y=x$. We have to check that $a_i+kb_{i+2}>a_{i+2}+kb_{i+1}$. If possible, let \begin{align*} & a_i+kb_{i+2}\leq a_{i+2}+kb_{i+1}\\ & a_{i+2}-a_{i}\geq k(b_{i+2}-b_{i+1}). \end{align*} Since the maximum value of $a_{i+2}-a_{i}$ is $2k$. Therefore, our assumption is incorrect, leading to the confirmation and proof of our claim. Thus in each case, we get two distinct elements of $A+k\cdot B,$ which are not in~\eqref{eq:3} and~\eqref{eq:4}. Hence, we get $|A|-2$ extra integers of $A+k\cdot B,$ which are not included in~\eqref{eq:1}. Consequently, $|A+k\cdot B|\geq 3|A|+|B|-4$. \subsection*{Inverse Problem for \texorpdfstring{$|A+k\cdot B|$}{|A+k. B|}} Let us begin with the case $|A|=|B|=r$ and assume that $A=\{a_0\ell(A_0^{*})$ and $\ell(B)\geq \ell(A)>\ell(A_1^{*})$, therefore $\delta_{B,A_0^{*}}=\delta_{B,A_1^{*}}=0$. Let's consider the case where $m\leq n$. Assuming Claim~\ref{cla1} is false, then $\ell(B)\geq l+n-1=|B+|A_1^{*}|-1\geq l+m-1=|B|+|A_0^{*}|-1$ and $d(B)=1$. Thus by Theorem~\ref{LSS}, $|A_0^{*}+B|\geq l+2|A_0^{*}|-2=l+2m-2$ and $|A_1^{*}+B|\geq l+2|A_1^{*}|-2=l+2n-2$ Hence $|A+3\cdot B|\geq 2l+2r-4,$ which contradicts to our hypothesis. In the case where $n\leq m$, we can obtain the result by following the same approach as described earlier. Thus $\ell(B)\leq l+\max{(m,n)}-2$. Since $r=m+n$ and $\max(m,n)\leq r-1$, therefore $\ell(B)\leq l+\max{(m,n)}-2\leq l+r-3$. This completes the proof of Claim~\ref{cla1}. \end{proof} \begin{enonce}{Claim}\label{cla2} $|A+3\cdot B|\geq |A|+2(|B|-1)+h_{B}$. \end{enonce} \begin{proof}[Proof of Claim~\ref{cla2}] Assume the case $m\leq n$. According to Claim~\ref{cla1}, it is evident that $\ell(B)\leq l+n-2$. Additionally, referring to Theorem~\ref{LSS}, we have $|A_1^{*}+B|\geq (n+l-1)+h_{B}$. Consequently, \begin{align*} |A+3\cdot B|&=|A_0^{*}+B|+|A_1^{*}+B|\\ &\geq (|A_0^{*}|+|B|-1)+(n+l-1)+h_{B}\\ &\geq (m+l-1)+(n+l-1)+h_{B}\\ &=2l+r-2+h_{B}. \end{align*} Similarly for the remaining case $(n\leq m)$, $|A+3\cdot B|\geq 2l+r-2+h_{B}$. Thus, we obtain that $h_B$ satisfies $0\leq h_B\leq |A+3\cdot B|-(2l+r-2)=h\leq r-3$. Therefore, $B\subseteq \{b_0,b_0+1,b_0+2,\dots,b_{l-1}\}\subseteq \{0,1,\dots,b_{l-1}\}$ and $B$ is an arithmetic progression of length at most $b_{l-1}+1=l+h_{B}\leq l+h\leq r+l-3$. As $\ell(A)\leq \ell(B),$ the set $A$ is also contained in A.P. of length at most $r+l-3$. The result can be easily verified for the case $|A_0|=m\geq 1$, $|A_1|=n=0$ and $|A_2|=p\geq 1$. \end{proof} \let\qed\relax \end{proof} \begin{proof}[Case 