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Let the common ratio/difference of the arithmetic progression be a number d. The exercise forces $d=/=1$ (d cannot be one). Is there any sort of proof for this exercise or am I supposed to play a guessing game? For example:

We say $15=8+k*d$, where $k$ is a natural number and $d$ is the common difference which is a real number. We also know $15 = 8+7$ therefore $k*d = 7$. We can maybe consider $k = 14$ and $d = 1/2$. We know $24 = 15+9 = 15 + x*d$.
$x$ is also a natural number and $d=1/2$ from our guess a few rows above. $x*d=9 => x = 9/d => x = 18$.
Therefore there is a $14$ term difference between $8$ and $15$ (including $15$) and a $18$ term difference between $15$ and $24$ (including $24$) and the common difference can be $1/2$.

The answer to the exercise question, in this case, would be yes. But this takes a bit of work (sometimes you may not get the correct guess from the start) if I may say so and maybe there is a more thorough proof.

questionasker22
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    My guess is that the problem is requiring the common difference to be a whole number. Or else you are right that taking $d = 1/2$ would work. You could also take $d = 1/n$ for any positive integer $n$. – Pedro Amaral Jan 17 '21 at 17:49
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    If the common difference is an integer $d$, then $d$ divides $15-8=7$ and $24-15=9$ – J. W. Tanner Jan 17 '21 at 17:58
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    It applies to all real numbers, see this: post, even if you use Rationals for d, then you still have infinite solutions – Steve237 Jan 17 '21 at 21:37

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