Solution:

Here,
\(t_1 = 2, t_2 = 4, t_3 = 6, t_4 = 8\)

Now,
\(t_2 - t_1 = 4 - 2 = 2\)
\(t_3 - t_2 = 6 - 4 = 2\)
\(t_4 - t_3 = 8 - 6 = 2\)

∴ The difference between the consecutive terms is 2 which is a constant.

∴ The given sequence is an A. P. and the common difference \(d\) is 2.

Solution:

Here,
\(\displaystyle t_1 = 2, t_2 = \frac {5}{2}, t_3 = 3, t_4 = \frac {7}{2}\)

Now,

\(\displaystyle t_2 - t_1 = \frac {5}{2} - 2 = \frac {5 - 4}{2} = \frac {1}{2}\)

\(\displaystyle t_3 - t_2 = 3 - \frac {5}{2} = \frac {6 - 5}{2} = \frac {1}{2}\)

\(\displaystyle t_4 - t_3 = \frac {7}{2} - 2 = \frac {7 - 6}{2} = \frac {1}{2}\)

∴ The difference between the consecutive terms is \(\displaystyle \frac {1}{2}\) which is a constant.

∴ The given sequence is an A. P. and the common difference \(d\) is \(\displaystyle \frac {1}{2}\).

Solution:

Here,
\(t_1 = - 10, t_2 = - 6, t_3 = - 2, t_4 = 2\)

Now,
\(t_2 - t_1 = - 6 - (- 10) = - 6 + 10 = 4\)
\(t_3 - t_2 = - 2 - (- 6) = - 2 + 6 = 4\)
\(t_4 - t_3 = 2 - (- 2) = 2 + 2 = 4\)

∴ The difference between the consecutive terms is 4 which is a constant.

∴ The given sequence is an A. P. and the common difference \(d\) is 4.

Solution:

Here,
\(t_1 = 0.3, t_2 = 0.33, t_3 = 0.333\)

Now,
\(t_2 - t_1 = 0.33 - 0.3 = 0.03\)
\(t_3 - t_2 = 0.333 - 0.33 = 0.003\)

∴ The difference between the consecutive terms is not constant.

∴ The given sequence is not an A. P.

Solution:

Here,
\(t_1 = 0, t_2 = - 4, t_3 = - 8, t_4 = - 12\)

Now,
\(t_2 - t_1 = - 4 - 0 = - 4\)
\(t_3 - t_2 = - 8 - (- 4) = - 8 + 4 = - 4\)
\(t_4 - t_3 = - 12 - (- 8) = - 12 + 8 = - 4\)

∴ The difference between the consecutive terms is − 4 which is a constant.

∴ The given sequence is an A. P. and the common difference \(d\) is − 4.

Solution:

Here,
\(\displaystyle t_1 = - \frac {1}{5}, t_2 = - \frac {1}{5}, t_3 = - \frac {1}{5}\)

Now,

\(\displaystyle t_2 - t_1 = - \frac {1}{5} - \left(- \frac {1}{5}\right) = - \frac {1}{5} + \frac {1}{5} = 0\)

\(\displaystyle t_3 - t_2 = - \frac {1}{5} - \left(- \frac {1}{5}\right) = - \frac {1}{5} + \frac {1}{5} = 0\)

∴ The difference between the consecutive terms is 0 which is a constant.

∴ The given sequence is an A. P. and the common difference \(d\) is 0.

Solution:

Here,
\(t_1 = 3, t_2 = 3 + \sqrt {2}, t_3 = 3 + 2\sqrt {2}, t_4 = 3 + 3\sqrt {2}\)

Now,
\(t_2 - t_1 = \cancel {3} + \sqrt {2} - \cancel {3} = \sqrt {2}\)
\(t_3 - t_2 = 3 + 2\sqrt {2} - (3 + \sqrt {2}) = \cancel {3} + 2\sqrt {2} - \cancel {3} - \sqrt {2} = \sqrt {2}\)
\(t_4 - t_3 = 3 + 3\sqrt {2} - (3 + 2\sqrt {2}) = \cancel {3} + 3\sqrt {2} - \cancel {3} - 2\sqrt {2} = \sqrt {2}\)

∴ The difference between the consecutive terms is \(\sqrt {2}\) which is a constant.

∴ The given sequence is an A. P. and the common difference \(d\) is \(\sqrt {2}\).

Solution:

Here,
\(t_1 = 127, t_2 = 132, t_3 = 137\)

Now,
\(t_2 - t_1 = 132 - 127 = 5\)
\(t_3 - t_2 = 137 - 132 = 5\)

∴ The difference between the consecutive terms is 5 which is a constant.

∴ The given sequence is an A. P. and the common difference \(d\) is 5.

