Question

The sum of three numbers in GP is 56. If we subtract 1, 7, 21 from these numbers in that order, we get an AP. Find the numbers.

Answer 1

Let the required numbers be a, ar and $a{r}^2$. Then,

a + ar + $a{r}^2$ = 56 . ... (i)

Also, (a – 1), (ar – 7) and $(a{r}^2–21)$ are in AP.

$\therefore$ 2 (ar – 7) = (a – 1) + ($a{r}^2$ – 21) $\Rightarrow$ a + $a{r}^2$ = 2ar + 8. . . . (ii)

Using (ii) in (i), we get

$ar+(2ar+8)=56\Rightarrow 3ar=48\Rightarrow ar=16r=\frac{16}{a}$

Putting $r=\frac{16}{a}$ in (i), we get

$a+16+\frac{256}{a}=56\Rightarrow a^2+16a+256=56a$

$\Rightarrow a^2-40a+256=0$

$\Rightarrow a^2-8a-32a+256=0$

$\Rightarrow$ a(a-8)-32 (a-8)=0

$\Rightarrow$ (a-8)(a-32) = 0

$\Rightarrow$ a = 8 or a = 32

Now, $a=8\Rightarrow r=\frac{16}{8}\Rightarrow r=2$

And, $a=32\Rightarrow r=\frac{16}{32}\Rightarrow r=\frac{1}{2}$

Hence, the required numbers are (8, 16, 32) or (32, 16, 8)