Three numbers are in
a.p. and their sum is 15. if 1, 4, and 19 are added to these numbers respectively, the resulting numbers are in g.p. find the numbers.
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Let the original numbers be
a, a+d, and a+2d
3a + 3d = 15 or a+d = 5 ----> d = 5-a
after the addition, the three numbers are:
a+1, a+d+4, and a+2d+19
they are now in GP, that is ....
(a+d+4)/(a+1) = (a+2d+19)/(a+d+4)
(a + 5-a + 4)/(a+1) = (a + 10-2a + 19)/(a + 5-a + 4)
9/(a+1) = (-a + 29)/9
81 = -a^2 + 28a + 29
a^2 - 28a + 52 = 0
(a - 26)(a - 2) = 0
a = 26 or a = 2
if a = 26, then d = 5-26 = -21
and the original 3 numbers were:
26, 5, and 16
a, a+d, and a+2d
3a + 3d = 15 or a+d = 5 ----> d = 5-a
after the addition, the three numbers are:
a+1, a+d+4, and a+2d+19
they are now in GP, that is ....
(a+d+4)/(a+1) = (a+2d+19)/(a+d+4)
(a + 5-a + 4)/(a+1) = (a + 10-2a + 19)/(a + 5-a + 4)
9/(a+1) = (-a + 29)/9
81 = -a^2 + 28a + 29
a^2 - 28a + 52 = 0
(a - 26)(a - 2) = 0
a = 26 or a = 2
if a = 26, then d = 5-26 = -21
and the original 3 numbers were:
26, 5, and 16
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