Verify that each of the following is an AP, and then write its next three terms.$5, \frac{14}{3}, \frac{13}{3}, 4, \ldots$

Given:

Given sequence is $5, \frac{14}{3}, \frac{13}{3}, 4, \ldots$

To do:

We have to verify whether the given sequence is an AP and write its next three terms.

Solution:

In the given sequence,

$a_1=5, a_2= \frac{14}{3}, a_3=\frac{13}{3}, a_4=4$

$a_2-a_1=\frac{14}{3}-5=\frac{14-3(5)}{3}=\frac{14-15}{3}=\frac{-1}{3}$

$a_3-a_2=\frac{13}{3}-\frac{14}{3}=\frac{13-14}{3}=\frac{-1}{3}$

$a_4-a_3=4-\frac{14}{3}=\frac{4(3)-13}{3}=\frac{12-13}{3}=\frac{-1}{3}$

Therefore,

$a_2-a_1=a_3-a_2=a_4-a_3$

The given sequence is an AP.

$d=\frac{-1}{3}$

$a_5=a_4+d=4+\frac{-1}{3}=\frac{4(3)-1}{3}=\frac{12-1}{3}=\frac{11}{3}$

$a_6=a_5+d=\frac{11}{3}+\frac{-1}{3}=\frac{11-1}{3}=\frac{10}{3}$

$a_7=a_6+d=\frac{10}{3}+\frac{-1}{3}=\frac{10-1}{3}=\frac{9}{3}=3$

The next three terms of the given sequence are $\frac{11}{3}, \frac{10}{3}$ and $3$.

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Updated on: 10-Oct-2022

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