#### i)Which of the following form an AP? Justify your answer.–1, –1, –1, –1, ...ii) Which of the following form an AP? Justify your answer. 0, 2, 0, 2, ...iii)Which of the following form an AP? Justify your answer.1, 1, 2, 2, 3, 3,...iv)Which of the following form an AP? Justify your answer.  11, 22, 33,...v)Which of the following form an AP? Justify your answer.$\frac{1}{2}, \frac{1}{3}, \frac{1}{4},......................$vi)Which of the following form an AP? Justify your answer.$2, 2^2, 2^3, 2^4, ...$vii)Which of the following form an AP? Justify your answer.$\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots$

i)

Solution.

$-1, -1, -1, -1, \ldots \quad \text{Here } a_{1} = -1, \quad a_{2} = -1, \quad a_{3} = -1, \quad a_{4} = -1 \\ a_{2} - a_{1} = -1 - (-1) = -1 + 1 = 0 \\ a_{3} - a_{2} = (-1) - (-1) = -1 + 1 = 0 \\ a_{4} - a_{3} = -1 - (-1) = -1 + 1 = 0 \\$

Here the difference of the successive term is the same hence it is an AP.

ii)

Solution.

$0, 2, 0, 2, \ldots \quad \text{Here } a_{1} = 0, \quad a_{2} = 2, \quad a_{3} = 0, \quad a_{4} = 2 \\ a_{2} - a_{1} = 2 - 0 = 2 \\ a_{3} - a_{2} = 0 - 2 = -2 \\ a_{4} - a_{3} = 2 - 0 = 2 \\$

Here the difference in successive terms is not the same. Hence it is not an AP.

iii)

Solution.

$1, 1, 2, 2, 3, 3, \ldots \quad \text{Here } a_{1} = 1, \quad a_{2} = 1, \quad a_{3} = 2, \quad a_{4} = 2 \\ a_{2} - a_{1} = 1 - 1 = 0 \\ a_{3} - a_{2} = 2 - 1 = 1 \\ a_{4} - a_{3} = 2 - 2 = 0 \\$

Here the difference in successive term is not the same. Hence it is not an AP.

iv)

Solution.

$\text{The given series is } \{11, 22, 33,\ldots\} \quad \text{here } a_{1} = 11, \quad a_{2} = 22, \quad a_{3} = 33 \\ a_{2} - a_{1} = 22 - 11 = 11 \\ a_{3} - a_{2} = 33 - 22 = 11 \\$

Here the difference between the successive terms is the same. Hence it is an AP.

v)

Solution.

\begin{aligned} &\text { Here the given series is } \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots \ldots \ldots \text { here } \mathrm{a}_{1}=\frac{1}{2} \mathrm{a}_{2}=\frac{1}{3} \mathrm{a}_{3}=\frac{1}{4}\\ &\mathrm{a}_{2}-\mathrm{a}_{1}=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6}\\ &a_{3}-a_{2}=\frac{1}{4}-\frac{1}{3}=\frac{3-4}{12}=\frac{-1}{12} \end{aligned}

Here the difference between the successive terms is not the same. Hence it is not an AP.

vi)

Solution.

$\text{Here the given series is } 2, 2^2, 2^3, 2^4, \ldots \text{ here } a_1 = 2, \quad a_2 = 2^2, \quad a_3 = 2^3, \quad a_4 = 2^4 \\ a_2 - a_1 = 2^2 - 2 = 4 - 2 = 2 \\ a_3 - a_2 = 2^3 - 2^2 = 8 - 4 = 4 \\ a_4 - a_3 = 2^4 - 2^3 = 16 - 8 = 8 \\$

Here the difference of the successive terms is not the same. Hence it is not an AP.

vii)

Solution.

\begin{aligned} &\text { Here the given series is } \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots . \text { here } \mathrm{a}_{1}=\sqrt{3} \mathrm{a}_{2}=\sqrt{12} \mathrm{a}_{3}=\sqrt{27}\\ &a_{4}=\sqrt{48}\\ &a_{2}-a_{1}=\sqrt{12}-\sqrt{3}=2 \sqrt{3}-\sqrt{3}=\sqrt{3}=\sqrt{3}(2-1)=\sqrt{3}\\ &a_{3}-a_{2}=\sqrt{27}-\sqrt{12}=3 \sqrt{3}-2 \sqrt{3}=\sqrt{3}\\ &a_{4}-a_{3}=\sqrt{48}-\sqrt{27}=4 \sqrt{3}-3 \sqrt{3}=\sqrt{3} \end{aligned}

Here the difference of the successive term is the same. Hence it is an AP.