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?
Best Answer
$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 $