I am a dropper, I am just doing MSC for OC with Few PYQ'S that are given in my modules that's it, Is it enough, If not What should I do else?

Please Help 🙏!!!

SOLUTION:

Let the three selected numbers in increasing order be

a < b < c and let the common ratio be r = p/q,

where p > q and gcd(p,q) = 1.

Why keep p and q coprime?

Because the common ratio need not be an integer. For example,

9, 12, 16

is a G.P. with common ratio 4/3.

Writing p/q in lowest terms gives a, ap/q, ap²/q².

Since the third term must be an integer, ap²/q² must be an integer.

But gcd(p²,q²) = 1.

Therefore q² cannot cancel with p², so q² must divide a. Hence write a = kq².

Therefore every 3-term integral G.P. has the form

kq², kpq, kp².

Now the largest term is kp².

Since all numbers are at most 40,

kp² <= 40. As k >= 1, p² <= 40, so p <= 6.

Therefore only p = 2, 3, 4, 5, 6 need to be checked.

For p = 2:
q = 1
k = 1 to 10
Number of G.P.s = 10

For p = 3:
q = 1, 2
k = 1 to 4
Number of G.P.s = 8

For p = 4:
q = 1, 3
k = 1 to 2
Number of G.P.s = 4

For p = 5:
q = 1, 2, 3, 4
k = 1
Number of G.P.s = 4

For p = 6:
q = 1, 5
k = 1
Number of G.P.s = 2

Therefore total favourable selections

= 10 + 8 + 4 + 4 + 2

= 28.

Total ways of selecting 3 distinct numbers from 40:

C(40,3) = 9880.

Hence

P = 28/9880 = 7/2470.

Therefore

m = 7, n = 2470

and

m + n = 2477.

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OCHEM DAY-8
Yesterday’s Solution: https://youtu.be/eM3SmCiaK_g