Statement 1: The number of ways in which three distinct numbers can be selected from  the set $\{3^1,3^2,3^3,\dots$,  $3^{100},3^{101}\}$ so that they form a G.P. is 2500 .
Statement 2: If a, b, c are in A.P., then $3^a,3^b,3^c$  are in G.P.
 

Statement 2 is correct as when  $3^a,3^b,3^c$ are in G.P., we have  ${\left(3^b\right)}^2=\left(3^a\right)\left(3^c\right)\Rightarrow 2b=a+c\Rightarrow a,b,c$  are in A.P. Thus, selecting three numbers in G.P. from   $\{3^1,3^2,3^3,\dots ,3^{100},3^{101}\}$ is equivalent to selecting 3 numbers from $\{1,2,3,\dots ,101\}$  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 selection ways of 2 odd numbers is ${ }^{51}{C}_2$ . Number of selection ways of 2 even  numbers is ${ }^{50}{C}_2$ . Hence, total number of ways is  ${ }^{51}{C}_2{+}^{50}{C}_2=1275+1225=2500$.