Solution:

Here,
\(a = \bbox[white, 5pt, border: 3px solid red]{\ 1\ }, t_1 = \bbox[white, 5pt, border: 3px solid red]{\ 1\ }, t_2 = \bbox[white, 5pt, border: 3px solid red]{\ 8\ }, t_3 = \bbox[white, 5pt, border: 3px solid red]{\ 15\ },\)

\(t_2 - t_1 = \bbox[white, 5pt, border: 3px solid red]{\ 8\ } - \bbox[white, 5pt, border: 3px solid red]{\ 1\ } = \bbox[white, 5pt, border: 3px solid red]{\ 7\ }\)

\(t_3 - t_2 = \bbox[white, 5pt, border: 3px solid red]{\ 15\ } - \bbox[white, 5pt, border: 3px solid red]{\ 8\ } = \bbox[white, 5pt, border: 3px solid red]{\ 7\ }\)

∴ \(d = \bbox[white, 5pt, border: 3px solid red]{\ 7\ }\)

Solution:

Here,
\(\displaystyle t_1 = \bbox[white, 5pt, border: 3px solid red]{\ 3\ }, t_2 = \bbox[white, 5pt, border: 3px solid red]{\ 6\ }, t_3 = \bbox[white, 5pt, border: 3px solid red]{\ 9\ }, t_4 = \bbox[white, 5pt, border: 3px solid red]{\ 12\ }\)

\(t_2 - t_1 = \bbox[white, 5pt, border: 3px solid red]{\ 6\ } - \bbox[white, 5pt, border: 3px solid red]{\ 3\ } = 3\)

\(t_3 - t_2 = \bbox[white, 5pt, border: 3px solid red]{\ 9\ } - \bbox[white, 5pt, border: 3px solid red]{\ 6\ } = 3\)

∴ \(d = \bbox[white, 5pt, border: 3px solid red]{\ 3\ }\)

Solution:

Here,
\(t_3 = \bbox[white, 5pt, border: 3px solid red]{\ -\:13\ }, t_2 = \bbox[white, 5pt, border:3px solid red]{\ -\:8\ }, t_4 = \bbox[white, 5pt, border: 3px solid red]{\ -\:18\ }, t_1 = \bbox[white, 5pt, border: 3px solid red]{\ -\:3\ }\),

\(t_2 - t_1 = \bbox[white, 5pt, border: 3px solid red]{\ -\:5\ }\)

\(t_3 - t_2 = \bbox[white, 5pt, border: 3px solid red]{\ -\:5\ }\)

∴ \(a = \bbox[white, 5pt, border: 3px solid red]{\ -\:3\ },\ d = \bbox[white, 5pt, border: 3px solid red]{\ -\:5\ }\)

Solution:

Here,
\(t_1 = \bbox[white, 5pt, border: 3px solid red]{\ 70\ }, t_2 = \bbox[white, 5pt, border:3px solid red]{\ 60\ }, t_3 = \bbox[white, 5pt, border: 3px solid red]{\ 50\ }\) , ...

∴ \(a = \bbox[white, 5pt, border: 3px solid red]{\ 70\ },\ d = \bbox[white, 5pt, border: 3px solid red]{\ -\:10\ }\)

Solution:

Here,
\(t_1 = -\:2, t_2 = -\:5, t_3 = 2, t_4 = 9,\)
\(t_4 = 9, t_5 = 16, t_6 = 23, t_7 = 30\)

Now,
\(t_2 - t_1 = - 5 - (- 12) = - 5 + 12 = 7\)
\(t_3 - t_2 = 2 - (- 5) = 2 + 5 = 7\)
\(t_4 - t_3 = 9 - 2 = 7\)
\(t_5 - t_4 = 16 - 9 = 7\)
\(t_6 - t_5 = 23 - 16 = 7\)
\(t_7 - t_6 = 30 - 23 = 7\)

∴ The difference between the consecutive terms is 7 which is a constant.
∴ The given sequence is an A. P.

Now, \(t_n = a + (n - 1)d\)
∴ \(t_{20} = - 12 + (20 - 1)\times 7\)
∴ \(t_{20} = - 12 + 19\times 7\)
∴ \(t_{20} = - 12 + 133\)
∴ \(t_{20} = 121\)

∴ The given sequence is an A. P. and its 20^th term is 121.

Solution:

Here,
\(t_1 = a = 12, t_2 = 16, t_3 = 20, t_4 = 24\)

Also,
\(d = t_2 - t_1 = 16 - 12 = 4\)

Now, \(t_n = a + (n - 1)d\)
∴ \(t_{24} = 12 + (24 - 1)\times 4\)
∴ \(t_{24} = 12 + 23\times 4\)
∴ \(t_{24} = 12 + 92\)
∴ \(t_{24} = 104\)

The 24^th term of this progerssion is 104.

Solution:

Here,
\(t_1 = a = 7, t_2 = 13, t_3 = 19, t_4 = 25\)

Also,
\(d = t_2 - t_1 = 13 - 7 = 5\)

Now, \(t_n = a + (n - 1)d\)
∴ \(t_{19} = 7 + (19 - 1)\times 6\)
∴ \(t_{19} = 7 + 18\times 6\)
∴ \(t_{19} = 7 + 108\)
∴ \(t_{19} = 115\)

The 19^th term of this A. P. is 115.

