Find the sum of even numbers below 281.?

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

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

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

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

Given, F = 2E + 4Y ...(I)
?EY = 4?3
? EY = 48 ...(II)
and 2EY/ E + Y = 6 ? E + Y = 16 ...(III)
Now, (E -Y)² = (E + Y)² - 4EY
= (16)² - 4 x 48
= 256 - 192 = 64
? E - Y = 8 ...(iv)
From Eqs. (iii) and (iv), we get
E = 12 and Y = 4
From Eq. (i) F = 2 x 12 + 4 x 4 = 40 yr

The nth term of a GP is ar^{n - 1} ,
5th term = ar^{5 - 1} = ar⁴ = 81
1st term = a = 16
? r⁴ = 81/16
? r = ?81/16 = 3/2
? 4th term = ar^{4 - 1} = ar³ = 16 x 3/2 x 3/2 x 3/2 = 54