Statement 1: The number of ways in which three distinct numbers can be selected from the set {31,32,33,…, 3100,3101} so that they form a G.P. is 2500 .Statement 2: If a, b, c are in A.P., then 3a,3b,3c are in G.P.

Statement 2 is correct as when 3a,3b,3c are in G.P., we have (3b)2=(3a)(3c)⇒2b=a+c⇒a,b,c are in A.P. Thus, selecting three numbers in G.P. from {31,32,33,…,3100,3101} 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 selection ways of 2 odd numbers is 51C2 . Number of selection ways of 2 even numbers is 50C2 . Hence, total number of ways is 51C2+50C2=1275+1225=2500.

Statement 1: The number of ways in which three distinct numbers can be selected from the set {31,32,33,…, 3100,3101} so that they form a G.P. is 2500 .Statement 2: If a, b, c are in A.P., then 3a,3b,3c are in G.P.

Statement 2 is correct as when 3a,3b,3c are in G.P., we have (3b)2=(3a)(3c)⇒2b=a+c⇒a,b,c are in A.P. Thus, selecting three numbers in G.P. from {31,32,33,…,3100,3101} 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 selection ways of 2 odd numbers is 51C2 . Number of selection ways of 2 even numbers is 50C2 . Hence, total number of ways is 51C2+50C2=1275+1225=2500.

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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.

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STATEMENT 1 is TRUE and STATEMENT 2 is FALSE.

Both the statements are TRUE but STATEMENT 2 is NOT the correct explanation of STATEMENT 1.

Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1.

STATEMENT 1 is FALSE and STATEMENT 2 is TRUE.

answer is A.

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Detailed Solution

Statement 2 is correct as when $3^{a}, 3^{b}, 3^{c}$ are in G.P., we have ${(3^{b})}^{2} = (3^{a}) (3^{c}) \Rightarrow 2 b = a + c \Rightarrow a, b, c$ 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 $^{51} C_{2} +^{50} C_{2} = 1275 + 1225 = 2500$ .