AP or GP?

As B , C , D form an A.P with common difference as 6 ,

B=B , C = B+6 , D = B+12

Also , A = D = B+12

As A , B , C form a G.P ,

B/A = C/B ...............[ Common Ratio ]

B^2 = A * C

B^2 = (B + 12) * (B + 6)

B^2 = B^2 + 18B + 72

18B = -72

B = -4

C = -4 + 6 = 2

A = D = -4 +12 = 8

So , A + B + C + D = 8 - 4 + 2 + 8 = 14 :)

Let first term be $a$ , common ratio be $r$ .

Since first 3 terms are in GP, second term is $ar$ and third would be $ar^{2}$ .

Now, we are given that last 3 are in AP, so $ar^{2}-ar=6$ and $ar+12=a$ $[4^{th} term=1^{st} term]$

$ar+12=a$ $\Rightarrow r=\dfrac{a-12}{a}$

$ar^{2}-ar=6$ $\Rightarrow a×\Big (\dfrac{a-12}{a}\Big )^2-(a-12)=6$ $\Rightarrow (a-12)^{2}=a(a-6)$

Solving this linear equation gives $a=8$ .

$ar+12=a$ $\Rightarrow r=-\dfrac{1}{2}$

So the sequence is $\color{#D61F06}{\text{8,-4,2,8}}$ which adds up to give $\color{#3D99F6}{\boxed{14}}$

Let $A, B, C, A$ be our desired sequence, where $A, B,$ and $C$ are real numbers. Since $A, B, C$ is a geometric sequence, we have $B = \sqrt{AC}.$ Since $B, C, A$ is an arithmetic sequence, we have $C = \dfrac{B+A}{2}.$ Substituting this value of $C$ into our first equation, we get

$\begin{aligned} B &= \sqrt{A\left(\frac{B+A}{2}\right)} \\ 2B &= \sqrt{2A(B+A)} \\ 4B^{2} &= 2AB + 2A^{2} \\ A^{2} + AB - 2B^{2} &= 0\\ (A + 2B)(A - B) &= 0\\ A &= -2B \quad \text{or} \quad B. \\ \end{aligned}$

We discard $A = B$ , since that does not satisfy the common difference of $6$ criteria, so $A = -2B,$ and our sequence becomes $-2B, B, -\dfrac{1}{2}B, -2B.$

Finally, we solve $B + 6 = -\dfrac{1}{2}B,$ which yields $B = -4.$ Thus, our sequence is $8, -4, 2, 8,$ which indeed satisfy the criteria given in the problem. The sum is $\boxed{14}.$

Let the Arithmetic Progression be D-12, D-6 and D

Since the 1st number = 4th number, then, the numbers is: D, D-12, D-6, D with a sum of 4D-18

by using the ratio of a geometric progression,

(D-12)/D = (D-6)/(D-12)

D^2 - 6D = D^2 - 24D + 144

18D = 144 ---> 4D = 32

Therefore, the sum of the 4 numbers is 4D - 18 = 32 - 18 = 14 (ans)..

we have : $A$ , $B=qA$ , $C=q^2A=qA+6$ and $D=A=qA+12$ . From last equality, $A(1-q)=12$ . From third equality, $qA(q-1)=6$ so with the precedent we get, $-12q=6$ , hence, $q =-\dfrac{1}{2}$ . Replace $q$ in last we get, $A=\dfrac{2 \times 12}{3}=8$ then, $A+B+C+D=14$