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) ₹ $\Big(16(4^n - 1)\Big)$

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.