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 = \frac3r`

`b = \frac3r \cdot r = 3`

`c = \frac3r \cdot r^2 = 3r`

It follows that

`c = 2b-a \iff 3r = 6 - \frac3r`

Solve for
`r`.

`3r - 6 + \frac3r = 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, …}.