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What is the greatest three-digit number "abc'' such that \(4, a, b\) forms a geometric sequence and \(b, c, 5\) forms an arithmetic sequence?

 May 5, 2019
 #1
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c-b=5-c

2c+b=5

a*a/4=b

2c+a^2 /4=5

we try to make a as big as possible while still being an interger

a = 4 

c is not interger

a=2 is the biggest 

so c=2

and b=1

 May 5, 2019
 #2
avatar+101872 
+1

Here's my best attempt....

 

4, a, b       b, c, 5                                                         

 

 

b + (c - b) + (c - b) = 5              

2c - b  = 5

b = 2c - 5

4b  = 8c - 20    (1)

 

4 * (a/4)^2  = b            

a^2/4 = b

a^2 = 4b    (2)       equate (1) and (2)

 

a^2  = 8c - 20               

 

c        a         b

9        x

8        x

7        6        9

6        x

5        x

4        x

3        2        1

 

The largest possible  "abc"  is 697

 

Proof

 

4, 6 , 9     is geometrc if  r  = 1.5

9, 7,  5      is  arithmetic if  d  = -2

 

 

 

cool cool cool

 May 5, 2019

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