FIND THE GEOMETRIC MEAN OF THE GIVEN EXTREMES

Problem 1 :

Find the geometric mean of the following numbers

a) 5 and 25

b) 7 and 63

c) -2 and -8

Solution :

a) 5 and 25

Geometric mean = √(x₁ ⋅ x₂)

= √(5 ⋅ 25)

= √125

= √(25 × 5)

= 5√5

So, geometric mean is 5√5.

b) 7 and 63

Geometric mean = √(x₁ ⋅ x₂)

= √(7 ⋅ 63)

= √441

= 21

So, geometric mean is 21.

c) -2 and -8

Geometric mean = √(x₁ ⋅ x₂)

= √((-2) ⋅ (-8))

= √16

= 4

So, geometric mean is 4.

Problem 2 :

Insert 3 numbers between 4 and 324 such that the resulting sequence in a G. P.

Solution :

Let the numbers be G₁, G₂, and G₃.

4, G₁, G₂, G₃, 324 are in G.P.

T₅ = 324

t_n = ar^{(n  - 1)}

ar⁴ = 324

4 × r⁴ = 324

Dividing 4 on both sides.

(4 × r⁴)/4 = 324/4

r⁴ = 81

r⁴ = 3⁴

r = 3

G₁ = 4 × 3 = 12

G₂ = 4 × 3² = 36

G₃ = 4 × 3³ = 108

So, the numbers be 12, 36, 108.

Problem 3 :

A G.P is given 1/2, 1, 2, 4, … If a constant k is multiplied in each term of the G.P., then find the 10^th term of the G.P.

Solution :

Accordingly the properties of geometric progression, if every term is multiplied by some non zero constant, then the new progression created is also a geometric progression with same common ratio.

To find the10^th term of the G.P :

.a_n = a₁r^{n  - 1}

a₁₀ = 1/2 × (2)^{10  - 1}

= 1/2 × 2⁹

= 1/2 × 256

a₁₀ = 128

So, 10th term of the G.P is a₁₀ = 128.