Question

2. Which is larger?
The common ratio, r , in a geometric sequence whose second term is
24 and whose fifth term is 1536
or
The common difference, d , in an arithmetic sequence whose fourth
term is 16 and whose seventh term is 31.

Answer 1

Answer:

The common difference d is larger than the common ratio r

Step-by-step explanation:

• The common difference in the arithmetic sequence  $d=u_{n}-u_{n-1}$

• The nth term in the arithmetic sequence is $a_{n}=a+(n-1)d$ , where a is the first term

• The common ratio in the geometric sequence $r=\frac{u_{n}}{u_{n-1}}$

• The nth term in the geometric sequence is $a_{n}=a(r)^{n-1}$ , where a is the first term

Geometric sequence

∵ The second term is 24

∴ $u_{2}$ = 24

∵ $u_{2}=a(r)^{2-1}=ar$

- Equate it by its value

∴ ar = 24 ⇒ (1)

∵ The fifth term is 1536

∴  $u_{5}$ = 1536

∵ $u_{5}=a(r)^{5-1}=ar^{4}$

- Equate it by its value

∴ $ar^{4}$ = 1536 ⇒ (2)

Divide (2) by (1)

∴ $\frac{ar^{4}}{ar}=\frac{1536}{24}$

- Divide up and down by ar

∴ r³ = 64

- Take ∛  for both sides

∴ r = 4

Arithmetic sequence

∵ The fourth term is 16

∴ $u_{4}$ = 16

∵ $u_{4}$ = a + (4 - 1)d

∴ $u_{4}$ = a + 3 d

- Equate it by its value

∴ a + 3d = 16 ⇒ (1)

∵ The seventh term is 31

∴ $u_{7}$ = 31

∵ $u_{7}$ = a + (7 - 1)d

∴ $u_{7}$ = a + 6 d

- Equate it by its value

∴ a + 6 d = 31 ⇒ (2)

Subtract equation (1) from equation (2) to eliminate a and find d

∵ (a - a) + (6 d - 3 d) = (31 - 16)

∴ 3 d = 15

- Divide both sides by 3

∴ d = 5

∵ r = 4 and d = 5

∴ d > r

∴ The common difference d is larger than the common ratio r