Bunuel wrote:
What is 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 geometric progression ? (A geometric progression, also known as a geometric sequence, is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.)
A. 500
B. 625
C. 1250
D. 1500
E. 2500
Are You Up For the Challenge: 700 Level Questions: 700 Level QuestionsA GP is a series where each term is multiplied by the same ratio. For example 3,6,12..., here the ratio is 2.
Now if we are looking at GP with three terms of power of 3, the power of middle term should be average of the outer terms. For example 3^1, 3^9,3^17...2*9=1+17.
This means we look for two outer terms that can have a middle term too.
For this it is important that the outer terms are of same type, when it comes to even or odd.
Two even powers will always have a middle term, 3^2a and 3^(2a+6) will have 2^(2a+3), and two odd powers will also have a middle term, 3^(2a+1) and 3^(2a+7) will have 3^(2a+4) as the middle term.
Choose two even powers from 50 even numbers (2,4,...100) in 50C2 ways =50*49/2=25*49
Choose two odd powers from 51 odd numbers (1,3,5...101) in 51C2 ways = 51*50/2=25*51
Total = 25*49+25*51=25(49+51)=25*100=2500
E
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