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 arithmetic progression. Find the numbers.

Answer 1

Let a, ar, $a{r}^2$ be three numbers in G.P.

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

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

According to the given condition if we subtract 1, 7, 21 from $a,ar,a{r}^2$ respectively then resulting numbers are also in A.P.

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

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

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

$\Rightarrow ar-a-6=a{r}^2-21-ar+7$

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

$\Rightarrow a(r^2-2r+1)=8\dots \dots \left(2\right)$

Dividing (1) by (2) we get

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

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

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

$\Rightarrow 2r^2-5r+2=0$

$\Rightarrow r=\frac{-(-5)\pm \sqrt{(-5{)}^2-4\times 2\times 2}}{2\times 2}$

$\Rightarrow r=\frac{5\pm \sqrt{25-16}}{4}\Rightarrow r=\frac{5\pm 3}{4}$

Either $r=\frac{5+3}{4}$ i.e. $r=\frac{8}{4}\Rightarrow r=2$

or $r=\frac{5-3}{4}i.e.r=\frac{2}{4}\Rightarrow r=\frac{1}{2}$

Now, putting r = 2 in (1), we get

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

$\therefore$ The required numbers are: 8, 16, 32

Putting $r=\frac{1}{2}$ in (1), we get

$a(1+\frac{1}{2}+\frac{1}{4})=56$

$\Rightarrow a\left(\frac{4+2+1}{4}\right)=56$

$\Rightarrow a\left(\frac{7}{4}\right)=56$

$\Rightarrow a=56\times \frac{4}{7}=32$

$\therefore$ The required numbers are: 32, 16, 8