Question -
Answer -
Let the first term ofan A.P. be ‘a’ and its common difference be‘d’.
a1 + a2 +a3 = 21
Where, the threenumber are: a, a + d, and a + 2d
So,
3a + 3d = 21 or
a + d = 7.
d = 7 – a …. (i)
Now, according to thequestion:
a, a + d – 1, and a +2d + 1
they are now in GP,that is:
(a+d-1)/a =(a+2d+1)/(a+d-1)
(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
Then,
For a = 3 and d = 4,the A.P is 3, 7, 11
For a = 12 and d = -5,the A.P is 12, 7, 2
∴ The numbers are3, 7, 11 or 12, 7, 2