proving that an arithmetic sequence is equal to a geometric sequence

Question

how can I prove that an arithmetic sequence is equal to a geometric sequence if and only if the initial values of the sequences are the same and the common difference of the arithmetic sequence is 0; the common ratio of the geometric sequence is 1?

Answer 1

Hint:

If $a-d,a,a+d,a+2d,\cdots$ are also in Geometric progression,

$(a-d)(a+d)=a^2\implies d=?$

What will be common ratio?

Alternatively, if $b,br,br^2,\cdots$ are also in arithmetic progression,

$b+br^2=2br\implies b(r-1)^2=0$

For non-trivial cases, $b\ne0$

Answer 2

The difference between any two consecutive terms of an arithmetic sequence is a constant $d$. Then the ratio between any two consecutive terms is $\frac {a_k + d}{a_k} = 1 + \frac d{a_k}$ which is in general not constant unless the $a_k$s are all equal. (in which case $a_{k+1} -a_k = d =0$ and $a_{k+1} = a_k$ and the sequence is constant.)

The ratio between any two consecutive terms of a geometric sequence is a constant $d$. Then the difference between any two consecutive terms is $b*a_k - a_k = (b-1)*a_k$ which is in general not constant unless either $b =1 $ (in which case $a_{k+1} = a_k*1 = a_k$ and the sequence is constant) or the $a_k$ are all equal (in which case the sequence is constant and $b = 1$... unless all the terms are $0$ in which case the value of $b$ is irrelevent).

So the only way a sequence can only be both arithmetic and geometric if both the difference and the ratios of consecutive terms are constant and the only way that can occur is if the sequence is constant and $d = 0$ and $b =0$ and the sequence is $a_0, a_0, a_0,.......$