Plz Help

\(a)~a_n = a_0 + n d,~n \in \mathbb{Z^+},~d,a_0 \in \mathbb{R}\\ \text{Suppose there are 2 or more integers in the sequence, at say $k_1,~k_2$}\\ a_0 + k_1 d \in \mathbb{Z} \text{ and }a_0 + k_2 d \in \mathbb{Z}\\ (k_1-k_2)d \in \mathbb{Z}\\ m(k_1 - k_2)d \in \mathbb{Z},~m \in \mathbb{Z}\\ a_0 + k_1 d + m(k_1-k_2)d \in \mathbb{Z}\\ \text{This corresponds to index }\\ (m+1)k_1 - mk_2\\ \text{and thus there are infinitely many integers in the sequence as $m \in \mathbb{Z}$}\)

 

\(\text{On the other hand we can easily form a sequence with 1 integer by having $\\a_0 \in \mathbb{Z},~d \not \in \mathbb{Q}$}\)

.

\(b)~a_n = a_0 d^n,~n \in \{0\}\cup \mathbb{N},~a_0, d\in \mathbb{R}\\ \text{If we let $a_0 = \left(\dfrac 1 d\right)^{r-1},~\dfrac 1 d \in \mathbb{N}$}\\ \text{we end up with exactly $r$ integers in the sequence}\)

.