## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 5 - Number Theory and the Real Number System - 5.7 Arithmetic and Geometric Sequences - Exercise Set 5.7 - Page 332: 151

#### 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.

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