Let the first term and the common difference of the given AP be a and d, respectively.
According to the question,
Third term + Seventh term = 6
\(\Rightarrow\) (a + 2d) + (a + 6d) = 6
\(\Rightarrow\)  a + 4d = 3  ....(i)
and Third term x Seventh term = 8
\(\Rightarrow\) (a + 2d) (a + 6d) =8
\(\Rightarrow\) {(a + 4d) - 2d}{(a + 4d) + 2d} = 8
\(\Rightarrow\) (3- 2d)(3 + 2d) = 8    [using Eq. (i)]
\(\Rightarrow\) 9 - 4d² = 8  [\(\because\)(a - b) (a + b) = a² - b²]
\(\Rightarrow\) 4d² = 9 - 8
\(\Rightarrow \quad d^2=\frac{1}{4}\)
\(\Rightarrow \quad d= \pm \frac{1}{2}\)
When, d = \(\frac{1}{2}\), then from Eq. (i), we get
\(\begin{aligned} a+4\left(\frac{1}{2}\right) & =3 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & a+2 & =3 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & a & =3-2 \Rightarrow a=1 \end{aligned}\)
Now, sum of first sixteen terms of the AP,
\(\begin{aligned} S_{16} & =\frac{16}{2}[2 a+(16-1) d] \quad\left[\because S_n=\frac{n}{2}\{2 a+(n-1) d\}\right] \\ \end{aligned}\)
\(\begin{aligned} =8[2 a+15 d]=8\left[2(1)+15\left(\frac{1}{2}\right)\right]\left[\because a=1, d=\frac{1}{2}\right] \\ \end{aligned}\)
\(\begin{aligned} =8\left(2+\frac{15}{2}\right)=8\left(\frac{19}{2}\right)=76 \end{aligned}\)
When, d = -\(\frac{1}{2}\), then from Eq. (i), we get
\(a+4\left(-\frac{1}{2}\right)=3 \Rightarrow a-2=3 \Rightarrow a=5\)
Now, sum of first sixteen terms of this AP,
\(\begin{aligned} S_{16} & =\frac{16}{2}[2 a+(16-1) d]\left[\because S_n=\frac{n}{2}\{2 a+(n-1) d\}\right] \\ \end{aligned}\)
\(\begin{aligned} =8(2 a+15 d) \\ \end{aligned}\)
\(\begin{aligned} =8\left[2(5)+15\left(-\frac{1}{2}\right)\right]=8\left(10-\frac{15}{2}\right)=8\left(\frac{5}{2}\right)=20 \end{aligned}\)
Hence, the sum of first sixteen terms, S₁₆ = 20 or 76.

Here, S₁₄ = 1050, n = 14, a = 10.
As  \(\begin{aligned} \mathrm{S}_n & =\frac{n}{2}[2 a+(n-1) d] \\ \end{aligned}\)
So, \(\begin{aligned} 1050 & =\frac{14}{2}[20+13 d]=140+91 d \end{aligned}\)
i.e., 910 = 91d
or, d = 10
Therefore, a₂₀ = 10 + (20 - 1) \(\times\) 10 = 200, i.e. 20th term is 200.

Let the first three terms of an A.P. be a - d, a, a + d
\(\therefore\)  a - d + a + a + d = 33
\(\Rightarrow\)     3a = 33
a = 11
Now, according to the given condition|
( a - d ) ( a + d ) = a + 29
( 11 - d ) ( 11 + d ) = 11 + 29
\(\Rightarrow\)     121 - d1 = 40
\(\Rightarrow\)    d² = 81
\(\Rightarrow\)  d = \(\pm\) 9
\(\therefore\)  The required A.P is 2, 11, 20, ... or 20, 11, 2, ...

