Let the 1st term or number of the AP be x
.
The second number of the AP = x+6 by
data.
By another condtion, sum of the 1st 2nd and third
numbers = 54.
Threfore the 3rd number = 54 -(first
+second) = 54 -(x+x+6) = 54-2x-6 = 48-2x.
Since x , x+6
and 48-2x are inAP, they must have the equal common increment between any two
successive terms. Or
2nd term -1st term = 3rd term -2nd
term
x+6-x = 48-2x -(x+6)
6 =
42 -3x.
3x= 42-6 =36
3x/x =
36/3
x = 12.
So the three
numbers in AP are x, x+6 and x+12
and they are 12,18, 24.
Obviously they satisfy the condtion. of their sum = 54 .
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