a1 + a2 + a3 = 69



Since we are given a1,a2,a3 form an AP, we can write:



a1 + (a1+k) + (a1+2k) = 69    where k is the common difference



3a1 + 3k = 69

a1 + k = 23   



Trying  a1 = 21,  a1+k = 23, and a1+2k = 25,  we see this AP meets the requiremnt that the first two terms multiply out to 483,  thus  {21,23,25} satisfies the condition.



----

Alternate solution:

a1 * (a1+k) = 483





Expanding it back to a quadratic:

  so we see  a1=21, k=2 works, resulting in {21,23,25} as above



  

Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
 Split 69 into three parts

a + b + c = 69

 such that they are in A. P.

Assume this means Arithmetic Progression. (dif between two Nos. is the same)

let d = difference between consecutive numbers

Then

b = a + d

c = a + 2d 

the product of the two smaller parts is 483

ab = 483

replace b with a+d

a(a+d) = 483

a^2 + ad = 483

:

in the first equation write it

a + (a+d) + (a+2d) = 69

3a + 3d = 69

simplify, divide by 3

a + d = 23

d = (23-a)

:

Back to a^2 + ad = 83, replace d with (23-a)

a^2 + a(23-a) = 483

a^2 + 23a - a^2 = 483

a^2 drops out

23a = 483

a = 21

Find d

23-21 = 2 is the difference

The numbers then are:

21, 23, 25

:

Check: 

21 * 23 = 483

and

21 + 23 + 25 = 69 

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!
 


Given three terms in AP with a sum of 69, we know the middle term is 69/3=23.  Then if d is the common difference, the other two numbers are 23-d and 23+d.


The product of the two smaller parts is 483:











The middle term is 23, and the common difference is 2, so


ANSWER: 21, 23, and 25


 

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