Case Study

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, then

Based on the above given information, answer the following questions:

1. The number of bacteria that would be present at the end of 4th hour is :
   (a) 240
   (b) 480
   (c) 960
   (d) 450

2. The number of bacteria that would be present at the end of 10th hour is :
   (a) 30720
   (b) 15360
   (c) 61440
   (d) 27540

3. The number of bacteria that would be present at the end of nth hour is :
   (a) 30 × 2^{(n - 1)}
   (b) 30 × 2^{(n + 1)}
   (c) 30 × 2ⁿ
   (d) none of these

4. What would be the sum of the first 5 terms of the G.P. that is formed by the number of bacteria after the completion of each hour?
   (a) 450
   (b) 930
   (c) 1890
   (d) none of these

5. What would be the sum of the first n terms of the G.P. that is formed by the number of bacteria after the completion of each hour?
   (a) 30 × 2⁽ⁿ⁾ - 1
   (b) 30 × (2^{(n - 1)} - 1)
   (c) 30 × (2^{(n + 1)} - 1)
   (d) 30 × 2ⁿ - 30

The number of bacteria in a certain culture doubles every hour. this forms G.P.

30, 60, 120, 240…

1. a = 30

r = $\dfrac{60}{30}$ = 2

n = 5 [At the end of 4th hour, means starting of 5th hour]

We know that,

T_n = ar^{n - 1}

Substituting values we get,

T₅ = 30(2)^{5 - 1}

= 30.(2)⁴

= 480.

Hence, option (b) is the correct option.

2. Given,

n = 11 [At the end of 10th hour, means starting of 11th hour]

a = 30

r = 2

We know that,

T_n = ar^{n - 1}

⇒ T₁₁ = 30(2)^{11 - 1}

= 30.(2)¹⁰

= 30.(1024)

= 30720.

Hence, option (a) is the correct option.

3. Given,

a = 30

r = 2

n = n + 1 [At the end of nth hour, means starting of (n + 1) th hour]

We know that,

⇒ T_n = ar^{n - 1}

⇒ T_n = 30.(2)^{n + 1 - 1}

= 30 × (2)ⁿ.

Hence, option (c) is the correct option.

4. Formula for sum of n terms of a G.P.,

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Given,

a = 30

r = 2

n = 5

$\Rightarrow S_5 = \dfrac{30(2^5 - 1)}{2-1} \\[1em] = \dfrac{30(32 - 1)}{1} \\[1em] = 30.(31) \\[1em] = 930.$

Hence, option (b) is the correct option.

5. Formula for sum of n terms of a G.P.

$S_n = \dfrac{a(r^n-1)}{r-1}$ [r > 1]

Given,

a = 30

r = 2

$\Rightarrow S_5 = \dfrac{30(2^n - 1)}{2-1} \\[1em] = \dfrac{30(2^n - 1)}{1} \\[1em] = 30.(2^n - 1) \\[1em] = 30 \times 2^n - 30 .$

Hence, option (d) is the correct option.