n = 30

\(a=\frac{1}{5},d=\frac{2}{5}\)

9

k = 4

a_n = 1 - 4n
\(\Rightarrow\)  a₁ = 1 - 4 x 1 = -3
a₂ = 1 - 4 x 2 = -7
d = a₂ - a₁
= - 7 - ( - 3 ) = - 4
a₂₅ = a + 24d
- 3 = 24 x ( - 4 ) = - 99
Now,    S₂₅ = \(\frac{25}{2}({a}_{1}+{a}_{25})\)
\(=\frac{25}{2}(-3-99)\)
= 25 x ( - 51 ) = - 1275

Given, the penalty for each succeeding day is Rs 50 more than the preceding day, therefore the penalties for the first day,the second day, the third day etc., will form an AP. Here, a= 200,d= 250 - 200 = 50 and n = 30. Clearly, the money required by the contractor to pay as penalty, if he delayed the work by 30 days, will be S₃₀ We know that
\(\begin{aligned} S_n & =\frac{n}{2}[2 a+(n-1) d] \end{aligned}\)
\(\begin{aligned} \therefore \quad S_{30} & =\frac{30}{2}[2 \times 200+(30-1) 50] \\ \end{aligned}\)
= 15 (400 + 1450) = 15 \(\times\) 1850 = 27750
Hence, the contractor has to pay Rs 27750, if he delayed the work.

Here, a₁₂ = 37 and d = 3
then, a₁₂ = 37
\(\Rightarrow\) a + 11d = 37   [\(\because\) a_n = a + (n - 1) d]
\(\Rightarrow\) a + 11(3) = 37   [\(\because\) d = 3]
\(\Rightarrow\) a = 37 - 33 = 4
On putting n = 12, a = 4 and l = a₁₂ = 37 in
\(\begin{aligned} & S_n=\frac{n}{2}(a+l) \text {, we get } \\ \end{aligned}\)
\(\begin{aligned} & \qquad S_{12}=\frac{12}{2}(4+37)=6 \times 41=246 \end{aligned}\)
Hence, a = 4 and S₁₂ = 246.

Let Is term = a, common difference = d.
a₈ = 0 \(\Rightarrow\) a + 7d = 0
\(\Rightarrow\)   a = -7d
Now,   a₁₈ = a + 17
= -7d + 17d = 10d 
\(\Rightarrow\)   3 X a₁₈ = 30d          ..(i)
Also,  a₃₈ = a + 37d   
= -7d + 37d = 30d      ...(ii)
From (i) and (ii), we get
a₃₈ = 3 x a₁₈ Hence proved.

a = 3, d = 15 - 3 = 12.
a_n = a + ( n - 1)d
\(\Rightarrow\)   a₂₁ = 3 + ( 21 - 1 ) X 12 = 243
Let  a_m = a21 + 120
\(\Rightarrow\)   a_m = 243 + 120 = 363
\(\therefore\)      363 = a + ( m - 1 ) d
\(\Rightarrow\)    363 = 3 + ( m - 1 ) X 12
\(\Rightarrow\)     360 = ( m - 1) X 12
\(\Rightarrow\)    m - 1 = 30
\(\Rightarrow\)     m = 31
\(\therefore\)    31st term is 120 more than 21st term.

Let Is term = a, common difference = d.
a₁₀ = a + 9d, a₂₄ = a + 23d
According to the question, a₂₄ = 2 X a₁₀
\(\Rightarrow\) a + 23d = 2 ( a + 9d ) \(\Rightarrow\) a + 23d = 2a + 18d \(\Rightarrow\) a = 5d
Now,  a₁₆ = a + 15d = 5d + 15d = 20d     ..(i)
a₃₆ = a + 35d = 5d + 35d = 40d         ..(ii)
From (i) and (ii), we get
a₃₆ = 2 x a₁₆                 
 Hence proved.

First AP: 63, 65, 67, ....
Here,    a = 63, d = 65 - 63 = 2
a_n = a + ( n - 1 ) d = 63 + ( n - 1 ) 2 = 63 + 2n - 2 \(\Rightarrow\) a_n = 61 + 2n
Second AP: 3, 10, 17.......
Here,    a = 3,   d = 10 - 3 = 7
a_n = a + ( n - 1 ) d = 3 + ( n - 1 ) 7 = 3 + 7n - 7 = 7n - 4
Now,    a_n of first AP = a_n of second AP
\(\Rightarrow\)     61 + 2n = 7n - 4 \(\Rightarrow\) 61 + 4 = 7n - 2n
\(\Rightarrow\)   65 = 5n  \(\Rightarrow\) n = 13

Here, we have
a₂ - a₁ = a² - a = 1 (a - 1),
and a₃ - a₂ = a³ - a² = a²(a - 1)
Since, a₂ - a₁ \(\neq\) a₃ - a₂
Therefore, the given list of numbers does not form an AP.

Here, we have   
\(a_{2}-a_{1}=\frac{5}{2}-2=\frac{5-4}{2}=\frac{1}{2}\),
\(a_{3}-a_{2}=3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}\)
\(a_{4}-a_{3}=\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\) and so on.
Since, the difference of any two consecutive terms is same. therefore, the given list of numbers forms an AP and its common difference (d) is \(\frac{1}{2}\).
Now, next three terms of this AP are,
a₅ = a₄ + d = \(\frac{7}{2}+\frac{1}{2}\)
[\(\because\) a₅ = a + 4d = a + 3d + d = a₄ + a₅]
\(=\frac{7+1}{2}=\frac{8}{2}=4\)
\(a_{6}=a_{5}+d=4+\frac{1}{2}=\frac{9}{2}\)
and \(a_{7}=a_{6}+d=\frac{9}{2}+\frac{1}{2}=\frac{9+1}{2}=\frac{10}{2}=5\)

a₁ = -1, d = \(\frac{1}{2}\) and a₂ = \(\frac{-1}{1}+\frac{1}{2}+\frac{-1}{2}\)
a₃ =\(\frac{1}{2}\)+ d = \(\frac{-1}{2}+\frac{1}{2}=0\) and a₄ = \(0+d=0+\frac{1}{2}=\frac{1}{2}\)
The first four terms of the AP are -1, \(-\frac{1}{2}\),  \(0,\frac{1}{2}\)