goformit100
  • goformit100
Find all triples of natural numbers (a,b,c) such that a, b and c are in geometric progession, and a+b+c=111.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathslover
  • mathslover
If a,b , c are in GP then : b = ar c = br \(b^2 = ac\) --(1)
goformit100
  • goformit100
ok
mathslover
  • mathslover
We have to find, all the solns for a, b and c such that a, b and c are in GP , right?

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More answers

goformit100
  • goformit100
yes..
mathslover
  • mathslover
Any ideas?
mathslover
  • mathslover
@UnkleRhaukus @.Sam. @Callisto @mathstudent55 - any of u can help?
goformit100
  • goformit100
No idea but can we solve it using pascal triangle ?
mathslover
  • mathslover
It is for determining Binomial coefficients... what is the use of binomial theorem related topics in this question?
goformit100
  • goformit100
ok
mathslover
  • mathslover
I got something : \(\bf{a + c = 111 - \sqrt{ac} \\ (a+c)^2 = (111-\sqrt{ac})^2 }\)
goformit100
  • goformit100
Ya ok
mathslover
  • mathslover
\(\cfrac{111}{2} \ge b^{\frac{3}{2}} \)
goformit100
  • goformit100
ok
goformit100
  • goformit100
@UnkleRhaukus
mathslover
  • mathslover
@hartnn
mathslover
  • mathslover
Correction : \(\cfrac{a+b+c}{3} \ge( {\sqrt{abc}})^{\frac{1}{3}}\)
goformit100
  • goformit100
ok
mathslover
  • mathslover
37+37+37 = 111 and they all r in GP too :P :)
mathslover
  • mathslover
a=b=c=37
goformit100
  • goformit100
Ok Thank you
ganeshie8
  • ganeshie8
1, 10, 100
mathslover
  • mathslover
^
mathslover
  • mathslover
Many numbers are there
mathslover
  • mathslover
\(a(1+r+r^2) = 3 *37\) a = 3 , then 1 + r + r^2 = 37 similarly factorize 111 and do that
ganeshie8
  • ganeshie8
yea and this is a olympiad problem it wud be tough
mathslover
  • mathslover
Yep.
ganeshie8
  • ganeshie8
thats a brilliant logic :)
goformit100
  • goformit100
Thank you for the help :)
ganeshie8
  • ganeshie8
111, -111, 111
mathslover
  • mathslover
@ganeshie8 , -111 is not a natural number. So, 111,-111,111 is not possible. We only have 2 sets of solns : (37,37,37) and (1,10,100) \(a+ar + ar^2 = 111\) \(a(1 + r + r^2) = 111\) Case 1 : \(a(1+r+r^2) = 1 \times 111\) a = 1 or \(1 + r + r^2 = 1 \) and a = 111 or \(1+ r + r^2 = 111\) \(1 + r + r^2 \) can not be equal to 1. So, a = 1 and \(1 + r + r^2 = 111\) \(1 + r + r^2 = 111\) \(r^2 + r -110 = 0\) This gives, r = 10 (r can not be negative) If r = 10, then we have : a = 1 , b = ar = 10, c = ar^2 = 100 \(\bf{(1,10,100)}\) Case 2 : \(a(1+r+r^2) = 3 \times 37\) Either a =3 or \(1 + r+r^2 =3\) and a = 37 or \(1 + r + r^2 = 37\) \(1 + r +r^2 \ne 37\) as , r can not be zero. Therefore, a = 37 and \(1 + r + r^2 =3\) \(r^2 + r + 1 = 3\) Which gives, r = 1 (r can not be negative) Therefore , we get : a = 37, b = ar = 37 , c = ar^2 = 37 \(\bf{(37,37,37)}\)

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