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

Correct Option: D

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

Correct Option: C

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.

In the given AP,
we have a = 3, d = (9 - 3) = 6, n =15
T₁₅ = a + (n - 1)d

Correct Option: B

In the given AP,
we have a = 3, d = (9 - 3) = 6, n =15
T₁₅ = a + (n - 1)d
= 3 + (15 -1)6
= 3 + 84 = 87

Given, F = 2E + 4Y ...(I)
√EY = 4√3
⇒ EY = 48 ...(II)
and 2EY/ E + Y = 6 ⇒ E + Y = 16 ...(III)

Correct Option: C

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

Correct Option: B

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