)/(1+|a/(bx)|)` for `(a+bx)^(n)`. For `x = 2` and `r = 8` (as `T(9)` will be largest). `(n)/(1+|a/(bx)|) le 8 le (n+1)/(1+3/10)` `rArr 10n le 104 le 10(n+1)` `rArr n = 10`