In the given AP,
we have a = 13 , d = (8 - 13) = -5 and n = 10
T_n = a + (n - 1)d
T₁₀ = 13 + (10 - 1) - 5
= 13 + 9 x (-5) = 13 - 45 = -32

We have, a = 10 , d = (8 - 10) = -2, T_n = - 28
T_n = a + (n - 1)d
? -28 = 10 + (n - 1) x (- 2)
? -28 = 10 - 2n + 2
? 2n = 12 + 28
? 2n = 40
? n = 20

T₉ = a + (9 - 1)d = a + 8d
T₅ = a + (5 - 1)d = a + 4d
T₉ - T₅ = 32
? (a + 8d) - (a + 4d) = 32
? 4d = 32
? d = 8
T₉ + T₅ = (a + 8d) + (a + 4d)
= 2a + 12d
T₉ + T₅ = 114
? 114 = 2(a + 6d)
? a + 6d = 57
? T₈ = a + 7d
= a + 6d + d
= 57 + 8 = 65

This is an AP with a = 1 and d = 7
? 11th term = a + (n - 1) x d = 1 + (11 - 1) x 7
= 1 + 10 x 7 = 71

The first 100 terms of this series is
(1 - 2 - 3) + (2 - 3 - 4) + ... + (33 - 34 - 35) + 34
The first 33 terms of the above series will given an AP as
-4, -5, -6, ... -36
? Answer = 33 x (-20 ) + 34 = - 626

The least number between 200 and 400, which is divisible by 8 is 200. The last term less than 400, which is divisible by 8 is 400 .
? The series is 200, 208, 216,... , 400.
The number of terms in the AP is
(400 - 200/ 8) + 1 = 26
n = 26
d = 8
? Sum = n/2[2a + (n - 1) d]
= 26/2 [2x 200 + (26 -1) x 8]
= 13[400 + 200] = 7800

S_n = 120, a = -20, d = 4
S_n = n/2[2a + (n - 1)d ]
? 120 = n/2 [2 x (-20) + (n - 1) x 4]
? 120 = -20n + 2(n -1)n
? 120 = -20n + 2n² - 2n
? 120 = 2n² - 22n
? 2n² - 22n - 120 = 0
? n²-11n - 60 = 0
? n² - 15n + 4n - 60 = 0
? n(n - 15) + 4(n - 15) = 0
? (n + 4) (n - 15) = 0
? n = -4 or 15
? n = 15

Sum of even numbers = n/2[(n/2) + 1)] = (1672/2) x [(1672/2) + 1]
= 699732

As, n is odd.
? Sum of even number = [(n - 1)/2 ] [ (n + 1 )/2]
= 280/2 x 282/2 = 19740

As, S₁₂ = S₁₈
So, S₁₁ = S₁₉
S₁₀ = S₂₀
S₉ = S₁₉
.........
.........
.........
S₀ = S₃₀
As, S₀ = 0
So, S₃₀ = 0