Verify that each of the following is an AP, and then write its next three terms.
$ 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots $

To do:

We have to verify whether the given sequences are APs and write their next three terms.

Solution: 

(i) In the given sequence,

$a_1=0, a_2= \frac{1}{4}, a_3=\frac{1}{2}, a_4=\frac{3}{4}$

$a_2-a_1=\frac{1}{4}-0=\frac{1}{4}$

$a_3-a_2=\frac{1}{2}-\frac{1}{4}=\frac{2-1}{4}=\frac{1}{4}$

$a_4-a_3=\frac{3}{4}-\frac{1}{2}=\frac{3-2}{4}=\frac{1}{4}$

Therefore,

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

The given sequence is an AP.

$d=a_2-a_1=\frac{1}{4}-0=\frac{1}{4}$

$a_5=a_4+d=\frac{3}{4}+\frac{1}{4}=\frac{3+1}{4}=\frac{4}{4}=1$

$a_6=a_5+d=1+\frac{1}{4}=\frac{1\times4+1}{4}=\frac{5}{4}$

$a_7=a_6+d=\frac{5}{4}+\frac{1}{4}=\frac{5+1}{4}=\frac{6}{4}$

The next three terms of the given sequence are $1, \frac{5}{4}$ and $\frac{6}{4}$.   

(ii) 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 $3$.   

(iii) In the given sequence,

$a_1=\sqrt{3}, a_2= 2\sqrt{3}, a_3=3\sqrt{3}$

$a_2-a_1=2\sqrt{3}-\sqrt{3}=\sqrt{3}$

$a_3-a_2=3\sqrt{3}-2\sqrt{3}=\sqrt{3}$

Therefore,

$a_2-a_1=a_3-a_2$

The given sequence is an AP.

$d=\sqrt{3}$

$a_4=a_3+d=3\sqrt{3}+\sqrt{3}=4\sqrt{3}$

$a_5=a_4+d=4\sqrt{3}+\sqrt{3}=5\sqrt{3}$

$a_6=a_5+d=5\sqrt{3}+\sqrt{3}=6\sqrt{3}$

The next three terms of the given sequence are $4\sqrt3, 5\sqrt3$ and $6\sqrt3$.

(iv) In the given sequence,

$a_1=a+b, a_2= (a+1)+b, a_3=(a+1)+(b+1)$

$a_2-a_1=(a+1)+b-a+b=1$

$a_3-a_2=(a+1)+(b+1)-[(a+1)+b]=a+b+2-a-1-b=1$

Therefore,

$a_2-a_1=a_3-a_2$

The given sequence is an AP.

$d=1$

$a_4=a_3+d=(a+1)+(b+1)+1=(a+2)+(b+1)$

$a_5=a_4+d=(a+2)+(b+1)+1=(a+2)+(b+2)$

$a_6=a_5+d=(a+2)+(b+2)+1=(a+3)+(b+2)$

The next three terms of the given sequence are $(a+2)+(b+1), (a+2)+(b+2)$ and $(a+3)+(b+2)$.   

(v) In the given sequence,

$a_1=a, a_2= 2a+1, a_3=3a+2, a_4=4a+3$

$a_2-a_1=(2a+1)-a=a+1$

$a_3-a_2=(3a+2)-(2a+1)=a+1$

$a_4-a_3=(4a+3)-(3a+2)=a+1$

Therefore,

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

The given sequence is an AP.

$d=a+1$

$a_5=a_4+d=(4a+3)+(a+1)=5a+4$

$a_6=a_5+d=(5a+4)+(a+1)=6a+5$

$a_7=a_6+d=(6a+5)+(a+1)=7a+6$

The next three terms of the given sequence are $(5a+4), (6a+5)$ and $(7a+6)$.