Geometric sequence $a,b,c,d$ and arithmetic sequence $a, \frac{b}{2},\frac{c}{4}, d-70$

Question

The first four terms, given in order, of a geometric sequence $a,b,c,d$ and arithmetic sequence $a, \frac{b}{2},\frac{c}{4}, d-70$, find the common ratio $r$ and the values of each $a,b,c,d$.

What I have tried:
$b=ar, c=ar^2,d=ar^3$, we have find the value of $b$ by calculating the mean from the values $1$ and $3$.$$ar=a+\frac{ar^2}{4} \implies 0 = a(r^2-4r+4a) \\0 = a(r-2)(r+2)$$ To calculate for $c$ we take the second and fourth term of the sequence. $$\frac{ar^2}{2}=\frac{ar}{2}+ar^3-70 \implies 70 = a(r^3-\frac{r^2}{2}+\frac{r}{2})$$

However, this seems a little messy because it involves complex numbers - am I still on the right track?

Answer 1

$b = a r, c = a r^2, d = a r^3 $

Using a constant common difference,

$\frac{1}{2} a r - a = \frac{1}{4} a r^2 - \frac{1}{2} a r = a r^3 - 70 - \frac{1}{4} a r^2 $

Re-arranging

$ r^2 - 4 r + 4 = 0\hspace{25pt}(1) $

and

$ a r^3 - \dfrac{1}{2} a r^2 + \dfrac{1}{2} a r = 70\hspace{25pt}(2) $

From $(1)$ , $ r = 2 $

Substitute this into $(2)$:

$ a ( 8 - 2 + 1 ) = 70 $

So $a = 10 $

Answer 2

There is a concealed "joke" in this problem which is obscured by choosing to solve directly for $ \ r \ \ . $ If we instead seek the common difference $ \ \Delta \ $ between the terms of the arithmetic sequence, we might start by writing those terms as $ \ a \ \ , \ \ a + \Delta \ = \ \frac{b}{2} \ \ , \ \ a \ + \ 2 \Delta \ = \ \frac{c}{4} \ \ , \ \ a \ + \ 3 \Delta \ = \ d - 70 \ \ . $ In the geometric sequence then, the common ratio between terms is found from $$ r \ \ = \ \ \frac{2a \ + \ 2 \Delta}{a} \ \ = \ \ \frac{4a \ + \ 8 \Delta}{2a \ + \ 2 \Delta} \ \ = \ \ \frac{a \ + \ 3 \Delta \ + \ 70}{4a \ + \ 8 \Delta} \ \ . $$

Cross-multiplication of the first two ratios (assuming $ \ a \neq 0 \ $ and $ \ a \ \neq \ -\Delta \ , \ $ which in fact cannot work for both sequences simulatneously) produces $$ a·(4a + 8 \Delta) \ \ = \ \ (2a + 2 \Delta)^2 \ \ \Rightarrow \ \ 4a^2 \ + \ 8a \Delta \ \ = \ \ 4a^2 \ + \ 8a \Delta \ + \ 4 \Delta^2 \ \ \Rightarrow \ \ 4 \Delta^2 \ = \ 0 \ \ (!) \ \ . $$ [Yeah, I didn't believe this the first time I ran through a solution either...]

Since $ \ a \ = \ 0 \ $ will not "work" in the arithmetic sequence, we must indeed have $ \ \Delta \ = \ 0 \ \ , $ which of course means that all of the arithmetic sequence terms are identical. The equation for the geometric series ratio "collapses" to $$ r \ \ = \ \ \frac{2a }{a} \ \ = \ \ \frac{4a }{2a } \ \ = \ \ 2 \ \ = \ \ \frac{a \ + \ 70}{4a } \ \ , $$ with the equality of the last two segments of the equation yielding $ \ 8a \ = \ a + 70 \ \Rightarrow \ a \ = \ 10 \ \ . $ Hence, the geometric sequence is $ \ 10 \ , \ 20 \ , \ 40 \ , \ 80 \ \ , $ making the arithmetic sequence $ \ 10 \ \ , \ \ \frac{20}{2} \ \ , \ \ \frac{40}{4} \ \ , \ \ 80 - 70 \ \ . $ (Very amusing, Problem-Poser Person...)