Let's analyze the math problems shown in the image.

Activity 5: Arithmetic Series

I. Find the sum of the following arithmetic sequences:

1. $3, 8, 13, \ldots, A_n$

2. $-32, -12, \ldots, A_n$

3. $238, 232, 226, \ldots, A_n$

II. Solve the following and show your solution:

1. Problem Statement: The Department of Health is asking for an additional fund in the procurement of PPE. The DOH has a budget of PHP 93,312,000 as of today. If the department has to spend PHP 972,000 on day 1, PHP 1,960,000 on day 2, PHP 2,948,000 on day 3, and so on, in how many days will the fund of the department last?

Let's start by solving these problems step-by-step:

Part I: Finding the Sum of the Arithmetic Sequence

1. Sequence: $3, 8, 13, \ldots, A_n$

   • This is an arithmetic sequence with a common difference $d = 8 - 3 = 5$.
   • The first term $a = 3$.

   The sum of the first $n$ terms of an arithmetic sequence is given by: $S_n = \frac{n}{2} \times (2a + (n - 1) \cdot d)$ To find the sum, we need to know the value of $n$ or the last term $A_n$.

2. Sequence: $-32, -12, \ldots, A_n$

   • First term $a = -32$.
   • Common difference $d = -12 - (-32) = 20$.
   • Apply the same sum formula.

3. Sequence: $238, 232, 226, \ldots, A_n$

   • First term $a = 238$.
   • Common difference $d = 232 - 238 = -6$.
   • Apply the same sum formula.

Part II: Solving the Fund Problem

1. Determine the number of days until the funds are exhausted:

The sequence of spending forms an arithmetic sequence where:

• First term $a = 972,000$ (Day 1 expenditure)
• Second term $a_2 = 1,960,000$
• Third term $a_3 = 2,948,000$

Find the common difference $d = a_2 - a_1 = 1,960,000 - 972,000 = 988,000$.

The total expenditure forms an arithmetic series where we need to find $n$ (number of days) such that:

$S_n = \frac{n}{2} \times (2a + (n - 1) \cdot d) \leq 93,312,000$

Let's solve these step-by-step. Do you want me to solve all the problems in detail or focus on specific ones?