Arithmetic and Geometric Progression problem

Question

We know that $a,b,c,d\in \mathbb{R}$ form a geometric sequence in that order and $a,(b/2),(c/4),(d-140)$ form an arithmetic sequence in that order. Find the value $(d-b)$.

For as simple as it seems, I got completely stuck! Any help?

Thanks in advance.

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EDIT

The alleged correct solution is $120$

Answer 1

Let the common ratio of the GP be $2r$.

Hence GP is $(a, b,c,d)=(a,2ar, 4ar^2, 8ar^3)$

and AP is $\left(a,\frac b2, \frac c4, d-140\right)=(a, ar, ar^2, 8ar^3-140)$.

As $(a, ar, ar^2)$ are in AP, hence $\color{blue}{r=1}$

which means the AP is $(a,a,a,a)$, i.e. $8a-140=a$ giving $\color{blue}{a=20}$, i.e. the AP is $(20,20,20,20)$.

Hence $$d-b=8a-2a=6a=\color{red}{120}$$

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NB - as the common ratio is $2$, the GP is $(20,40, 80,160)$.

Answer 2

Let $a,b,c,d=b-r,b,b+r,b+2r$.

Then, $\left(\dfrac b2\right)^2 = a\left(\dfrac c4\right)$.

Substitute: $\left(\dfrac{b}2\right)^2 = (b-r)\left(\dfrac{b+r}4\right)$.

$b^2 = (b-r)(b+r)$

$b^2 = b^2 - r^2$

Thus $r^2=0$, and $r=0$, so $d-b = (b+2r)-(b) = 0$.

Answer 3

As $a,(b/2),(c/4),(d-140)$ is an arithmetic progression, we can say that:

$$\frac{b}{2}=\frac{a+\frac{c}{4}}{2}$$

At the same time, as $a,b,c,d$ configure a geometric progression, it is safe to say that:

$$\frac{aq}{2}=\frac{a+\frac{aq^2}{4}}{2}$$

Which leads us to:

$$ \begin{align} 4aq&=4a+aq^2 \\ aq^2-4aq+4a&=0\\ a(q^2-4q+4)&=0\\ a(q-2)^2=0 \end{align} $$

Giving as possible solutions $a=0$ and $q=2$.

Now:

$$\frac{c}{4}=\frac{\frac{b}{2}+(d-140)}{2}$$

Meaning that:

$$\frac{aq^2}{4}=\frac{\frac{aq}{2}+(aq^3-140)}{2}$$

Which clearly does not hold for $a=0$.

$$ \begin{align} aq^2&=aq+2aq^3-280\\ 2aq^3-aq^2+aq&=280\\ a(2q^3-q^2+q)&=280\\ a&=\frac{280}{(2q^3-q^2+q)}=\frac{280}{14}=20\\ \end{align} $$

Then $\{a,b,c,d\}=\{a,aq,aq^2,aq^3\}=\{20,40,80,160\}$

Finally,

$$d-b=160-40=120$$