Question

The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21, from these numbers in that order, we obtain an A.P. Find the numbers.

Answer 1

Let the first term of a G.P. ba a and its common ratio be r.

$\therefore a_1+a_2+a_3=56$

$\Rightarrow a+ar+a{r}^2=56$

$\Rightarrow a(1+r+r^2)=56$

$\Rightarrow a=\frac{56}{1+r+r^2}$ .........(i)

Now, according to the question :

$a-1,ar-7$ and $a{r}^2-21$ are in A.P.

$\therefore 2(ar-7)=a-1+a{r}^2-21$

$\Rightarrow 2ar-14=a{r}^2+a-22$

$\Rightarrow a{r}^2-2ar+a-8=0$

$\Rightarrow a(1-r{)}^2=8$

$\Rightarrow a=\frac{8}{(1-r{)}^2}$ ........ (ii)

Equating (i) and (ii):

$\Rightarrow \frac{8}{(1-r{)}^2}=\frac{56}{1+r+r^2}$

$\Rightarrow 8(1+r+r^2)=56(1+r^2-2r)$

$\Rightarrow 1+r+r^2=7(1+r^2-2r)$

$\Rightarrow 1+r+r^2=7+7r^2-14r$

$\Rightarrow 6r^2-15r+6=0$

$\Rightarrow 3(2r^2-5r+2)=0$

$\Rightarrow 2r^2-4r-r+2=0$

$\Rightarrow 2r(r-2)-1(r-2)=0$

$\Rightarrow (r-2)(2r-1)=0$

$\Rightarrow r=2,\frac{1}{2}$

When r = 2, a = 8 [Using (ii)]

And, the required numbers are 32, 16 and 8.