Let d(n) denote the number of divisors of n including 1 and itself. Then `d(225), d(1125) and d(640)` are (a) In AP (6) In HP(c) In GP (d) Consecutive integers

Let d(n) denotes the number of divisors of n including 1 and itself. Then d(225), d(1125) and d(640) are

Let d(n) denote the number of divisors of n inciuding 1 and iteslf . Then d(225), d(1125) and d(640) are

Let d(n) denote the number of positive divisors of a positive integer n. Which of the following are correct? 1. d(5) = d(11) 2. d(5).d(11) = d(55) 3.d(5) + d(11) = d(16) Select the correct answer using the code given below :

For a positive integer n, define d(n) - the number of positive divisors of n. What is the value of d(d(d(12))) ?

For any integer nge1 , the number of positive divisors of n is denoted by dựn). Then for a prime p, d(d(d(p^(7)))) is

For any integer n ge1 , the number of positive divisors of n is denoted by d(n). Then for a prime P, d(d(d(p^(7))) is equal to

Let the positive numbers a ,b ,ca d nd be in the A.P. Then a b c ,a b d ,a c d ,a n db c d are a. not in A.P. /G.P./H.P. b. in A.P. c. in G.P. d. in H.P.

Let the positive numbers a ,b ,ca d nd be in the A.P. Then a b c ,a b d ,a c d ,a n db c d are a. not in A.P. /G.P./H.P. b. in A.P. c. in G.P. d. in H.P.

Let the positive numbers a,b,c,d be in AP. Then abc,abd,acd,bcd are (2001,1M) not in AP/GP/HP ( b ) in AP in GP (d) in HP