Question

Find three numbers in G.P. such that their
sum is 21 and sum of their squares is 189.​

Answer 1

Answer:

The numbers in G.P are 12 , 6 and 3

Step-by-step explanation:

As per the question,

Let the three numbers in G.P are $a\ , \ ar\ and \ ar^{2}$

where a= first term

and r = common ratio

Since it is given in question that the,

sum of three terms is 21 that is,

$a+ar+ar^{2}=21$

$a(1+r+r^{2}) =21$

Squaring both sides we get,

$(a(1+r+r^{2}) )^{2}=21^{2}$

$a^{2}(1+r+r^{2}) ^{2}=441$ say this equation as (i)

Also given that sum of their squares is 189 that is,

$a^{2}+(ar)^{2}+(ar^{2})^{2}=189$ say this equation as (ii)

Now, divide the equation (i) by (ii) we get,

$\frac{a^{2}(1+r+r^{2}) ^{2}}{a^{2}+(ar)^{2}+(ar^{2})^{2}}=\frac{441}{189}$

On solving this equation a quadratic equation is obtained that is,

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

On further solving we get

r = 1/2 or 2

Now,

when r = 1/2

a = 12

and when r = 2

a = 3

Hence a= 12

ar=12×1/2 =6

$ar^{2}$ = 12×1/4=3

Therefore, the numbers we get is 12 , 6 and 3