Math Problem Statement

If pt+1 = 0.96pt and p0 = 14, what is p69? If pt+1 = pt + 0.13 and p0 = 20, what is p100?

Solution

Let’s analyze the problem:

Part (a)

We are given the recursive relation:

pt+1=0.96pt,with p0=14.p_{t+1} = 0.96p_t, \quad \text{with } p_0 = 14.

We need to find p69p_{69}. This is a geometric progression where:

pt=p0(0.96)t.p_t = p_0 \cdot (0.96)^t.

Substitute p0=14p_0 = 14 and t=69t = 69:

p69=14(0.96)69.p_{69} = 14 \cdot (0.96)^{69}.

Let’s calculate this.

Part (b)

We are given the recursive relation:

pt+1=pt+0.13,with p0=20.p_{t+1} = p_t + 0.13, \quad \text{with } p_0 = 20.

This is an arithmetic progression. The general formula for an arithmetic progression is:

pt=p0+td,p_t = p_0 + t \cdot d,

where dd is the common difference. Here, d=0.13d = 0.13.

Substitute p0=20p_0 = 20 and t=100t = 100:

p100=20+1000.13.p_{100} = 20 + 100 \cdot 0.13.

Let me compute these values.ations for the given problem

Part (a) calculations

p0_a = 14 # Initial value for part (a) decay_factor = 0.96 # Decay factor t_a = 69 # Time index for part (a) p69 = p0_a * (decay_factor ** t_a)

Part (b) calculations

p0_b = 20 # Initial value for part (b) increment = 0.13 # Increment factor

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Math Problem Analysis

Mathematical Concepts

Sequences
Geometric Progression
Arithmetic Progression

Formulas

General formula for geometric progression: pt = p0 * (common ratio)^t
General formula for arithmetic progression: pt = p0 + t * common difference

Theorems

None explicitly mentioned, but utilizes properties of progressions

Suitable Grade Level

Grades 9-12