Solution:

Here,
\(a = 10\text { and } d = 5\)

Now, \(t_1 = a = 10\)
\(t_2 = a + d = 10 + 5 = 15\)
\(t_3 = a + 2d = 10 + 2 \times 5 = 10 + 10 = 20\)
\(t_4 = a + 3d = 10 + 3 \times 5 = 10 + 15 = 25\)

∴ That A. P. is 10, 15, 20, 25, ...

Solution:

Here,
\(a = -\:3\text { and } d = 0\)

Now, \(t_1 = a = -\:3\)
\(t_2 = a + d = -\:3 + 0 = -\:3\)
\(t_3 = a + 2d = -\:3 + 2 \times 0 = -\:3 + 0 = -\:3\)
\(t_4 = a + 3d = -\:3 + 3 \times 0 = -\:3 + 0 = -\:3\)

∴ That A. P. is − 3, − 3, − 3, − 3, ...

Solution:

Here,
\(\displaystyle a = -\:7\text { and } d = \frac {1}{2}\)

Now, \(t_1 = a = -\:7\)

\(\displaystyle t_2 = a + d = -\:7 + \frac {1}{2} = \frac {-\:14 + 1}{2} = \frac {-\:13}{2} = -\:6.5\)

\(\displaystyle t_3 = a + 2d = -\:7 + \cancel {2} \times \frac {1}{\cancel {2}} = -\:7 + 1 = -\:6\)

\(\displaystyle t_4 = a + 3d = -\:7 + 3 \times \frac {1}{2} = \frac {-\:14 + 3}{2} = \frac {-\:11}{2} = -\:5.5\)

∴ That A. P. is − 7, − 6.5, − 6, − 5.5, ...

Solution:

Here,
\(a = -\:1.25\text { and } d = 3\)

Now, \(t_1 = a = -\:1.25\)
\(t_2 = a + d = -\:1.25 + 3 = 1.75\)
\(t_3 = a + 2d = -\:1.25 + 2 \times 3 = -\:1.25 + 6 = 4.75\)
\(t_4 = a + 3d = -\:1.25 + 3 \times 3 = -\:1.25 + 9 = 7.75\)

∴ That A. P. is − 1.25, 1.75, 4.75, 7.75, ...

Solution:

Here,
\(a = 6\text { and } d = -\:3\)

Now, \(t_1 = a = 6\)
\(t_2 = a + d = 6 -\: 3 = 3\)
\(t_3 = a + 2d = 6 + 2 \times -\:3 = 6 -\:6 = 0\)
\(t_4 = a + 3d = 6 + 3 \times -\:3 = 6 -\:9 = -\:3\)

∴ That A. P. is 6, 3, 0, − 3, − 6, ...

Solution:

Here,
\(a = -\:19\text { and } d = -\:4\)

Now, \(t_1 = a = -\:19\)
\(t_2 = a + d = -\:19 -\: 4 = -\:23\)
\(t_3 = a + 2d = -\:19 + 2 \times -\:4 = -\:19 -\:8 = -\:27\)
\(t_4 = a + 3d = -\:19 + 3 \times -\:4 = -\:19 -\:12 = -\:31\)

∴ That A. P. is − 19, − 23, − 27, − 31, ...

Solution:

Here,
\(t_1 = a = 5, t_2 = 1, t_3 = - 3, t_4 = - 7\)

Now,
d = \(t_2 - t_1 = 1 - 4 = - 4\)

∴ The first term \((a)\) is 5 and the common difference \((d)\) is \(-\:4\).

Solution:

Here,
\(t_1 = a = 0.6, t_2 = 0.9, t_3 = 1.2, t_4 = 1.5\)

Now,
\(d = t_2 - t_1 = 0.9 - 0.6 = 0.3\)

∴ The first term \((a)\) is 0.6 and the common difference \((d)\) is 0.3.

Solution:

Here,
\(t_1 = a = 127, t_2 = 135, t_3 = 143, t_4 = 151\)

Now,
\(d = t_2 - t_1 = 135 - 127 = 8\)

∴ The first term \((a)\) is 127 and the common difference \((d)\) is 8.

Solution:

Here,
\(\displaystyle t_1 = a = \frac {1}{4}, t_2 = \frac {3}{4}, t_3 = \frac {5}{4}, t_4 = \frac {7}{4}\)

Now,
\(\displaystyle d = t_2 - t_1 = \frac {3}{4} - \frac {1}{4} = \frac {3 - 1}{4} = \frac {\cancelto {1}{2}}{\cancelto {2}{4}} = \frac {1}{2}\)

∴ The first term \((a)\) is \(\displaystyle \frac {1}{4}\) and the common difference \((d)\) is \(\displaystyle \frac {1}{2}\).