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.
 A: That $S_4,S_5,S_7$ are in geometric progression means $S_4S_7=S_5^2$. Now $S_5=5b$ and $S_7=\frac{21b-7a}2$, so
$$(a+3b)\frac{21b-7a}2=(5b)^2$$
$$7(a+3b)\frac{3b-a}2=25b^2$$
$$\frac72(9b^2-a^2)=25b^2$$
$$63b^2-7a^2=50b^2$$
$$7a^2=13b^2$$
A: Hint
Find $S_5$ and $S_7$ using your formula $S_n = \frac{n}{2}(U_1 + U_n)$. Then if the common ratio of the geometric series is $k$:
$$k^2 S_4 = kS_5  = S_7$$
Please edit your question explaining what you need if you find that one part of the calculations is too tedious.
A: 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$$\frac{S(7)}{S(5)}=\frac{S(5)}{S(4)}\iff\frac{13 b^2-7 a^2}{10 b (a+3 b)}=0\iff13b^2=7a^2.$$
A: I know that:
$$\left\{\begin{matrix}
c_1=a+3b
\\c_1\cdot q=5b
\\ c_1\cdot q^2=\frac{7}{2}(3b-a)
\end{matrix}\right.$$
For simplicity, I put the first term $c_1$ equal to $S_4$.
From this I have $q=\frac{5b}{c_1}$. Substituting in the third equation: $$(a+3b)\cdot \frac{25b^2}{(a+3b)^2}=\frac{7}{2}(3b-a)$$
In other words: $50b^2=7(3b-a)(3b+a)$; so $-13b^2=-7a^2$ and $13b^2=7a^2$