Let the three numbers in A.P. be a - d, a, a + d
\(\therefore\)    Sum is   a - d + a + a + d =15
\(\Rightarrow\)  3a = 15
\(\Rightarrow\)  a = 5
Also, product is ( a - d ) (a) ( a + d ) = 80
( 5 - d ) (5) ( 5 + d ) = 80
5 ( 25 - d² ) = 8
\(\Rightarrow\)      25 - d² = \({80 \over 5}\)
\(\Rightarrow\)     25 - d2 = 16
\(\Rightarrow\)          d² = 9
\(\Rightarrow\)    d = \(\pm\) 3 
\(\therefore\) Three numbes in A.P. are 5, -3, 5, 5 + 3
i.e.,   2, 5, 8
or  5 + 3,  5, 5 - 3
i.e.,  8, 5, 2

Let the three numbers in A.P. be a - d, a, a + d
Sum is a - d + a + d = 21
3a  = 21
a = 7
Product is ( a - d ) (a) ( a + d ) = 231
( 7 - d ) (7) ( 7 + d ) = 231
49 - d² = 33
d² = 16
d = \(\pm\) 4
\(\therefore\)  Numbers are 7- 4, 7, 7 + 4
i.e.,        3, 7, 11
or      7 + 4, 7, 7 - 4
i.e.,   11 7, 3

13, 17, 21

Let d be the common difference of the AP.
LHS = \({1 \over{T}_{1}{T}_{2}}+{1 \over{T}_{2}{T}_{3}}+{1 \over{T}_{3}.{T}_{4}}+ ...{1 \over{T}_{n-1}.{T}_{n}}\)
\(={1\over d}\left[ {d \over {T}_{1}{T}_{2}}+{d \over {T}_{2}{T}_{3}}+{d \over {T}_{3}.{T}_{4}}+...+{d \over {T}_{n-1}{T}_{n}} \right]\)
\(={1\over d}\left[ {{T}_{2}-{T}_{1}\over{T}_{1}{T}_{2}}+{{T}_{3}-{T}_{2}\over{T}_{2}.{T}_{3}}+{{T}_{4}-{T}_{3}\over{T}_{3}{T}_{4}} +...+{{T}_{n}-{T}_{n-1}\over{T}_{n-1}{T}_{n}}\right]\)
\(\because\)   d = T₂ - T₁ = T₃ - T₂ = ...    
\(={1\over d}\left[ {{T}_{2}\over{T}_{1}{T}_{2}}-{{T}_{1}\over{T}_{1}{T}_{2}}+{{T}_{3}\over{T}_{2}{T}_{3}} +{{T}_{4}\over{T}_{3}{T}_{4}}- ... +{{T}_{n}\over{T}_{n-1}{T}_{n}}-{{T}_{n-1}\over{T}_{n-1}{T}_{n}} \right]\)
\(={1\over d}\left[ {1\over{T}_{1}}+{1\over{T}_{2}}+{1\over{T}_{2}}+{1\over{T}_{3}}+{1\over{T}_{3}}-{1\over{T}_{4}}+...+{1\over{T}_{n-1}}-{1\over{T}_{n}} \right]\)
\(={1\over d }\left[ {1 \over {T}_{1}}-{1 \over {T}_{n}} \right]\)
\(={1 \over d}\left[ {{T}_{n}-{T}_{1}\over{T}_{1}{T}_{n}} \right]={1 \over d}\left[ {{T}_{1}+(n-1)d-{T}_{1}\over{T}_{1}.{T}_{n}} \right]\)
\(={n-1\over {T}_{1}.{T}_{n}}\) = RHS

Let a be the Ist term and d be common difference
\(\therefore\)         d = T_n - T_n-1           ....(i)
  [Tn \(\longrightarrow\) nth term]
Tn = S_n - S_n-1
and T_n-1 = S_n-1 - S_n-2
Putting the value of Tn and Tn-1 in equation (i), we get
d = S_n - S_{n -1} - S_{n -1} + S_{n -2}
= S_n - 2_Sn-1 + S_n-2

Let Ist term be a and common difference be d.
Now, a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₄ = 300
\(\Rightarrow\)  a + ( a + 4d ) + ( a + 9d ) + ( a + 14d ) + ( a + 19d )+ ( a + 23d ) = 300
\(\Rightarrow\)    6a + 69d = 300
\(\Rightarrow\)  3 ( 2a + 23d ) = 300
\(\Rightarrow\)  2a + 23d = 100              ...(i)
Now,   S₂₄ = \({24 \over 2}[2a+(24-1)d]\)
= 12 [ 2a + 23d ]
= 12 x 100 = 1200       [ Using (i) ]

