Since 18, a, b, and - 3 are in A.P., Then
a-18 = -3-b
\(\Rightarrow\)a + b = - 3 + 18 -
\(\Rightarrow\) a + b = 15 

3y - 1, 3y + 5 and 5y + 1 in AP.
\(\therefore\)(3y + 5)-(3y-1) = (5y + 1)-(3y + 5)
\(\Rightarrow\) 3y + 5 - 3y + 1 = 5y + 1- 3y - 5
\(\Rightarrow\) 6 = 2y-4
\(\Rightarrow\) 2y = 6 + 4
\(\Rightarrow\)y = 10/2
= 5

Let S_n = 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n.
Therefore,   \(\mathrm{S}_{n}=\frac{n(1+n)}{2} \text { or } \mathrm{S}_{n}=\frac{n(n+1)}{2}\)
So, the sum of first n positive integers is given by
\(\mathrm{S}_{n}=\frac{n(n+1)}{2}\) 

  a = p and d = q
 a_n = a + (n - 1)d
a₁₀ = p + 9q

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.

a_n = a + ( n - 1 ) d; a_k = a + ( k - 1 )d
Now,     a_n - a_k = [ a + ( n - 1 ) d ] - [ a + ( k - 1 ) d ] = ( n - 1 ) d - ( k - 1 ) d = ( n - 1 - k + 1 )d
\(\Rightarrow\)  a_n - a_k = ( n - k ) d  
(i) Here a₁₁ = 5, a₁₃ = 79
Taking n = 13, k = 11, eq. (i) becomes
a₁₃ - a₁₁ = ( 13 - 11 )d \(\Rightarrow\) 79 - 5 = 2d \(\Rightarrow\) d = 37
(ii) Let a₁₈ = x
\(\therefore\)      a₂₀ = x + 10
Taking n = 20 and k = 18, equation (i) becomes
a₂₀ - a₁₈ = (  20 - 18 ) d \(\Rightarrow\) ( x + 10 ) -x = 2d \(\Rightarrow\) d = 5

a = 25, d = 50 - 25 = 25, x = 1000
A.T.Q.,         a_k = x
\(\Rightarrow\)    a + (k - 1)d = 1000
\(\Rightarrow\) 25 + (k - 1)25 = 1000
\(\Rightarrow\)  (k - 1)25 = 975
\(\Rightarrow\)   k - 1  = \(\frac{975}{25}\)
\(\Rightarrow\)   k - 1 = 39
\(\Rightarrow\)  k = 40

Let Ist term of AP = a and common difference = d.
Now,    a₆ = -10 \(\Rightarrow\)  a +  5d  = -10                 ..(i)
Also,     a₁₀ = -26  \(\Rightarrow\)    a + 9d = -26           ...(ii)
Subtract (i) from (ii),    
a + 9d = - 26
a + 5d = -10
-  -          +
 4d = -16   \(\Rightarrow\) d = - 4
Substituting in (i), we get
a + 5 x ( -4 ) = - 10  \(\Rightarrow\) a = 10
Now,  a₁₅ = a + 14d = 10 + 14 X  - 4 = - 46

Here, we have a₂ - a₁ = 3² - 1² = 9 - 1 = 8
and a₃ - a₂ =5² - 3² = 25 - 9 =16
Since, a₂ -  a₁ \(\neq\) a₃ - a₂, therefore the given list of numbers does not form an AP.

Here, we have a₂ - a₁ = 0.22 - 0.2 = 0.02
and a₃ - a₂ = 0.222 - 0.22 = 0.002
Since, a₂ - a₁ \(\neq\) a₃ - a₂, therefore the given list of numbers does not form an AP.

0.6, 1.7, 2.8, 3.9 Here,   a₁ = 0.6, a₂ = 1.7
d = a₂ - a₁ = 1.7 - 0.6 = 1.1
So, first term a₁ = 0.6 and common difference, d = 1.1

a₁ = 4, d = -3 and a₂ = 4 + d = 4 - 3 = 1
a₃ = 1 + d = 1 - 3 = -2 and a₄ = -2 + d = -2 - 3 = -5
The first four terms are 4, 1, -2 and -5.

Given, a = 10, d = 10
We know that a, a + d, a + 2d, a + 3d, ... are in AP.
Here, a = 10
a + d = 10 + 10 = 20
a + 2d = 10 + 2 \(\times\) 10 = 30
and a + 3d = 10 + 3 \(\times\) 10 = 40
Thus, the first four terms of the AP are 10, 20, 30 and 40.

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 =  a, common difference = d.
\(\therefore\)  a_m = a + ( m - 1 )d and a_n = a + ( n - 1 )d
A.T.Q.,
m.a_m = n.a_n                      \(\Rightarrow\) m { a + ( m - 1 )d} = n{ a + ( n - 1 )d }
\(\Rightarrow\)    ma - na = n( n -1 )d - m ( m - 1)d - m( m - 1 )d   \(\Rightarrow\) ( m - n ) a = ( n² - n - m² + m )d
\(\Rightarrow\)   ( m - n ) a = ( n - m ) ( m + n - 1 )d   \(\Rightarrow\) a = - ( m + n - 1 )d
Now    a_{m + n} = a + ( m + n - 1)d = - ( m + n - 1 )d + ( m + n - 1 )d = 0

n^th term of A.P. is
t_n = a + (n - 1)d
\({ S }_{ n }=\frac { n }{ 2 } [2a+(n-1)d\)
\(\Rightarrow\) S₁₀ = 5(2a + 9d)
\(\Rightarrow\) 210 = 5(2a + 9d)
\(\Rightarrow\) 2a + 9d = 42            ....(i)
Also, 15^th term from last
= (50 - 15 + 1)^th
= 36^th term.
\(\therefore\) Sum of last 15 terms= \(\frac { 15 }{ 2 } \)[2a₃₆ +(15-1)^th]
2565 = 15a + 42d         ....(ii)
\(\Rightarrow\) a = 3,d = 4
\(\therefore\)  A.P.is 3, 7, 11, 15 and 199.

a₃ + a₇ = 6; a₃ \(\times\) a₇ = 8
\(\Rightarrow\) 2a + 8d = 6; (a + 2d)(a + 6d) = 8. 
\(\Rightarrow\) a + 4d = 3 \(\Rightarrow\) a = 3 - 4d.
(3 - 4d + 2d)(3 - 4d + 6d) = 8
\(\Rightarrow\) (3 + 2d)(3 - 2d) = 8 \(\Rightarrow\) 9 - 4d² = 8
\(\Rightarrow \quad d=\pm \frac { 1 }{ 2 } \)
Case (i): \(d=\frac { 1 }{ 2 } \Rightarrow a=1;{ S }_{ 20 }=115\)
Case (ii): \(d=-\frac { 1 }{ 2 } \Rightarrow a=5;{ S }_{ 20 }=5\)