Solution:
Consider the first term of an A.P. to be ‘a’ and let its common difference be‘d’.
Therefore, we can write: a1 + a2 + a3 = 21
Where the three numbers are as follows:
a, a + d, and a + 2d
So, we get:
3a + 3d = 21
Or, a + d = 7.
We have: d = 7 – a …. (i)
Now, we have:
a, a + d – 1, and a + 2d + 1
All these are in geometric progression
$ \left( a+d-1 \right)/a\text{ }=\text{ }\left( a+2d+1 \right)/\left( a+d-1 \right) $
$ {{\left( a\text{ }+\text{ }d-1 \right)}^{2}}~=~a\left( a\text{ }+\text{ }2d\text{ }+\text{ }1 \right) $
$ {{a}^{2}}~+\text{ }{{d}^{2}}~+\text{ }1\text{ }+\text{ }2ad-2d-2a\text{ }=\text{ }{{a}^{2}}~+\text{ }a\text{ }+\text{ }2da $
$ {{\left( 7-a \right)}^{2}}-3a\text{ }+\text{ }1-2\left( 7-a \right)\text{ }=\text{ }0 $
$ 49\text{ }+\text{ }{{a}^{2}}-14a-3a\text{ }+\text{ }1-14\text{ }+\text{ }2a\text{ }=\text{ }0 $
$ {{a}^{2}}-15a\text{ }+\text{ }36\text{ }=\text{ }0 $
$ {{a}^{2}}-12a-3a\text{ }+\text{ }36\text{ }=\text{ }0 $
$ a\left( a-12 \right)-3\left( a-12 \right)\text{ }=\text{ }0 $
$ a\text{ }=\text{ }3\text{ }or\text{ }a\text{ }=\text{ }12 $
$ d\text{ }=\text{ }7-a $
$ d\text{ }=\text{ }7-3\text{ }or\text{ }d\text{ }=\text{ }7-12 $
$ d\text{ }=\text{ }4\text{ }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 are 3, 7, 11 or 12, 7, 2