Answer
The provided statement is false and if\[\frac{{{d}_{n+1}}}{r}={{d}_{n}}\], the common difference of the two adjacent terms will be equal.
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.
\[\begin{align}
& r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
& =\frac{{{a}_{3}}}{{{a}_{2}}} \\
& =\frac{{{a}_{3}}}{{{a}_{4}}}
\end{align}\]
Now, find the difference of two adjacent terms in G.P.
\[\begin{align}
& {{d}_{1}}=ar-a \\
& =a\left( r-1 \right)
\end{align}\]
And,
\[\begin{align}
& {{d}_{2}}=a{{r}^{2}}-ar \\
& =ar\left( r-1 \right)
\end{align}\]
So, \[{{d}_{1}}\ne {{d}_{2}}\].
Hence, the common difference of two adjacent terms is not equal.
If\[\frac{{{d}_{n+1}}}{r}={{d}_{n}}\], the common difference of the two adjacent terms will be equal.
Hence, the provided statement is false.
