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

Correct Option: B

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 .

Correct Option: A

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 = 120, n/2[2a + (n - 1)d ]

Correct Option: C

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)]

Correct Option: D

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]

Correct Option: D

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