Here S₁₄ = 1050
n = 14
a = 10
S_n = \(\frac{n}{2}\) 2(2a + (n -1)d)
1050 = \(\frac{14}{2}\)(20 + 13d)
= 140 + 91d
910 = 91d
d = 10
a₂₀ = 10 + (20 - 1) x 10
= 20
∴ 20th term = 200.

The number of rose plants in the 1st, 2nd, 3rd, . . . rows are
23,21, 19, ... 5
It forms an A.P.
Let the number of rows in the flower bed be n.
Then a = 23, d = 21 - 23 = -2/a = 5.
As, a_n = a + (n - 1)d i.e. t_n = a + (n - 1)d
We have 5 = 23 + (n - 1)(-2)
i.e. -18 = (n - 1)(-2)
n = 10
ஃ There are 10 rows in the flower bed.

We have
a₃ = a + (3 - 1)d = a + 2d = 5  (1)
a₇ = a + (7 - 1)d = a + 6d = 9 (2)
(1) - (2) ⇒ -4d -4 ⇒ d = 1.
Sub, d = 1 in (1), we get
a + 2(1) = 5
a = 3
Hence the required A.P. is 3, 4, 5, 6, 7.

1,1,1,2,2,2,3,3,3
t₂ - t₁ = 1-1 = 0
t₃ - t₂ = 1-1 = 0
t₄ - t₃ = 2-1 = 1
Here t₂ - t₁ ≠ t₃ - t₂
ஃ It is not an A.P.

-2, 2, -2, 2, -2
t₂ - t₁ = 2-(-2) = 4
t₃ - t₂ = -2 -2 = -4
t₄ - t₃ = 2 - (-2) = 4
It is not an A.P.

1,-1,-3, -5, ...
t₂ - t₁ = -1 - 1 = -2
t₃ - t₂ = -3 - (-1)= -2
t₄ - t₃ = -5 - (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

4, 10, 16,22, ...
We have a² - a² = 10 - 4 = 6
a³ - a² = 16 -10 = 6
ஃ It is an A.P. with common difference 6
ஃ The next two terms are, 

The natural numbers between 300 and 600 which are divisible by 7 are 301, 308, 315, …, 595.
The sum of all natural numbers between 300 and 600 is 301 + 308 + 315 +...+ 595
The terms of the above series are in A.P.
First term a = 301; common difference d = 7; Last term l = 595.
\(n=\left( \frac { l-a }{ d } \right) +1=\left( \frac { 595-301 }{ 7 } \right) +1=43\)
Since, \({ S }_{ n }=\frac { n }{ 2 } \left[ a+l \right] \), we have \({ s }_{43 }=\frac { 43 }{ 2 } \left[ 301+595 \right] \) = 19264 

Given the 13^th term = 3 so, t₁₃ = a + 12d = 3..... (1)
Sum of first 13 terms = 234 gives \(\frac { 13 }{ 2 } \) [2a + 12d] = 234
2a + 12 = 36...(2)
Solving (1) and (2) we get , a = 33, d = \(\frac { -5 }{ 2 } \)
Therefore, common difference is \(\frac { -5 }{ 2 } \).
Sum of first 21 terms S₂₁ = \(\frac { 21 }{ 2 } \left[ 2\times 33+\left( 21-1 \right) \times \left( -\frac { 5 }{ 2 } \right) \right] =\frac { 21 }{ 2 }\)[66 - 50] = 168.

Here we have to find the value of m, such that S_n = 190.
Sum of first n terms of an A.P
S_n = \(\frac { n }{ 2 } \)[2a + (n - 1)d] = 190
\(\frac { n }{ 2 } \)[2 x 1 + (n - 1) x 4 ] = 190
n [4n - 2] = 380
2n² -n - 190 = 0
(n - 10) (2n + 19) = 0
But n = 10 as n = \(\frac { 19 }{ 2 } \) is impossible , therefore, n = 10