Question -
Answer -
Let the three numbersbe a, ar, ar2
According to thequestion
a + ar + ar2┬а=56 тАж (1)
Let us subtract 1,7,21we get,
(a тАУ 1), (ar тАУ 7), (ar2┬атАУ21)
The above numbers arein AP.
If three numbers arein AP, by the idea of the arithmetic mean, we can write 2b = a + c
2 (ar тАУ 7) = a тАУ 1 +ar2┬атАУ 21
= (ar2┬а+a) тАУ 22
2ar тАУ 14 = (56 тАУ ar) тАУ22
2ar тАУ 14 = 34 тАУ ar
3ar = 48
ar = 48/3
ar = 16
a = 16/r тАж. (2)
Now, substitute thevalue of a in equation (1) we get,
(16 + 16r + 16r2)/r= 56
16 + 16r + 16r2┬а=56r
16r2┬атАУ40r + 16 = 0
2r2┬атАУ5r + 2 = 0
2r2┬атАУ4r тАУ r + 2 = 0
2r(r тАУ 2) тАУ 1(r тАУ 2) =0
(r тАУ 2) (2r тАУ 1) = 0
r = 2 or 1/2
Substitute the valueof r in equation (2) we get,
a = 16/r
= 16/2 or 16/(1/2)
= 8 or 32
тИ┤┬аThe threenumbers are (a, ar, ar2) is (8, 16, 32)