Here a = \(\frac{a-b}{a+b},d=\frac{3a-2b}{a+b}-\frac{a-b}{a+b}\)
\(={2a-b\over a+b}\) and n = 11.
S_n  = \({n\over2}[2a+(n-1)d]\)
\(\Rightarrow\) S₁₁ = \(\frac{11}{2}\left[ 2\left( a-b\over a+b \right)+(11-1)\left( 2a-b \over a+b \right)\right]\)
\(\Rightarrow\)  S₁₁ = \({11\over2}\times2\left[ {a-b\over a+b }+{5(2a-b)\over a+b} \right]\)
\(\Rightarrow\)   S₁₁ = \(11\left[ {a-b+10a-5b\over a+b} \right]\)
\(\Rightarrow\)   S₁₁ = \(11\left[ 11a-6b\over a+b \right]\)

Let a be the first term and d be the common difference of the AP.
Gien,         T₂ = \(7\frac{3}{4}\)
\(\Rightarrow\)    a + d = \(\frac{31}{4}\)           ...(i)
and         T₃₁ = \(\frac{1}{2}\)
\(\Rightarrow\)    a + 30d = \(\frac{1}{2 }\)
Subtracting (i) from (ii), we get
29d = \(\frac{1}{2}-\frac{31}{4}=-\frac{29}{4}\)   \(\Rightarrow\) d = - \(\frac{1}{4}\)
Putting the value of d in (i), we get
\(a-\frac{1}{4}=\frac{31}{4}\)      \(\Rightarrow\)  \(a=\frac{31}{4}+\frac{1}{4}=\frac{32}{4}=8\)
Let the number of terms be n, so that
T_n = \(-\frac{13}{2}\)
i.e.,  a + ( n - 1 )d = - \(\frac{13}{2}\)  \(\Rightarrow\)   \(8-\frac{n}{4}+\frac{1}{4}=-\frac{13}{2}\)
\(\Rightarrow\)   32 - n + 1 = - 26 \(\Rightarrow\) n = 59
Hence, first term = 8 and number of terms = 59.

Numbers should be of the form 4m + 3, i.e., 11, 15, 19,......, 299
 a = 11,  d = 4,    a_n = 299
 a_n = a + (n - 1)d
 299 = 11 + (n - 1)4
\(\Rightarrow\)  299 - 11 = (n - 1)4
\(\Rightarrow\)  \(\frac{288}{4}\) = n - 1
\(\Rightarrow\)  72 = n - 1
\(\Rightarrow\)   73 = n
\(\Rightarrow\)     n = 73

Given;     a₄ = 18
\(\Rightarrow\)  a + 3d = 18    ...... (i)
and   a₁₅ - a₉ = 30
\(\Rightarrow\)   (15 - 9)d = 30
\(\Rightarrow\)     6d = 30
\(\Rightarrow\)    d = 5
Putting the value of d in (i), we have
  a + 3d = 18
\(\Rightarrow\)  a + 3 x 5 = 18
\(\Rightarrow\)      a + 15 = 18
\(\Rightarrow\)     a = 3
Required AP is 3, 8, 13,......

Here, a = -6, d = - 2 +  6 and a_n = 58
a_n = 58
\(\Rightarrow\)    a + ( n - 1 )d = 58  \(\Rightarrow\) - 6 + ( n - 1 )4 = 58
\(\Rightarrow\)    ( n - 1 )4 = 64  \(\Rightarrow\) n - 1 = 16 \(\Rightarrow\) n = 17 (odd)
\(\therefore\)       Middle term = \(\frac{17+1}{2}=\frac{18}{2}\) = 9th term
\(\therefore\)   9th term is the middle term.
Now,    a₉ = a + 8d = - 6 + 8 X 4 = - 6 + 32 = 26

Let Ist term of AP be a and common difference be d.
Now, a₉ = 0
\(\Rightarrow\)     a + 8d = 0 \(\Rightarrow\) a = - 8d                              ...(i)
Now,    a₂₉ = a + 28d = -8d + 28d          [Using eq.(i)]
\(\Rightarrow\)   a₂₉ = 20d                                                 ...(ii)
Also,  a₁₉ = a + 18d = -8d + 18d = 10d
\(\Rightarrow\)    2 X a₁₉ = 2 X 10d = 20d                            ...(iii)
From (ii) and (iii), we have        
a₂₉ = 2 X a₁₉