This problem involves an arithmetic series where Junior starts saving 10 pesos on the first day (September 16, 2024), and adds 3 pesos more than the previous day's amount each day. Let's break this down and solve the questions step by step.

Part A: How much money will he have after 1 year (366 days)?

• First term (a): Junior saved 10 pesos on the first day.
• Common difference (d): He adds 3 pesos more each day, so the common difference is 3.
• Number of terms (n): Since the year is a leap year, $n = 366$.

The formula for the sum of an arithmetic series is:

$S_n = \frac{n}{2} \times (2a + (n - 1)d)$

Substitute the known values into the formula:

$S_{366} = \frac{366}{2} \times (2(10) + (366 - 1)(3))$

$S_{366} = 183 \times (20 + 1095)$

$S_{366} = 183 \times 1115$

$S_{366} = 204,045$

So, after 366 days, Junior will have 204,045 pesos in total.

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Part B: How much money did he put in his piggy bank on the 17th day of October 2024?

To find how much Junior saved on any specific day, we use the formula for the $n$-th term of an arithmetic sequence:

$a_n = a + (n - 1)d$

We need to count how many days it is from September 16 to October 17, 2024:

• September 16 to September 30 = 15 days
• October 1 to October 17 = 17 days

So, the 17th day of October 2024 is the 32nd day of savings. Now, apply the formula:

$a_{32} = 10 + (32 - 1)(3)$

$a_{32} = 10 + 31 \times 3$

$a_{32} = 10 + 93 = 103$

So, on the 17th of October 2024, Junior will have saved 103 pesos.

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Part C: After how many days did he reach a total savings of 2,250 pesos?

We use the sum formula again, but now we're solving for $n$ when the sum equals 2,250 pesos:

$S_n = \frac{n}{2} \times (2a + (n - 1)d)$

Substitute $S_n = 2250$, $a = 10$, and $d = 3$:

$2250 = \frac{n}{2} \times (20 + (n - 1)(3))$

Simplify the equation:

$2250 = \frac{n}{2} \times (20 + 3n - 3)$

$2250 = \frac{n}{2} \times (17 + 3n)$

Multiply both sides by 2 to eliminate the fraction:

$4500 = n(17 + 3n)$

Expand the right-hand side:

$4500 = 17n + 3n^2$

Rearrange into a standard quadratic equation:

$3n^2 + 17n - 4500 = 0$

Solve this quadratic equation using the quadratic formula:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 3$, $b = 17$, and $c = -4500$:

$n = \frac{-17 \pm \sqrt{17^2 - 4(3)(-4500)}}{2(3)}$

$n = \frac{-17 \pm \sqrt{289 + 54000}}{6}$

$n = \frac{-17 \pm \sqrt{54289}}{6}$

$n = \frac{-17 \pm 233}{6}$

Taking the positive root:

$n = \frac{-17 + 233}{6} = \frac{216}{6} = 36$

So, Junior reaches 2,250 pesos after 36 days.

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Summary of Answers:

• A: Junior will have 204,045 pesos after 1 year.
• B: On the 17th of October 2024, he will save 103 pesos.
• C: Junior will reach 2,250 pesos after 36 days.

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Do you want more details or have any questions?

Here are 5 related questions:

1. How much will Junior save on the 100th day?
2. How many days will it take to save 10,000 pesos?
3. What is the total saved after 6 months (183 days)?
4. How much more will Junior save on the 366th day compared to the first day?
5. If Junior starts adding 5 pesos instead of 3, how much will he save in 1 year?

Tip: In an arithmetic series, the constant difference makes the growth predictable, which allows you to solve for any term or sum efficiently!