Question

plzz solve this question.The sum of three numbers in G.P is 19 and the sum of their squares is 133.find the numbers​

Answer 1

Answer:

YOUR ANSWER IS HERE

sum of three numbers in G.p is 19 and the sum of their squares is 133 find the numbers.

Not sure what G.p means. How about:

4 + 6 + 9 = 19

16 + 36 + 81 = 133

Answer 2

The possible value of  three numbers are

4,6 and 9 or 9,6 and 4

Step-by-step explanation:

Let a be the first term of G.P and r be the common ratio of G.P

Three term of G. P are

$a,ar, ar^2$

According to question

$a+ar+ar^2=19$

$a(1+r+r^2)=19$

Squaring on both sides then we get

$a^2(1+r+r^2)^2=361$

Identity: $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$

By using the identity

$a^2(1+r^2+r^4+2r+2r^3+2r^2)=361$

$a^2(1+r^2+r^4+2r(1+r^2+r)=361$

$a^2(1+r^2+r^4)+2a^2r(1+r+r^2)=361$

$a^2+a^2r^2+a^2+r^4=133$

$a^2(1+r^2+r^4)=133$

Substitute the values then we geta

$133+2ar(19)=361$

$2ar(19)=361-133=228$

$ar=\frac{228}{2(19)}=6$

$a=\frac{6}{r}$..(1)

Substitute the value of a

$\frac{6}{r}(1+r+r^2}=19$

$6+6r+6r^2=19r$

$6r^2+6r-19r+6=0$

$6r^2-13r+6=0$

$r=\frac{13\pm \sqrt{169-144}}{2(6)}=\frac{13\pm 5}{12}$

By using quadratic formula:$D=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$r=\frac{13+5}{12}=\frac{3}{2}$

$r=\frac{13-5}{12}=\frac{2}{3}$

Substitute r=3/2

$a=\frac{6}{\frac{3}{2}}=4$

Substitute r=2/3

$a=\frac{6}{\frac{2}{3}}=9$

When we take r=3/2 and a=4

Then, $a_2=4\times \frac{3}{2}=6$

$a_3=4\times (\frac{3}{2})^2=9$

When we take a=9 and r=2/3

$a_2=9\times \frac{2}{3}=6$

$a_2=9(\frac{2}{3})^2=4$

Therefore, the possible value of  three numbers are

4,6 and 9 or 9,6 and 4

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