1\leq a<ar<ar^2\leq40
Let first term of GP = a
Common ratio = r
Case I: If r is natural number.
If r=2
1\leq a<2a<4a\leq40
a∈{1,…,10}
If r=3
1\leq a<3a<9a\leq40
a∈{1,2,3,4}
If r=4
1\leq a<4a<16a\leq40
a∈{1,2}
If r=5
1\leq a<5a<25a\leq40
a∈{1}
If r=6
1\leq a<6a<36a\leq40
a∈{1}
1\leq a<ar<ar^2\leq40
Let first term of GP = a
Common ratio = r
Case I: If r is natural number.
If r=2
1\leq a<2a<4a\leq40
a∈{1,…,10}
If r=3
1\leq a<3a<9a\leq40
a∈{1,2,3,4}
If r=4
1\leq a<4a<16a\leq40
a∈{1,2}
If r=5
1\leq a<5a<25a\leq40
a∈{1}
If r=6
1\leq a<6a<36a\leq40
a∈{1}
Case II: If r is fraction > 1
If r=23
then a, 23a, 49a
Case II: If r is fraction > 1
If r=23
then a, 23a, 49a
A must be multiple of 4
a∈{4,8,12,16}
(4,6,9)
(8,12,18)
(12,18,27)
(16,24,36)
If r=25, then a, 25a, 425a
a=4
r=34
ar2=916a→a=9k
(9,12,16), (18,24,32)
... (GP)
r=35
ar2=925a;a=9k
r=45
(9,15,25)
…(I)GP
ar2=1625a;a=16k
(16,20,25)
…(I)GP
r=56
ar2=2536a;a=25k
(25,30,36)
…(I)GP
Total =18+10=28
P=4028C33=988028=24707
m+n=2477
