Find the number of ways in which 3 distinct numbers can be selected from the set `{3^(1),3^(2),3^(3),..,3^(100),3^(101)}` so that they form a G.P.

Answer

2500

Explanation

Let three numbers selected be `3^(a),3^(b),3^(c )` which are in G.P.
`therefore (3^(b))^(2)=(3^(a))(3^(c ))`
`implies 2b=a+c`
`implies a,b,c ` are in A.P.
Thus, selecting three number in G.P. from given set is equivalent to selecting 3 numbers from {1,2,3,..,101} which are in A.P. Now, a,b,c are in A.P. if either a and c are odd or a and c are even. Number of ways of selecting two odd numbers is `. ^(51)C_(2)` and those of selecting two even numbers is `.^(50)C_(2)`.
Onece a and c are selected, b is fixed.
Hence total number of ways `= .^(51)C_(2)+ .^(50)C_(2)=1275+1225=2500`