Three numbers are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively,

Solution:

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

$a_{1}+a_{2}+a_{3}=15$

$\Rightarrow a+(a+d)+(a+2 d)=15$

$\Rightarrow 3 a+3 d=15$

$\Rightarrow a+d=5$                 ....(i)

Now, according to the question:

$a+1, a+d+3$ and $a+2 d+9$ are in G.P.

$\Rightarrow(\mathrm{a}+\mathrm{d}+3)^{2}=(\mathrm{a}+1)(\mathrm{a}+2 \mathrm{~d}+9)$

$\Rightarrow(5-\mathrm{d}+\mathrm{d}+3)^{2}=(5-\mathrm{d}+1)(5-\mathrm{d}+2 \mathrm{~d}+9) \quad[$ From $(\mathrm{i})]$

$\Rightarrow(8)^{2}=(6-\mathrm{d})(14+\mathrm{d})$

$\Rightarrow 64=84+6 d-14 d-d^{2}$

$\Rightarrow d^{2}+8 d-20=0$

$\Rightarrow(d-2)(d+10)=0$

$\Rightarrow d=2,-10$

Now, putting $\mathrm{d}=2,-10$ in equation $(\mathrm{i})$, we get, $a=3,15$, respectively.

Thus, for $a=3$ and $d=2$, the $A$. P. is $3,5,7$.

And, for $a=15$ and $d=-10$, the A.P. is $15,5,-5$.