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The positive integers A, B, and C form an arithmetic sequence while the integers B, C, and D form a geometric sequence. If (C/B) = (5/3), what is the smallest possible value of A + B + C + D?

Guest Mar 21, 2018

Best Answer 

 #2
avatar+92206 
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The positive integers A, B, and C form an arithmetic sequence while the integers B, C, and D form a geometric sequence. If (C/B) = (5/3), what is the smallest possible value of A + B + C + D?

 

arithmetic progression

A

B, 

C= A+2(B-A)

 

 

geometric progression

B      r=C/B=5/3

B, B(5/3), B(5/3)^2

B,  5B/3,     25B/9

so

C=5B/3

D=25B/9

 

A, B, 5B/3, 25B/9 

 

B - A = 5B/3 - B

2B-5B/3 = A

(2-5/3)B = A

A=B/3

 

\(A, B,C, D \\ \frac{B}{3},\;B, \;\frac{5B}{3},\;\frac{25B}{9} \)

 

These all have to be positive integers so B must be a multiple of 9, The smallest values are if B is 9

 

\( \frac{9}{3},\;9, \;\frac{45}{3},\;\frac{9*25}{9}\\ 3,\;9, \;15,\;25\\ \)

 

So the smallest possible value for A+B+C+D = 3+9+15+25  = 52

Melody  Mar 22, 2018
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3+0 Answers

 #1
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AP = 1, 3, 5...............etc.

GP = 3, 5, 8 1/3.........etc.

A + B + C + D =1 + 3 + 5 + 8 1/3 = 17 1/3

Guest Mar 21, 2018
 #2
avatar+92206 
+2
Best Answer

The positive integers A, B, and C form an arithmetic sequence while the integers B, C, and D form a geometric sequence. If (C/B) = (5/3), what is the smallest possible value of A + B + C + D?

 

arithmetic progression

A

B, 

C= A+2(B-A)

 

 

geometric progression

B      r=C/B=5/3

B, B(5/3), B(5/3)^2

B,  5B/3,     25B/9

so

C=5B/3

D=25B/9

 

A, B, 5B/3, 25B/9 

 

B - A = 5B/3 - B

2B-5B/3 = A

(2-5/3)B = A

A=B/3

 

\(A, B,C, D \\ \frac{B}{3},\;B, \;\frac{5B}{3},\;\frac{25B}{9} \)

 

These all have to be positive integers so B must be a multiple of 9, The smallest values are if B is 9

 

\( \frac{9}{3},\;9, \;\frac{45}{3},\;\frac{9*25}{9}\\ 3,\;9, \;15,\;25\\ \)

 

So the smallest possible value for A+B+C+D = 3+9+15+25  = 52

Melody  Mar 22, 2018
 #3
avatar+85673 
0

Nice, Melody  !!!

 

cool cool cool

CPhill  Mar 22, 2018

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