Geometric Progression Explorer

A sequence is in G.P. if the ratio between consecutive terms is constant.

Solutions:

(i) 9/3 = 3, 27/9 = 3, 81/27 = 3 → G.P. (r = 3)

(ii) 44/4 = 11, 444/44 ≈ 10.09 → Not G.P.

(iii) 0.05/0.5 = 0.1, 0.005/0.05 = 0.1 → G.P. (r = 0.1)

(iv) (1/6)/(1/3) = 0.5, (1/12)/(1/6) = 0.5 → G.P. (r = 0.5)

(v) -5/1 = -5, 25/-5 = -5, -125/25 = -5 → G.P. (r = -5)

(vi) 60/120 = 0.5, 30/60 = 0.5, 18/30 = 0.6 → Not G.P.

(vii) 4/16 = 0.25, 1/4 = 0.25, (1/4)/1 = 0.25 → G.P. (r = 0.25)

Using the formula: t₁ = a, t₂ = ar, t₃ = ar²

Solutions:

(i) 6, 6×3 = 18, 6×3² = 54

(ii) 2, 2×2 = 4, 2×2² = 8

(iii) 1000, 1000×(2/5) = 400, 1000×(2/5)² = 160

First, find the common ratio (r) = 243/729 = 1/3

First term (a) = 729

Using the formula: tₙ = arⁿ⁻¹

t₇ = 729 × (1/3)⁶ = 729 × (1/729) = 1

For three consecutive terms a, b, c in G.P., b² = a × c

Here: (x + 12)² = (x + 6)(x + 15)

Expanding: x² + 24x + 144 = x² + 21x + 90

Simplify: 3x = -54 → x = -18

Verification: Terms become -12, -6, -3 which is a G.P. with r = 0.5

Using the formula for n-th term: tₙ = arⁿ⁻¹

(i) Solution:

a = 4, r = 8/4 = 2, tₙ = 8192

4 × 2ⁿ⁻¹ = 8192 → 2ⁿ⁻¹ = 2048 → n-1 = 11 → n = 12

(ii) Solution:

a = 1/3, r = (1/9)/(1/3) = 1/3, tₙ = 1/2187

(1/3) × (1/3)ⁿ⁻¹ = 1/2187 → (1/3)ⁿ = 1/2187

3ⁿ = 2187 → 3ⁿ = 3⁷ → n = 7

Given: t₉ = ar⁸ = 32805, t₆ = ar⁵ = 1215

Divide the equations: (ar⁸)/(ar⁵) = 32805/1215 → r³ = 27 → r = 3

Now find a: a × 3⁵ = 1215 → a × 243 = 1215 → a = 5

Now find t₁₂ = ar¹¹ = 5 × 3¹¹ = 5 × 177147 = 885735

Given: r = 2, t₈ = ar⁷ = 768

Find a: a × 2⁷ = 768 → a × 128 = 768 → a = 6

Now find t₁₀ = ar⁹ = 6 × 2⁹ = 6 × 512 = 3072

Since a, b, c are in A.P., we have: 2b = a + c

Now consider 3ᵃ, 3ᵇ, 3ᶜ

For these to be in G.P., (3ᵇ)² should equal 3ᵃ × 3ᶜ

(3ᵇ)² = 3²ᵇ = 3ᵃ⁺ᶜ (from A.P. condition)

3ᵃ × 3ᶜ = 3ᵃ⁺ᶜ

Thus, (3ᵇ)² = 3ᵃ × 3ᶜ, proving they are in G.P.

Let the terms be a/r, a, ar

Product: (a/r) × a × ar = a³ = 27 → a = 3

Sum of products two at a time:

(a/r × a) + (a/r × ar) + (a × ar) = a²/r + a² + a²r = 57/2

9/r + 9 + 9r = 57/2 → Multiply by 2r: 18 + 18r + 18r² = 57r

18r² - 39r + 18 = 0 → 6r² - 13r + 6 = 0

Solving: r = [13 ± √(169-144)]/12 = [13 ± 5]/12 → r = 3/2 or 2/3

Thus, the terms are:

When r = 3/2: 2, 3, 9/2

When r = 2/3: 9/2, 3, 2

This is a G.P. problem where each year's salary is 105% of previous year's.

a = 60000, r = 1.05

Salary after 5 years = t₆ = ar⁵ = 60000 × (1.05)⁵

≈ 60000 × 1.27628 ≈ ₹76,577 (rounded to nearest rupee)

Both offers form G.P. sequences:

Offer A: a = 20000, r = 1.06

4th year salary = t₄ = ar³ = 20000 × (1.06)³ ≈ 20000 × 1.191016 ≈ ₹23,820

Offer B: a = 22000, r = 1.03

4th year salary = t₄ = ar³ = 22000 × (1.03)³ ≈ 22000 × 1.092727 ≈ ₹24,040

Since a, b, c are in A.P., we have: b - a = c - b = d (common difference)

Thus: b - c = -d, c - a = 2d, a - b = -d

Since x, y, z are in G.P., we can write: y² = xz

Let y = xr, z = xr² where r is common ratio

Now, the expression becomes:

x⁻ᵈ × (xr)²ᵈ × (xr²)⁻ᵈ = x⁻ᵈ × x²ᵈ × r²ᵈ × x⁻ᵈ × r⁻²ᵈ = x⁰ × r⁰ = 1