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Question
Assertion :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. Reason: If a, b, c are in A.P., then 3a,3b,3c are in G.P.
Assertion :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. Reason: If a, b, c are in A.P., then 3a,3b,3c are in G.P.
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Solution
The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
If we select 3 terms whose indices are in AP, we will get 3 corresponding terms in GP.
The indices are natural numbers from 1 to 101.
There are 50 even and 51 odd numbers.
If a, b, c are in AP, a + c = 2b = even. Thus, a dn c are either both even or both odd.
We select 2 even numbers in 50C2=1225 ways or select 2 odd numbers in 51C2=1275 ways.
Thus, total ways = 1225+1275=2500.
Hence, (A) is correct.
If we select 3 terms whose indices are in AP, we will get 3 corresponding terms in GP.
The indices are natural numbers from 1 to 101.
There are 50 even and 51 odd numbers.
If a, b, c are in AP, a + c = 2b = even. Thus, a dn c are either both even or both odd.
We select 2 even numbers in 50C2=1225 ways or select 2 odd numbers in 51C2=1275 ways.
Thus, total ways = 1225+1275=2500.
Hence, (A) is correct.
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