The sum of three numbers in G.P. is 21 and the sum of their squares is 189.

Question:

The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

Solution:

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

Sum of the numbers = 21

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

$\Rightarrow a\left(1+r+r^{2}\right)=21$    ...(1)

Sum of the squares of the numbers = 189

$\Rightarrow a^{2}+(a r)^{2}+\left(a r^{2}\right)^{2}=189$

$\Rightarrow a^{2}+(a r)^{2}+\left(a r^{2}\right)^{2}=189$

$\Rightarrow a^{2}\left(1+r^{2}+r^{4}\right)=189$    ....(2)

Now, $a\left(1+r+r^{2}\right)=21 \quad[$ From $(\mathrm{i})]$

Squaring both the sides

$\Rightarrow a^{2}\left(1+r+r^{2}\right)^{2}=441$

$\Rightarrow a^{2}\left(1+r^{2}+r^{4}\right)+2 a^{2} r\left(1+r+r^{2}\right)=441$

$\Rightarrow 189+2 a r\left\{a\left(1+r+r^{2}\right)\right\}=441 \quad[$ Using $($ ii $)]$

$\Rightarrow 189+2 a r \times 21=441 \quad[$ Using $(\mathrm{i})]$

$\Rightarrow a r=6$

$\Rightarrow a=\frac{6}{r}$ ....(3)

Putting $a=\frac{6}{r}$ in (i)

$\frac{6}{r}\left(1+r+r^{2}\right)=21$

$\Rightarrow \frac{6}{r}+6+6 r=21$

$\Rightarrow 6 r^{2}+6 r+6=21 r$

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

$\Rightarrow 3\left(2 r^{2}-5 r+2\right)=0$

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

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

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

Putting $r=\frac{1}{2}$ in $a=\frac{6}{r}$, we get $a=12$.

So, the numbers are 12,6 and 3 .

Putting $r=2$ in $a=\frac{6}{r}$, we get $a=3$.

So, the numbers are 3,6 and 12 .

Hence, the numbers that are in G.P are 3,6 and 12 .

 

 

 

 

 

 

 

 

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