<s>[INST] Show that $7a^{2} = 13b^{2}$ Here is another question which states that -

The first and the third terms of an arithmetic sequence are
  a and b respectively.    The sum of the first n terms is denoted by $S_n$.
(a) Find $S_4$ in terms of a and b.
  (b) Given that $S_4$, $S_5$,
  $S_7$ are consecutive terms of a geometric sequence, show that
  $$7a^{2} = 13b^{2}$$

My approach to part (a) -
$\Rightarrow\ U_1 = a$
$\Rightarrow\ U_3 = b$
$\Rightarrow\ a + 2d = b$
$\therefore\ d = \frac{b-a}{2}$
$\Rightarrow\ U_4 = b + \frac{b - a}{2}$
$\therefore\ U_4 = \frac{3b - a}{2}$
$\Rightarrow\ S_n = \frac{n}{2}(U_1 + U_4)$
$\Rightarrow\ S_4 = \frac{4}{2}(a + \frac{3b - a}{2})$
$\therefore\ S_4 = a+3b$ (Answer)

I do not know how to approach part (b) without the need for tedious calculations.
 [/INST] It is not hard to see that $S(n)=\frac{1}{4}\left(-a n^2+5 a n+b n^2-b n\right)$. So$$\frac{S(7)}{S(5)}=\frac{S(5)}{S(4)}\iff\frac{13 b^2-7 a^2}{10 b (a+3 b)}=0\iff13b^2=7a^2.$$
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