IIT JEE , AIEEE Forum - Mathematics , Algebra

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Since terms of GP are natural numbers, common ratios would also be natural numbers.

Lets take example of common ratio, r=2 and let the 1st term be a.

GP : a,2a,4a

Possible GPs : (1,2,4);(2,4,8);(3,6,12);........(25,50,100)

least value of a is 1 and maximum value is 4a=100 for which a=25.

so with common ratio 2, 25 GPs are possible (a varies from 1 to 25)

For common ratio 3 : GP is a,3a,9a

Max value of a is given by : 9a=100 for which a=11.11

since a is an integer we must take its integral value, so a=11 (a varies from 1 to 11)

Similarly when ratio is 4 : max(a) = [100/4²] = 6 ( a varies from 1 to 6)

And least value of a will always remain 1.

So we can deduce a formula :

[100/2²] + [100/3²] + [100/4²] + [100/5²] + ........

when denominator exceeds 10² terms after that are less than 1 and their integral values are 0.

so no. of possible GPs = 25+11+6+4+2+2+1+1+1 = 53

Editing : Similarly reverse of these GPs are also possible (4,2,1)......

so total possible GPs are 53x2 = 106

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