How many terms of the series -20, -16, -12, ... must be taken so that the sum is 120?

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

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

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

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

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

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

The number of bacteria at any given time forms a GP whose terms are given by 50, 100, 200, ... where
a = 50, r = 100/50 = 2
T_n = ar^{n -1}
? T₁₂ = 50(2)^{12 - 1}
= 50 x (2)¹¹
= 102400
So, the number of bacteria born in 12th hour is 102400.

This is a GP with a = 5 , r = 15/5 = 3, n = 7
S_n = a(rⁿ - 1 )/(r - 1)
? S₁₅ = 5(3⁷ - 1)/(3 - 1)
= 5/2(3⁷ - 1)
= 5/2 (2187 - 1)
= ? 5465