Question

The sum of three numbers in G.P. is 31 and the sum of their square is 651. find the numbers.

Answer 1

Please see the attachment

Answer 2

Answer:

The G.P form is 1,5,25,... and 25,5,1,...

Step-by-step explanation:

Given : The sum of three numbers in G.P. is 31 and the sum of their square is 651.

To find : The numbers ?

Solution :

Let the three numbers in G.P be $a,ar,ar^2$

The sum of three numbers in G.P. is 31.

$a+ar+ar^2=31$

$a(1+r+r^2)=31$ ....(1)

The sum of their square is 651.

$a^2+a^2r^2+a^2r^4=651$

$a^2(1+r^2+r^4)=651$ ....(2)

Squaring both side of (1),

$a^2(1+r+r^2)^2=(31)^2=961$ .....(3)

Dividing (2) and (3),

$\frac{a^2(1+r+r^2)^2}{a^2(1+r^2+r^4)}=\frac{961}{651}$

$\frac{(1+r+r^2)^2}{(r^2+1+r)(r^2+1-r)}=\frac{31}{21}$

$\frac{1+r+r^2}{r^2+1-r}=\frac{31}{21}$

$21r^2+21r+21=31r^2-31r+31$

$-10r^2+52r-10=0$

$10r^2-52r+10=0$

$10r^2-50r-2r+10=0$

$10r(r-5)-2(r-5)=0$

$(10r-2)(r-5)=0$

$r=5,\frac{1}{5}$

When r=5,

$a(1+r+r^2)=31$

$a(1+5+5^2)=31$

$a(1+5+25)=31$

$a(31)=31$

$a=1$

The G.P form is 1,5,25,...

When $r=\frac{1}{5}$,

$a(1+r+r^2)=31$

$a(1+\frac{1}{5}+\frac{1}{5}^2)=31$

$a(1+5+\frac{1}{25})=31$

$a(\frac{31}{25})=31$

$a=25$

The G.P form is 25,5,1,...