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 Arithmetic and Geometric Addle (Posted on 2014-10-10)
(A) Find two positive integers x and y, such that:
(i) y divides x, and:
(ii) x/y, x+y and x*y are in arithmetic sequence.

(B) Keeping all the other conditions in (A) unaltered, what would the value of x and y be, if clause (ii) is rewritten as:
x/y, x-y and x+y are in geometric sequence?

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution | Comment 2 of 3 |
Let x/y=k, and y=kx.

(A) If a,b,c form an arithmetic sequence then a+c=2b. Then

x/y+xy=2(x+y)   =>   k+ky^2=2y(k+1)

Solve for k:

k=2y/(y-1)^2=2/(y-1)*y/(y-1)

k is integer only if y=2, then k=4, x=8

Arithmetic sequence: 4, 10, 16

(B) If a,b,c form a geometric sequence then ac=b^2.  Then

x/y*(x+y)=(x-y)^2   =>   ky(k+1)=y^2(k-1)^2   =>

k(k+1)=y(k-1)^2

Solve for y

y=k(k+1)/(k-1)^2

y is integer only if

1) k=2, then y=6 and x=12

Geometric sequence: 2, 6, 18

2) For k>2, k and k-1 are relatively prime, so (k+1)/(k-1)^2
has to be integer, only solution is k=3, and y=3, x=9

Geometric sequence: 3, 6, 12
 Posted by Art M on 2014-10-10 22:17:10

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