Solution:
Consider the first term of an A.P. to be ‘a’ and let its common difference be‘d’.
We have, a1 + a2 + a3 = 15
Where the three numbers are as follows:
a, a + d, and a + 2d
So, we can write:
$ a\text{ }+\text{ }a\text{ }+\text{ }d\text{ }+\text{ }a\text{ }+\text{ }2d\text{ }=\text{ }15 $
$ 3a\text{ }+\text{ }3d\text{ }=\text{ }15\text{ } $
$ or\text{ }a\text{ }+\text{ }d\text{ }=\text{ }5 $
$ d\text{ }=\text{ }5-a\text{ }\ldots \text{ }\left( i \right) $
Now, according to the question:
a + 1, a + d + 3, and a + 2d + 9
they are in GP, that is:
$ \left( a+d+3 \right)/\left( a+1 \right)\text{ }=\text{ }\left( a+2d+9 \right)/\left( a+d+3 \right) $
$ {{\left( a\text{ }+\text{ }d\text{ }+\text{ }3 \right)}^{2}}~=~\left( a\text{ }+\text{ }2d\text{ }+\text{ }9 \right)\text{ }\left( a\text{ }+\text{ }1 \right) $
$ {{a}^{2}}~+\text{ }{{d}^{2}}~+\text{ }9\text{ }+\text{ }2ad\text{ }+\text{ }6d\text{ }+\text{ }6a\text{ }=\text{ }{{a}^{2}}~+\text{ }a\text{ }+\text{ }2da\text{ }+\text{ }2d\text{ }+\text{ }9 $
$ a\text{ }+\text{ }9{{\left( 5-a \right)}^{2}}~-4a\text{ }+\text{ }4\left( 5-a \right)\text{ }=\text{ }0 $
$ 25\text{ }+\text{ }{{a}^{2}}-10a-4a\text{ }+\text{ }20-4a\text{ }=\text{ }0 $
$ {{a}^{2}}-18a\text{ }+\text{ }45\text{ }=\text{ }0 $
$ {{a}^{2}}-15a-3a\text{ }+\text{ }45\text{ }=\text{ }0 $
$ a\left( a-15 \right)-3\left( a-15 \right)\text{ }=\text{ }0 $
$ a\text{ }=\text{ }3\text{ }or\text{ }a\text{ }=\text{ }15 $
$ d\text{ }=\text{ }5-a $
$ d\text{ }=\text{ }5-3\text{ }or\text{ }d=\text{ }5-15 $
$ d\text{ }=\text{ }2\text{ }or\text{ }\text{ }10 $
Then,
For a = 3 and d = 2, the A.P is 3, 5, 7
For a = 15 and d = -10, the A.P is 15, 5, -5
∴ The numbers are 3, 5, 7 or 15, 5, – 5