IIT JEE , AIEEE Forum - Mathematics , Algebra

djdylan2000

Find the number of Geometric Progressions having 3 terms, where all terms are natural numbers less than or equal to 100.

a)53

b)54

c)104

d)106

Please give the full solution+method that is reasonably fast.

Thank you

Greatdreams

It is given in the problem that all terms are

100

Now let us take the numbers ......

2,3,4,5,6,7,8,9,10......
So the terms of a G.P are n,2n,4n for common ratio = 2

similarly for others we have
n,3n,9n ; n,4n,16n ; n,5n,25n ....for common ratios respectively 3,4,5......

as it is said that the number cant be > 100,

for n =2...... 4n

100
thus no of terms = 25

for common ratio = 3
no of terms = 11

for common ratio = 4
no of terms = 6

for common ratio = 5
no of terms = 4

in these way

so no of geometric progressions = 25 + 11+ 6+4+2+2+1+1 = 53

edited.......

tarinbansal

The ans is 53.

The possible GPs are-

n,2n,4n=25 GPs for n=1,2,3,........25

n,3n,9n=11 GPs for n=1,2,3,.....11

n,4n,16n=6 GPs for n=1,2,3....6

n,5n,25n=4 GPs for n=1,2,3,4

n,6n,36n=2 GPs for n=1,2

n,7n,49n=2 GPs for n=1,2

n,8n,64n=1 GPs for n=1

n,9n,81n=1 GPs for n=1

n,10n,100n=1 GPs for n=1

Therefore total GPs=25+11+6+4+2+2+1+1+1=53

hsbhatt

If r>=2, then is the answer 74. The enumeration doesn't take long actually. Only numbers up to 25 and for many of them the no. of possible GPs is same

iitkgp_bipin

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

tarinbansal

I hav a doubt in this ques.
Why aren't we considering the GPs with common ratio=1.
There can be 100 such GPs.

srujana

@ tarin

if r=1 all the terms in the G.P would be equal......which means it is not a 'progression'.

and further the formula of sum upto n terms is not valid if r=1 as denominator would become 0

securitas2311

But I guess the number of GPs should double the no. you all are calculating, because 1,2,4 and 4,2,1 are different GPs as their common ratios are respectively 2 and 1/2.

Similarly, for all GPs (with common ratio r_i) ur considering, there is one corresponding GP each (with common ratio 1/r_i)...

So, answer must be 53 X 2 = 106.

tarinbansal

Hey, rightly said dude.
We forgot this thing.

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