Solution:

Here,
\(t_1 = a = 9, t_2 = 4, t_3 = -\:1, t_4 = -\:6, t_5 = -\:11\)

Also,
\(d = t_2 - t_1 = 4 - 9 = -\:5\)

Now, \(t_n = a + (n - 1)d\)
∴ \(t_{27} = 9 + (27 - 1)\times -\:5\)
∴ \(t_{27} = 9 + 26\times -\:5\)
∴ \(t_{27} = 9 - 130\)
∴ \(t_{27} = - 121\)

The 27^th term of this A. P. is − 121.

Solution:

Here,
\(t_1 = a = 100, d = 5, t_n = 995, n = ?\)

Now, \(t_n = a + (n - 1)d\)
∴ \(t_{27} = 9 + (27 - 1)\times -\:5\)
∴ \(995 = 100 + (n - 1)\times 5\)
∴ \(995 - 100 = (n - 1)\times 5\)
∴ \(895 = (n - 1)\times 5\)
∴ \((n - 1)\times 5 = 895\)

∴ \(\displaystyle n - 1 = \frac {\cancelto {179}{895}}{\cancelto {1}{5}}\)

∴ \(n - 1 = 179\)
∴ \(n = 179 + 1\)
∴ \(n = 180\)

∴ There are 180 three-digit natural numbers which are divisible by 5.

Solution:

Here,
\(t_{11} = 16, t_{21} = 29, t_{41} = ?\)

Now, \(t_n = a + (n - 1)d\)
∴ \(t_{11} = a + (11 - 1)d\)
∴ \(16 = a + 10d\) ... (Given)
i.e. \(a + 10d = 16\) ... (i)

Also, \(t_n = a + (n - 1)d\)
∴ \(t_{21} = a + (21 - 1)d\)
∴ \(29 = a + 20d\) ... (Given)
i.e. \(a + 20d = 29\) ... (ii)

Subtracting (i) from (ii),

∴ \(\displaystyle d = \frac {13}{10}\)

∴ \(d = 1.3\) ... (iii)

Substituting the value of \(d\) in (i),
 \(a + 10d = 16\) ... (i)
∴ \(a + 10 \times 1.3 = 16\)
∴ \(a + 13 = 16\)
∴ \(a = 16 - 13\)
∴ \(a = 3\) ... (iv)

Finally, \(t_n = a + (n - 1)d\)
∴ \(t_{41} = 3 + (41 - 1) \times 1.3\)
∴ \(t_{41} = 3 + 40 \times 1.3\)
∴ \(t_{41} = 3 + 52\)
∴ \(t_{41} = 55\) ... (v)

∴ The 41^st term of that A. P. is 55.

Solution:

Here,
\(t_1 = a = 11, t_2 = 8\),
∴ \(d = t_2 - t_1 = 8 - 11 = - 3\),
\(t_n = -\:151, n = ?\)

Now, \(t_n = a + (n - 1)d\)
∴ \(- 151 = 11 + (n - 1)\times - 3\)
∴ \(- 151 - 11 = (n - 1)\times - 3\)
∴ \(- 162 = (n - 1)\times - 3\)
i.e. \((n - 1)\times - 3 = - 162\)

∴ \(n - 1 = \displaystyle \frac {\cancelto {54}{- 162}}{\cancelto {1}{- 3}}\)

∴ \(n - 1 = 54\)
∴ \(n = 54 + 1\)
∴ \(n = 55\)

∴ The 55^thst term of that A. P. is − 151.

Solution:

Here,
\(t_1 = a = 12, d = 4\),
\(t_n = 248, n =\ ?\)

Now, \(t_n = a + (n - 1)d\)
∴ \(248 = 12 + (n - 1)\times 4\)
∴ \(248 - 12 = (n - 1)\times 4\)
∴ \(236 = (n - 1)\times 4\)
i.e. \((n - 1)\times 4 = 236\)

∴ \(n - 1 = \displaystyle \frac {\cancelto {59}{236}}{\cancelto {1}{4}}\)

∴ \(n - 1 = 59\)
∴ \(n = 59 + 1\)
∴ \(n = 60\)

∴ There are 60 natural numbers between 10 and 250 which are divisible by 4.

Solution:

Here,
\(t_{17} = t_{10} + 7, d =\ ?\)

Now, \(t_n = a + (n - 1)d\)
∴ \(t_{17} = a + (17 - 1)d\)
∴ \(t_{17} = a + 16d\) ... (i)

Also, \(t_n = a + (n - 1)d\)
∴ \(t_{10} = a + (10 - 1)d\)
∴ \(t_{10} = a + 9d\) ... (ii)

Now, \(t_{17} = t_{10} + 7\)
∴ \(\cancel {a} + 16d = \cancel {a} + 9d + 7\) ... From [(i) (ii)]
∴ \(16d = 9d + 7\)
∴ \(16d - 9d = 7\)
∴ \(7d = 7\)

∴ \(d = \displaystyle \frac {\cancelto {1}{7}}{\cancelto {1}{7}}\)

∴ \(d = 1\)

∴ The common difference is 1.