Simple Series Problem

Let the common ratio the GP $\{a,b,c,d,e\}$ be $r$ .

Then we have:

$a + e = \dfrac {c} {r^2} + cr^2 = c \left( \dfrac {1}{r^2} + r^2 \right) = 97$

Similarly,

$b + d = c \left( \dfrac {1}{r} + r \right) = 78$

Dividing the two equations, we have:

$\dfrac {a+e}{b+d} = \dfrac {\dfrac {1}{r^2} + r^2} {\dfrac {1}{r} + r} = \dfrac {\dfrac {1}{r^2} + 2+ r^2 -2} {\dfrac {1}{r} + r} = \dfrac {\left( \dfrac {1}{r} + r \right)^2 - 2} {\dfrac {1}{r} + r} = \dfrac {97}{78}$

Let $x = \dfrac {1}{r} + r$ , then we have:

$\dfrac {x^2 - 2} {x} = \dfrac {97}{78}\quad \Rightarrow 78x^2 - 97x -156 = 0\quad \Rightarrow x = \dfrac {1}{r} + r = - \frac {12}{13}$ or $\frac {13}{6}$

$\Rightarrow 13r^2+12r+13 = 0$ or $6r^2-13r+6 = 0$ .

We find that $13r^2+12r+13 = 0$ has no real roots.

And from $6r^2-13r+6 = 0$ we get $r = \frac {3}{2}$ or $\frac {2}{3}$ .

Substituting $r = \frac {3}{2}$ in $b + d = c \left( \dfrac {1}{r} + r \right) = 78$ , we get:

$\left( \dfrac {2}{3} + \dfrac {3}{2} \right) c = \dfrac {13}{6} c = 78\quad \Rightarrow c = \boxed {36}$

Note that we still get $c=36$ if we use $r = \frac {2}{3}$ .

let a=ar^-2,b=ar-1.c=a, d=ar,e=ar^2, a+e=a(r^-2+r^2) = 97, and b+d =(r^-1+r)=78, both two equation can be combined to 2a^2 +97a - 78^2=0,. Then a= 36

Let us call the common ratio $q$ :

$\large \displaystyle \color{#D61F06} \ (i) \color{#333333} \ \frac{c}{q^2} + cq^2 = 97$

$\large \displaystyle \color{#D61F06} \ (ii) \color{#333333} \ \frac{c}{q} + cq = 78$

Multiplying $(i)$ by $c$ :

$\large \displaystyle \color{#D61F06} \ (iii) \color{#333333} \ \frac{c^2}{q^2} + c^2q^2 = 97c$

Squaring $(ii)$ :

$\large \displaystyle \color{#D61F06} \ (iv) \color{#333333} \ \frac{c^2}{q^2} + c^2q^2 + 2c^2 = 6084$

Substituting $(iii)$ into $(iv)$ :

$\large \displaystyle 2c^2 + 97c - 6084 = 0$

This has roots $36$ and $-84.5$ . But if $c = -84.5$ we wouldn't have real $q$ (it's easy to check just by substituting $c$ into $(ii)$ ). Thus:

$\color{#3D99F6} \boxed{ \large \displaystyle c = 36}$