Answer
The provided statement is false. So, it can be say that those sequences which is not A.P is not necessary that it would be G.P.
Work Step by Step
It is known that the geometric sequence is;
\[a,ar,a{{r}^{2}},a{{r}^{3}},a{{r}^{4}},\ldots \]
Where a is first term and r is common ratio.
\[\begin{align}
& r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
& =\frac{{{a}_{3}}}{{{a}_{2}}} \\
& =\frac{{{a}_{3}}}{{{a}_{4}}}
\end{align}\]
It is known that the arithmetic sequence is;
\[a,a+d,a+2d,a+3d,\ldots \]
Where a is the first term and d is the common difference.
\[\begin{align}
& d={{a}_{2}}-{{a}_{1}} \\
& ={{a}_{3}}-{{a}_{2}}
\end{align}\]
Let’s take a sequence which is not A.P as;
\[1,3,4,6,9,10,\ldots \]
The common difference is not same in all the terms so it could not be A.P.
Let’s check it is G.P or not
If the common ratio is same the sequence will be G.P otherwise not.
\[\begin{align}
& {{r}_{1}}=\frac{3}{1} \\
& =3 \\
& {{r}_{2}}=\frac{4}{3} \\
& {{r}_{3}}=\frac{6}{4}
\end{align}\]
So,
\[{{r}_{3}}=\frac{3}{2}\]
So, \[{{r}_{1}}\ne {{r}_{2}}\ne {{r}_{3}}\].
Hence, the provided sequence is not in G.P.
So, it can be say that those sequence which is not A.P is not necessary that it would be G.P.
Hence, the provided statement is false. So, it can be say that those sequence which is not A.P is not necessary that it would be G.P.
