Find the numbers $a,b,c,d$ in a geometric sequence, knowing that $a+1$, $b+6$, $c+6$, $d-4$ are in an arithmetic sequence

Question

The exercise reads as follows:

Find the numbers $a,b,c,d$ in a geometric sequence, knowing that $a+1$, $b+6$, $c+6$, $d-4$ are in an arithmetic sequence.

I am interested in finding out the steps I should use to get to a solution. Or just a starting point, I've been thinking about this for some time and I wouldn't say I am getting close to the result soon.

I was thinking about:

$a+1 = b+6-r \implies a=b+5-r$

$c+6 = b+6+r \implies c=b+r$

And use $b^2=a*c$ from the geometric sequence.

The topic can be closed. Massive thanks to Rafaelle for the solution and also to Oscar Lanzi for explaining the process!

Answer 1

$$\begin{cases} \frac{a}{b}=\frac{b}{c}\\ \frac{b}{c}=\frac{c}{d}\\ -(a+1)+b+6=-(b+6)+c+6\\ -(b+6)+c+6=-(c+6)+d-4\\ \end{cases} $$ $$a= 5,b= 10,c= 20,d= 40$$

Answer 2

"Then a miracle occurs..."

Let's look inside Raffaele's solution process.

To implement the geometric sequence, render $b=ar,c=ar^2,d=ar^3$. Then use the fact that the second term in the arithmetic sequence must be the mean of the first and third terms to form tte equation

$2(ar+6)=(a+1)+(ar^2+6)$

which rearranges to

$a(r^2-2r+1)=a(r-1)^2=5$ Eq. 1

Similarly, the third term in the arithmetic sequence is the mean of the second and fourth ones giving

$2(ar^2+6)=(ar+6)+(ar^3-4)$

$a(r^3-2r^2+r)=ar(r-1)^2=10$ Eq. 2

Equations 1 and 2 differ only by a factor of $r$ on the left side, so dividing the latter by the former isolates $r=2$. Then $a=5$ is obtained by substitution into either equation leading to the geometric progression $(a,b,c,d)=(5,10,20,40)$ and the arithmetic progression $(6,16,26,36)$.

Answer 3

We might also start from the arithmetic sequence, which we will take to have a difference $ \ \Delta \ $ between successive terms ($ \ d \ $ is "already taken"). We have $$ \Delta \ \ = \ \ (b + 6) \ - \ (a + 1) \ \ = \ \ b - a + 5 \ \ = \ \ c - b \ \ = \ \ (d - 4) \ - \ (c + 6) \ \ = \ \ d - c - 10 \ \ . $$

The ratio between successive terms of the geometric series is given by $ \ r \ = \ \frac{b}{a} \ = \ \frac{c}{b} \ = \ \frac{d}{c} \ \ . $ We can use "componendo-dividendo" (which I learned about on MSE [!] -- they didn't teach it when I was in high-school algebra) to obtain $$ \frac{c}{b} \ = \ \frac{c \ - \ b}{b \ - \ a} \ \ = \ \ \frac{\Delta}{\Delta \ - \ 5} \ \ \ \text{and} \ \ \ \frac{d}{c} \ = \ \frac{d \ - \ c}{c \ - \ b} \ \ = \ \ \frac{\Delta \ + \ 10}{\Delta} \ \ . $$ Since both of these ratios are equal, we find $$ \ \Delta^2 \ \ = \ \ (\Delta \ - \ 5)·(\Delta \ + \ 10) \ \ \Rightarrow \ \ 5·\Delta \ - \ 50 \ \ = \ \ 0 \ \ \Rightarrow \ \ \Delta \ \ = \ \ 10 \ \ \Rightarrow \ \ r \ \ = \ \ 2 \ \ . $$

We now have our choice of equations to solve: we could just pick $$ b \ + \ 6 \ \ = \ \ a \ + \ 1 \ + \ \Delta \ \ = \ \ a \ + \ 11 \ \ \Rightarrow \ \ b \ \ = \ \ a \ + \ 5 \ \ = \ \ 2 · a \ \ \Rightarrow \ \ a \ \ = \ \ 5 \ \ , $$

at which point the entire problem "unravels". The geometric sequence is $ \ 5 \ , \ 10 \ , \ 20 \ , \ 40 \ \ $ and the arithmetic sequence is $ \ 5 + 1 \ = \ 6 \ \ , \ \ 10 + 6 \ = \ 16 \ \ , \ \ 20 + 6 \ = \ 26 \ \ , \ \ 40 - 4 \ = \ 36 \ \ . $