Find the number of such sequences

Algebra Level 5

Let a 1 , a 2 , a 3 , a 4 , a 5 { a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 },{ a }_{ 5 } be a 5 term geometric sequence satisfying 0 < a 1 < a 2 < a 3 < a 4 < a 5 < 100 0<{ a }_{ 1 }<{ a }_{ 2 }<{ a }_{ 3 }<{ a }_{ 4 }<{ a }_{ 5 }<100 , where each term is an integer. The number of such 5 terms are


The answer is 8.

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1 solution

Let the numbers be a,ar, ar^{2},ar^{3},ar^{4}.---------------------------------------------- Now a[r^{4}]<100. and a>=1 So r^{4}<100 r<\sqrt{10}. Now r>1.(increasing G.P) Two cases : 1>r is rational : Subcase: r is integer. r=2 or 3. for r=2 : 16a<100 So a=1,2,3,4,5,6.[ 6G.Ps ] for r=3 : 81a<100 So a=1.[ 1G.P ] if r is not integer r=p/q p,q>1 (p,q)=1 if q=2. p>=3.[since r>1] a(p^{4}/q^{4}) is an integer So q^{4} |ap^{4} q^{4} |a [Since q^{4} and p^{4} are co-prime. for q=2. a=16k. ar^{4} <100 => p=3,a=16. So another G.P with a=16 r=3/2. q>=3 =>a>=81 so ar>100 no solution 2>r is irrational ar is irrational so contradiction. So in total \boxed{8 G.P}

What the hell man!!!!!! Please any body help me to latex. I wrote that.

Chandrachur Banerjee - 6 years ago

L e t t h e n u m b e r s b e a , a r , a r 2 , a r 3 , a r 4 . N o w a [ r 4 ] < 100. a n d a > = 1 S o r 4 < 100 r < 10 . N o w r > 1. ( i n c r e a s i n g G . P ) T w o c a s e s A : 1 > r i s r a t i o n a l : S u b c a s e : r i s i n t e g e r . r = 2 o r 3. f o r r = 2 : 16 a < 100 S o a = 1 , 2 , 3 , 4 , 5 , 6. [ 6 G . P s ] f o r r = 3 : 81 a < 100 S o a = 1. [ 1 G . P ] B : I f r i s n o t i n t e g e r r = p / q p , q > 1 , ( p , q ) = 1 i f q = 2. p > = 3. [ s i n c e r > 1 ] a ( p 4 / q 4 ) i s a n i n t e g e r S o q 4 a p 4 q 4 a [ S i n c e q 4 a n d p 4 a r e c o p r i m e . F o r q = 2. a = 16 k . a r 4 < 100 p = 3 , a = 16. S o a n o t h e r G . P w i t h a = 16 r = 3 / 2. q > = 3 = > a > = 81 s o a r > 100 n o s o l u t i o n 2 > r i s i r r a t i o n a l a r i s i r r a t i o n a l s o c o n t r a d i c t i o n . S o i n t o t a l 8 G . P I hope it is correct Latex. It is almost your coding with  ( and  ) added. I used the same method, but not as logical as yours. up voted. Let~ the~ numbers~ be~ a,~ar,~ ar^{2},~ar^{3},~ar^{4}.-\\----- Now~ a[r^{4}]<100.~ and~ a>=1 ~So~ r^{4}<100 r<\sqrt{10}.\\ Now~ r>1.~(increasing G.P) ~Two~ cases\\ A : 1>r~ is~ rational :\\ Subcase:~ ~r~ is ~integer.~ r=2~ or~ 3.\\ for~ r=2 : 16a<100~ So~ a=1,2,3,4,5,6.[6G.Ps]\\ for~ r=3 : ~~81a<100 ~So ~a=1.[1G.P]~ \\ ~~\\B:If~ r~ is~ not ~integer~ r=p/q \\p,q>1,~~ (p,q)=1 ~if~ q=2.~ p>=3.~[since r>1]~ a(p^{4}/q^{4})~ is~ an ~integer \\So~ q^{4} |ap^{4}~~ q^{4} |a [~Since ~q^{4}~ and~ p^{4}~ are~ co-prime.\\ For ~~q=2.~~ a=16k. ~~ar^{4} <100 \leq p=3,~a=16.\\ So~ another~ G.P~ with ~a=16~ r=3/2.~ q>=3 =>a>=81 ~\\so~ ar>100~ no~ solution~ \\2>r ~is ~irrational~ ar~ is~ irrational~ so~ contradiction~.\\~ So~ in~ total~\large \boxed{8 G.P}\\\text{I hope it is correct Latex. It is almost your coding with \ ( and \ ) added.}\\ \text{ I used the same method, but not as logical as yours. up voted.}

Niranjan Khanderia - 5 years, 11 months ago

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Thanks. R u really 87???

Chandrachur Banerjee - 5 years, 11 months ago

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Born 1928.

Niranjan Khanderia - 5 years, 11 months ago

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