Finding first three terms of a GP whose sum and sum of squares is given? We have to find three numbers in G.P. such that their sum is $\frac{13}{3}$ and the sum of their squares is $\frac{91}{9}$.
Here's what I have tried to do:-
$$a+ar+ar^2=\frac{13}{3}$$
$$a^2+a^2r^2+a^2r^4=\frac{91}{9}$$
$$\frac{a^2+a^2r^2+a^2r^4}{a+ar+ar^2}=\frac{91}{9}*\frac{3}{13}$$
$$\frac{a+ar^2+ar^4}{1+r+r^2}=\frac{7}{3}$$
We have not been taught how to solve a biquadratic equation in school, so how should I proceed from here on?
 A: $a+ar+ar^2=\frac{13}{3}$ [equation 1]
$a^2+a^2r^2+a^2r^4=\frac{91}{9}$ [equation 2]
Square the first equation:
$a^2(1+r+r^2)^2 = \frac{169}{9}$
$a^2(1 + r^2 + r^4 + 2r + 2r^2 + 2r^3) = \frac{169}{9}$ [equation 3]
Now take equation 3 minus equation 2:
$2a^2r(1 + r + r^2) = \frac{26}{3}$
Using equation 1 again,
$2ar \frac{13}{3} = \frac{26}{3}$
So $ar = 1 \implies r = \frac{1}{a}$
Now put that back into equation 1 to get:
$a + 1 + \frac 1a = \frac{13}{3}$
Multiply throughout by $a$, rearrange:
$a^2 - \frac{10}{3}a + 1 = 0$
Solving that will give you the nice solutions $a = 3$ or $a = \frac 13$, which are both admissible, and the respective values of $r$ are the reciprocals of $a$, $\frac 13$ and $3$.
So the possible sequences are $3,1,\frac 13$ and $\frac 13, 1, 3$. Both possibilities are admissible because they haven't told you if the GP is increasing or decreasing.
A: Take 3 numbers $a/r$ ,a,ar $a/r$+a+ar=$13/3$ and $a^2/r^2$+$a^2$+$a^2r^2$=$91/9$ from this take $a^2$ common so on sim plifying we get $a^2/r^2$(1+$r^2+r^4$)=$(91/9$ Then square 1 st condition by using idw
Identity $(a+b+c)^2$=$a^2+b^2+c^2+2ab+2bc+2ac=$169/9$ from this we get $91/9$+$2a(a/r+a+ar)$=$169/9$ so $91/9$+2a$(13/3)$=$169/9$ from here a=1   plug  it in any of the condition you get r as 1/3 or 3 thus terms are 1,3,1/3.thanks!
