The sum of three numbers which are consecutive terms of an A.P. is 21.

Solution:

Let the first term of an A.P is a and its common difference be d.

$\therefore a_{1}+a_{2}+a_{3}=21$

$\Rightarrow a+(a+d)+(a+2 d)=21$

$\Rightarrow 3 a+3 d=21$

$\Rightarrow a+d=7$    ...(i)

Now, according to the question:

a , $a+d-1$ and $a+2 d+1$ are in $G . P .$

$\Rightarrow(\mathrm{a}+\mathrm{d}-1)^{2}=a(\mathrm{a}+2 \mathrm{~d}+1)$

$\Rightarrow(7+a-a-1)^{2}=a[a+2(7-a)+1]$

$\Rightarrow(6)^{2}=a(15-a)$

$\Rightarrow 36=15 a-a^{2}$

$\Rightarrow a^{2}-15 a+36=0$

$\Rightarrow(a-3)(a-12)=0$

$\Rightarrow a=3,12$

Now, putting $a=2,12$ in equation $($ i $)$, we get $d=5,-5$, respectively.

Thus, for $a=2$ and $d=5$, the numbers are 2,7 and 12 .

And, for $a=12$ and $d=-5$, the numbers are 12,7 and 2 .