Question:
Three positive integers x , y , z , are in AP such that sum of x , y , z is 33 and product of x , y , z is 1155 . Find the integers x , y , z .
Answer:
Given that x, y,z are in AP and
x+y+z =33............1
and x*y*z=1155............2
Now common difference d= y-x = z-y
=> 2y=x+z
=> y=(x+z)/2
Put value of y in equation 1, we get
x+(x+z)/2)+z=33
=> 2x+x+z+2z=66
=> 3x+3z=66
=> x+z=22
Put this value in eqyation 1, we get
22+y=33
=> y=33- 22
=> y = 11
Now, put y=11 in equation 2,
xz=1155/11
=> xz =105........3
We have x+z=22
=> x=22-z
Put value of x equation3, we het
(22-z)z=105
=> z2 - 22z + 105=0
=> z2 -15z -7z + 105=0
=> (z- 15)*(z - 7) = 0
=> z = 15, 7
Now when z=15 then from eqaution 3
x*15 = 105
=> x = 105/15
=> x = 7
Again now when z=15 then from eqaution 3
x*5 = 105
=> x = 105/5
=> x = 17
Since x, y and z are in AP
Hense the values of x, y, z are 7,11,15
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