Bump bump bump.

a) Let $a_k= a_1 + (k-1)d$ for $d$ be the common difference and $b_k = b_1.q^{k-1}$ for $q$ be the common ratio Since $(a_i)$ and $(b_1)$ are strictly increasing, we have $d \ge 1$, $q >1$ Then we have $$ a_1+(n-1)d=a_n=b_n=b_1.q^{n-1}=a_1.q^{n-1} \Leftrightarrow \frac{a_1}{d}=\frac{n-1}{q^{n-1}-1} $$ Now we want to prove that $a_k >b_k$ for all $k = 2,3,..., n -1$, means $$ a_1+(k-1)d > a_1.q^{k-1} \Leftrightarrow \frac{k-1}{q^{k-1}-1} > \frac{a_1}{d}=\frac{n-1}{q^{n-1}-1}$$Consider $f(x)= \frac{x}{q^{x}-1}$ for $x \in [1, +\infty)$ We have $$ f'(x)=\frac{q^x-1-q^x.{\ln{q^x}}}{(q^x-1)^2}=\frac{a-1-a.{\ln{a}}}{(a-1)^2} =\frac{g(a)}{(a-1)^2}$$For $a=q^x \in [1, +\infty )$ and $g(a)=a-1-a.{\ln{a}}$ We also have $g'(a)=-{\ln{a}}<0$, then $g(a)$ is decreasing in $[1, +\infty)$ or $g(a) \le g(1)=0 $ $\forall a \in [1,+\infty)$ Hence $f'(x) \le 0$ for all $x \in [1, +\infty)$ means $f(x)$ is decreasing in $[1, +\infty)$ Thus $f(x)>f(n-1)$ for all $x \in \{ 2,3,4, \ldots , n-2 \}$, which completes the problem immediately!! b) Let the common ratio of $(b_n)$ be $\frac{p+1}{p}$ for $p \in \mathbb{P}$ Take $b_1=p^n$, then $b_k=p^{n-k+1}(p+1)^{k-1}$ $\forall k =1,2,3, \ldots , n$ Take $a_1=b_2-1=p^{n-1}(p+1)-1$ and choose the common diffrence is $d=b_3-b_2=p^{n-2}(p+1)^2-p^{n-1}(p+1)$ Then since the difference between the $2$ consecutive numbers in $(b_n)$ forms a strictly increasing sequence (To be more specific, we have $b_2-b_1<b_3-b_2<b_4-b_3< \ldots < b^{n+1}-b^n$), which means $a_k < b_{k+1}$ $\forall k = 1,2,3, \ldots , n -1$ (Since $v_k=a_k-b_{k+1}$ forms a strictly increasing sequence and $v_1>0$) But since then $u_k=a_k-b_k$ is a decreasing sequence, we only have to confirm that $u_n=a_n-b_n>0$, means $$ a_n-b_n=p^{n-1}(p+1)-1+(n-1).d-p.(p+1)^{n-1}=p^{n-1}(p+1)-1+(n-1).(p^{n-2}(p+1)^2-p^{n-1}(p+1))-p.(p+1)^{n-1}>0$$Notice that LHS is a polynomial of degree $n-1$ with the leading coefficient greater than $0$, we complete the proof by choosing $p$ sufficiently large!!