Geometric Progression — Case-Study Based Questions

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

Answer

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.

Case Study

A man writes a letter to four of his friends. He asks each one of them to copy the letter and mail it to four different persons with the instruction that they move the chain similarly. Assume that the chain is not broken and it costs ₹4 to mail one letter.

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

1.The number of letters mailed in the 6th set of letters is :
(a) 2048
(b) 8192
(c) 4096
(d) 1024

2.The amount spent on the postage of 6th set of letters is :
(a) ₹ 16,384
(b) ₹ 8,192
(c) ₹ 32,768
(d) ₹ 4,096

3.The total number of letters mailed till the 5th set of letters is :
(a) 1024
(b) 4092
(c) 1236
(d) 1364

4.The amount spent on the postage till the 5th set of letters is :
(a) ₹ 16,368
(b) ₹ 4,096
(c) ₹ 5,456
(d) ₹ 4,944

5.The amount spent on the postage till the nth set of letters is :
(a) ₹ $\Big(\dfrac{4}{3}(4^n - 4)\Big)$

(b) ₹ $\Big(\dfrac{16}{3}(4^n - 1)\Big)$

(c) ₹ $\Big(\dfrac{2}{3}(4^n - 1)\Big)$

(d) ₹ $\Big(16(4^n - 1)\Big)$

Answer

Since the number of letters mailed increases by a factor of 4 at every step, the sequence of letters mailed forms a Geometric Progression.

4,16,64.....

1. In the given G.P.,

a = 4

r = 4

We know that,

T_n = ar^{n - 1}

⇒ T₆ = 4.(4)^{6 - 1}

= 4(4)⁵

= 4(1024)

= 4096.

Hence, option (c) is the correct option.

2. In the 6th set, 4096 letters are mailed and each letter costs ₹ 4.

Amount spent on the postage of 6th set of letters = 4096 × ₹ 4 = ₹ 16,384.

Hence, option (a) is the correct option.

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

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

Given,

a = 4

r = 4

n = 5

$\Rightarrow S_5 = \dfrac{4(4^5 - 1)}{4-1} \\[1em] = \dfrac{4(4^5 - 1)}{3} \\[1em] = \dfrac{4(1024 - 1)}{3} \\[1em] = \dfrac{4(1023)}{3} \\[1em] = \dfrac{4092}{3} \\[1em] = 1364.$

Hence, option (d) 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 = 4

r = 4

n = 5

$\Rightarrow S_5 = \dfrac{4(4^5 - 1)}{4-1} \\[1em] = \dfrac{4(4^5 - 1)}{3} \\[1em] = \dfrac{4(1024 - 1)}{3} \\[1em] = \dfrac{4(1023)}{3} \\[1em] = \dfrac{4092}{3} \\[1em] = 1364.$

Amount spent till the 5th set of letters postage = 1364 × ₹ 4 = ₹ 5,456

Hence, option (c) 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 = 4

r = 4

$\Rightarrow S_5 = \dfrac{4(4^n - 1)}{4-1} \\[1em] = \dfrac{4(4^n - 1)}{3} \\[1em]$

Amount spent till the nth set of letters postage = $\dfrac{4(4^n - 1)}{3}$ × ₹ 4 = $₹ \dfrac{16}{3}(4^n - 1)$.

Hence, option (b) is the correct option.