STATEMENT 1: The number of ways in which three distinct numbers can be selected from the set $\{3^1,3^2,3^3,{\mathrm{....3}}^{100},3^{101},\}$ 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.

Sol: 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 $\{3^1,3^2,3^3,\mathrm{...},3^{100}\}$ is equivalent to selecting 3 numbers from $\{1,2,3,\mathrm{...},101\}$ which are in A.P. Now, $a,b,c$ are in A.P. if either $aandc$ are odd or $aandc$ 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 ${ }^{51}{C}_2{+}^{50}{C}_2=1275+1225=2500.$