The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.
Let the original numbers be
a, a + d, and a + 2d
According to the question –
3a + 3d = 21 or a + d = 7.
⇒ d = 7 – a
After the addition, the three numbers are:
a, a + d – 1, and a + 2d + 1
they are now in GP, that is –
⇒ 
⇒ (a + d – 1)2 = a(a + 2d + 1)
⇒ a2 + d2 + 1 + 2ad – 2d – 2a = a2 + a + 2da
⇒ (7 – a)2 – 3a + 1 – 2(7 – a) = 0
⇒ 49 + a2 – 14a – 3a + 1 – 14 + 2a = 0
⇒ a2 – 15a + 36 = 0
⇒ a2 – 12a – 3a + 36 = 0
⇒ a(a – 12) – 3(a – 12) = 0
⇒ a = 3 or a = 12
∴ d = 7 – a
d = 7 – 3 or d = 7 – 12
d = 4 or – 5
∴ The numbers are 3,7,11 or 12,7,2
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