# Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.2 Arithmetic Sequences - 12.2 Assess Your Understanding - Page 814: 14

In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant. Hence here: $a_{n+1}-a_n=(\frac{2}{3}+\frac{1}{4}(n+1))-(\frac{2}{3}+\frac{1}{4}n)=(\frac{11}{12}+\frac{1}{4}n)-(\frac{2}{3}+\frac{1}{4}n)=\frac{1}{4}$, thus it is an arithmetic sequence. $a_1=\frac{11}{12}$ $a_2=\frac{7}{6}$ $a_3=\frac{17}{12}$ $a_4=\frac{5}{3}$

#### Work Step by Step

In order for a sequence to be arithmetic, the difference of all consecutive terms must be constant. Hence here: $a_{n+1}-a_n=(\frac{2}{3}+\frac{1}{4}(n+1))-(\frac{2}{3}+\frac{1}{4}n)=(\frac{11}{12}+\frac{1}{4}n)-(\frac{2}{3}+\frac{1}{4}n)=\frac{1}{4}$, thus it is an arithmetic sequence. $a_1=\frac{2}{3}+\frac{1}{4}(1)=\frac{11}{12}$ $a_2=\frac{2}{3}+\frac{1}{4}(2)=\frac{7}{6}$ $a_3=\frac{2}{3}+\frac{1}{4}(3)=\frac{17}{12}$ $a_4=\frac{2}{3}+\frac{1}{4}(4)=\frac{5}{3}$

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