Question

Determine three numbers a , b , c
such that a , b , c are three consecutive terms of a geometric sequence and an arithmetic sequence at the same time.
Note: i do not want the answer
d=0 and r=1, as in 2 , 2 , 2 , 2 , 2...
Given also:
abc=27 or a.b.c=27​

Answer 1

Since $a,b,c$ are in geometric progression, if $r$ is the common ratio between consecutive terms, then

$a=a$

$b = ar$

$c=ar^2$

Since $a,b,c$ are also in arithmetic progression, if $d$ is the common difference between consecutive terms, then

$a = a$

$b = a + d \implies d = b-a$

$c = b + d = a + 2d \implies c = a + 2(b-a) = 2b-a$

Given that $abc=27$, we have

$abc = a\cdot ar\cdot ar^2 = (ar)^3 = 27 \implies ar = 3 \implies a = \dfrac3r$

$b = \dfrac3r \cdot r = 3$

$c = \dfrac3r \cdot r^2 = 3r$

It follows that

$c = 2b-a \iff 3r = 6 - \dfrac3r$

Solve for $r$.

$3r - 6 + \dfrac3r = 0$

$3r^2 - 6r + 3 = 0$

$r^2 - 2r + 1 = 0$

$(r-1)^2 = 0$

$\implies r=1 \implies a=b=c=3$

so the only possible sequence is {3, 3, 3, …}.