#### i)Verify that each of the following is an AP, and then write its next three terms.  $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$ii)Verify that each of the following is an AP, and then write its next three terms.$5, \frac{14}{3}, \frac{13}{3}, 4, \ldots \ldots$iii)Verify that each of the following is an AP, and then write its next three terms.$\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \ldots$iv)Verify that each of the following is an AP, and then write its next three terms.a + b, (a + 1) + b, (a + 1) + (b + 1), ...v)Verify that each of the following is an AP, and then write its next three terms.a, 2a + 1, 3a + 2, 4a + 3,...

i)

Here the given series is,

$0,\frac{1}{4},\frac{1}{2},\frac{3}{4}$

$\\ a\textsubscript{1} = 0, a\textsubscript{2}= \frac{1}{4} {{a}_{3}} = \frac{3}{4} \\ a\textsubscript{2} - a\textsubscript{1} = \frac{1}{4}-0=\frac{1}{4} \\$

$\\a_3-a_2=\frac{1}{4}\\a_4-a_3=\frac{1}{4}$

Here common difference is same. Hence given series is an AP

$\\a\textsubscript{5} = a\textsubscript{1}+4d (using a\textsubscript{n} = a\textsubscript{1 }+ (n-1).d)\\ \vspace{\baselineskip} a\textsubscript{6} = a\textsubscript{1}+ 5d \\ \vspace{\baselineskip} a\textsubscript{7} = a\textsubscript{1}+ 6d \\$
\begin{aligned} & =0+6\left( \frac{1}{4} \right) \\ & =\frac{3}{2} \ \end{aligned} \\

The next three terms are

$1,\frac{5}{4},\frac{3}{2} \\$

ii)

$\\Here the given series is, 5, \frac{14}{3},\frac{13}{3},4,..... \\ Here common difference is same. \\ Hence the given series is an AP\\ a\textsubscript{5} = {{a}_{1}} +4d (using a\textsubscript{n} = a\textsubscript{1} + (n-1).d)\\ a\textsubscript{6 }= {{a}_{1}} +5d \\ a\textsubscript{7} = {{a}_{1}} + 6d\\ \vspace{\baselineskip} Next terms are \frac{11}{3} , \frac{10}{3} ,3 \\$

iii)

$\text{Here the given series is } \sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, \ldots . \\ \text{Here, the common difference is the same. Hence, the given series is an AP.} \\ a_{4} = a_{1} + 3d \quad (\text{using } a_{n} = a_{1} + (n-1)d) \\ = \sqrt{3} + 3\sqrt{3} = 4\sqrt{3} \\ a_{5} = a_{1} + 4d \\ = \sqrt{3} + 4\sqrt{3} = 5\sqrt{3} \\ a_{6} = a_{1} + 5d \\ = \sqrt{3} + 5\sqrt{3} = 6\sqrt{3} \\ \text{Next terms are } 4\sqrt{3}, 5\sqrt{3}, 6\sqrt{3}.$

iv)

$\text{Here the given series is } a + b, (a + 1) + b, (a + 1) + (b + 1), \ldots . \\ a_{1} = a + b, \quad a_{2} = (a + 1) + b, \quad a_{3} = (a + 1) + (b + 1) \\ a_{2} - a_{1} = (a + 1) + b - a - b = 1 \\ a_{3} - a_{2} = a + 1 + b + 1 - a - 1 - b = 1 \\ \text{Here, the common difference is the same, hence the given series is an AP.} \\ a_{4} = a_{1} + 3d \quad (\text{using } a_{n} = a_{1} + (n-1)d) \\ = (a + b) + 3(1) \\ = (a + 1) + (b + 1) + 1 \\ a_{5} = a + 4d \\ = (a + b) + 4(1) \\ = (a + 1) + (b + 1) + 1 + 1 \\ a_{6} = a + 5d \\ = (a + b) + 5(1) \\ = (a + 1) + (b + 1) + 1 + 1 + 1 \\ \text{Next three terms are:} \\ (a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1 + 1, (a + 1) + (b + 1) + 1 + 1 + 1.$

v)

$\text{Here the given series is } a, 2a + 1, 3a + 2, 4a + 3, \ldots . \\ a_{1} = a , \quad a_{2} = 2a+1 , \quad a_{3}=3a + 2 \\ a_{2} - a_{1}= 2a + 1 - a = a + 1 \\ a_{3} - a_{2} = 3a + 2 - (2a + 1) = a + 1 \\ \text{Here, the common difference is the same, hence the given series is an AP.} \\ a_{5} = a_{1}+4d \quad (\text{using } a_{n} = a_{1}+ (n-1)d) \\ = a + 4(a + 1) \\ = a + 4a + 4 \\ = 5a + 4 \\ a_{6} = a_{1} + 5d \\ = a + 5(a + 1) \\ = a + 5a + 5 \\ = 6a + 5 \\ a_{7} = a_{1} + 6d \\ = a + 6(a + 1) \\ = a + 6a + 6 \\ = 7a + 6 \\ \text{Next three terms are:} \\ 5a + 4, 6a + 5, 7a + 6, \ldots .$