Question 1083357: The sum of three numbers which are consecutive terms of an A.P is 21.If the second number is reduced by 1 & the third is increased by 1, we obtain three consecutive terms of a G.P,find the numbers?
Answer by natolino_2017(77)   (Show Source):
You can put this solution on YOUR website! Let the A.P : a, a + d , a + 2d.
1) a + (a + d) + (a + 2d) = 3(a + d) = 21.
so a + d = 7
so: a, a + d -1, a + 2d +1 is a G.P
So using the ratio condition between the terms, we get a second equation:
2) (a + d -1)/a = (a + 2d +1)/(a +d -1)
using 1) 36 = a(8 + d) = a(15 - a)
a^2 -15a +36 = 0 (quadratic equation can have one solution, two solutions or none)
using the formula 
a = {12,3} (two solutions)
so if a = 12 then d = -5 (using 1))
and if a = 3 the d = 4. (using 1))
We have two solutions of the numbers of the original A.P
(both solutions sharing the second term):
Solution i) 12, 7, 2
Solution ii) 3, 7, 11
@natolino_
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