The sum of three numbers in GP is 56.

To find: Three numbers

Given: Three numbers are in G.P. Their sum is 56

Formula used: When $a, b, c$ are in $G P, b^{2}=a c$

Let the three numbers in GP be $a, a r, a r^{2}$

According to condition :-

$a+a r+a r^{2}=56$

$a\left(1+r+r^{2}\right)=56 \ldots$ (i)

1, 7, 21 be subtracted from them respectively, we obtain the numbers as :-

$a-1, a r-7, a r^{2}-21$

According to question the above numbers are in AP

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

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

$\Rightarrow a r-a-6=a r^{2}-a r-14$

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

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

Multiplying the above eqn. with 7

$\Rightarrow 56=7 \mathrm{a}\left(\mathrm{r}^{2}-2 \mathrm{r}+1\right)$

$\Rightarrow \mathrm{a}\left(1+\mathrm{r}+\mathrm{r}^{2}\right)=7 \mathrm{a}\left(\mathrm{r}^{2}-2 \mathrm{r}+1\right)$

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

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

$\Rightarrow 6 \mathrm{r}^{2}-12 \mathrm{r}-3 \mathrm{r}+6=0$

$\Rightarrow 6 \mathrm{r}(\mathrm{r}-2)-3(\mathrm{r}-2)=0$

$\Rightarrow(6 \mathrm{r}-3)(\mathrm{r}-2)=0$

$\Rightarrow r=\frac{3}{6}=\frac{1}{2}$ Or $r=2$

Putting $r=\frac{1}{2}$ in eqn. (i)

$a\left(1+r+r^{2}\right)=56$

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

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

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

a = 32

The numbers are $a, a r, a r^{2}$

$\Rightarrow 32,32 \times \frac{1}{2}, 32 \times \frac{1}{2^{2}}$

⇒ 32, 16, 8

Putting r = 2 in eqn. (i)

$a\left(1+r+r^{2}\right)=56$

$a\left(1+2+2^{2}\right)=56$

$a(1+2+4)=56$

$a(7)=56$

a = 8

The numbers are $a, a r, a r^{2}$

$\Rightarrow 32,32 \times \frac{1}{2}, 32 \times \frac{1}{2^{2}}$

⇒ 8, 16, 32

Ans) We have two sets of triplet as 32, 16, 8 and 8, 16, 32.