2. $|A_0|=m\geq 1$, $|A_1|=n\geq 1$, $|A_2|=p\geq 1$ i.e. $c_3(A)=3$] Assume that \begin{align*} A_0&=\{0=3x_0<3x_1<\cdots<3x_{m-1}\}, \\ A_0^{*}&=\frac{1}{3}\cdot A_0=\{0=x_0 \ell(A_0^{*})$, $\ell(B)\geq \ell(A)>\ell(A_1^{*})$ and $\ell(B)\geq \ell(A)>\ell(A_2^{*}),$ therefore $\delta_{B,A_0^{*}}=\delta_{B,A_1^{*}}=\delta_{B,A_2^{*}}=0$. Let's start by considering the case where $m\leq n\leq p$. Assuming that Claim~\ref{cla3} is false, we can deduce that $\ell(B)\geq l+p-1=|B|+|A_2^{}|-1\geq l+n-1=|B|+|A_1^{}|-1\geq l+m-1=|B|+|A_0^{*}|-1$, while $d(B)=1$. Thus by Theorem~\ref{LSS} \begin{multline} |A_0^{*}+B|\geq l+2|A_0^{*}|-2=l+2m-2,\quad |A_1^{*}+B|\geq l+2|A_1^{*}|-2=l+2n-2\\ \text{and}\quad |A_2^{*}+B|\geq l+2|A_2^{*}|-2=l+2p-2. \end{multline} Hence $|A+3\cdot B|\geq 3l+2r-6,$ which contradicts our hypothesis. For all the remaining cases we obtain the result by proceeding like above. Thus $\ell(B)\leq\linebreak l+\max{(m,n,p)}-2$. Since $r=m+n+p$ and $\max(m,n,p)\leq r-1,$ therefore, $\ell(B)\leq l+\linebreak \max{(m,n,p)}-2\leq l+r-3$. This completes the proof of Claim~\ref{cla3}. \end{proof} \begin{enonce}{Claim}\label{cla4} $|A+3\cdot B|\geq |A|+3(|B|-1)+h_{B}$. \end{enonce} \begin{proof}[Proof of Claim~\ref{cla4}] Assume the case $m\leq n\leq p$. By Claim~\ref{cla3}, observe that $\ell(B)\leq l+p-2$. Also the By Theorem~1.3, $|A_2^{*}+B|\geq (p+l-1)+h_{B}$ and thus \begin{align*} |A+3\cdot B|&=|A_0^{*}+B|+|A_1^{*}+B|+|A_2^{*}+B|\\ &\geq (|A_0^{*}|+|B|-1)+(|A_1^{*}|+|B|-1)+|A_2^{*}+B|\\ &\geq (m+l-1)+(n+l-1)+(p+l-1)+h_{B}\\ &=3l+r-3+h_{B}. \end{align*} Similarly for all the remaining cases $|A+3\cdot B|\geq 3l+r-3+h_{B}$. Therefore, we can deduce that $h_B$ satisfies the inequality $0\leq h_B\leq |A+3\cdot B|-(3l+r-3)=h\leq r-3$. Consequently, it follows that $B\subseteq \{b_0,b_0+1,b_0+2,\dots,b_{l-1}\}\subseteq \{0,1,\dots,b_{l-1}\}$ and $B$ forms an arithmetic progression with a length of at most $b_{l-1}+1=l+h_{B}\leq l+h\leq r+l-3$. Since $\ell(A)\leq \ell(B)$, the set $A$ is also contained within an arithmetic progression with a length of at most $r+l-3$. This establishes the desired result. By combining both cases, we obtain the overall result. \end{proof} \let\qed\relax \end{proof} \let\qed\relax \end{proof} \subsection*{Acknowledgments} %The authors are thankful to Dr MSG for their guidance. The authors are thankful to Dr MSG for their guidance. % and the reseach of second author is supported by NBHM. \bibliographystyle{crplain} \bibliography{CRMATH_Singh_20220879} \end{document}