Question

Find the smallest number of three numbers constituting a G.P. If it is known that the sum of the numbers is equal to 26 and that when 1,6 and 3 are added to them respectively, the new numbers are obtained which form an A.P.

Answer 1

Let $a,ar,a{r}^2$ be in G.P.,
$\therefore a(1+r+r^2)=26$ ...(1)
Also $a+1,ar+6,a{r}^2+3$ are in A.P.
$\therefore 2(ar+6)=(a+1)+(a{r}^2+3)$ or
$a(r^2-2r+1)=8$...(2)
Dividing (1) and (2) and simplifying , we get
$18(r^2+1)=60r$
or
$3r^2-10r+3=0\therefore (r-3)(3r-1)=0\therefore r=3,\frac{1}{3},$
Putting in (1), $a=2,18.$
$\therefore$The numbers are $2,6,18$ or $18,6,2$