Solve the following word problems related to arithmetic series:

23. A formation of a marching band has 14 marchers in the front row, 16 in the second row, 18 in the third row, and so on for 25 rows. How many marchers are in the last row? How many marchers are there altogether?
24. How many poles will be in a pile of telephone poles if there are 50 in the first layer, 49 in the second layer, and so on, until there are 6 in the last layer?
25. A family saves money in an arithmetic sequence: Rs. 6,000 in the first year, Rs. 7,000 in the second year, and so on for 20 years. How much do they save in all?
26. Mr. Saleem saves Rs. 500 on October 1, Rs. 550 on October 2, and Rs. 600 on October 3, and so on. How much is saved during the whole of October? (October has 31 days).

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Background and Explanation

These problems can be modeled using arithmetic sequences and series. An arithmetic sequence is a list of numbers with a constant common difference (d) between terms. The total of these terms is an arithmetic series.

To solve these, we'll need to translate the word problems into the language of arithmetic series by identifying key values:

• $a_1$: The first term (e.g., marchers in the first row, initial savings).
• $d$: The common difference (the amount by which the value changes each time).
• $n$: The number of terms (e.g., number of rows, years, or days).
• $a_n$: The last term (e.g., marchers in the last row).
• $S_n$: The total sum of all terms.

By identifying the knowns and unknowns in each problem, we can select the appropriate formula to find the solution.

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Solution

Problem 23: Marching Band

This problem has two parts: first, find the number of marchers in the 25th row ($a_{25}$), and second, find the total number of marchers ($S_{25}$).

1. Identify the knowns:

   • First term ($a_1$): 14 (marchers in the front row).
   • Common difference ($d$): $16-14=2$.
   • Number of terms ($n$): 25 (rows).

2. Find the marchers in the last row ($a_{25}$): We use the formula for the n-th term, $a_n=a_1+(n-1)d$.

   • $a_{25}=14+(25-1)(2)$
   • $a_{25}=14+(24)(2)$
   • $a_{25}=14+48=62$. There are 62 marchers in the last row.

3. Find the total number of marchers ($S_{25}$): Now that we know the first and last terms, we can use the sum formula $S_n=\frac{n}{2}\left(a_1+a_n\right)$.

   • $S_{25}=\frac{25}{2}(14+62)$
   • $S_{25}=\frac{25}{2}(76)$
   • $S_{25}=25(38)=950$. There are 950 marchers altogether.

Problem 24: Telephone Poles

We need to find the total number of poles ($S_n$). First, we must determine how many layers ($n$) there are in the pile.

1. Identify the knowns:

   • First term ($a_1$): 50 (poles in the first layer).
   • Last term ($a_n$): 6 (poles in the last layer).
   • Common difference ($d$): $49-50=-1$ (since the number of poles is decreasing).

2. Find the number of layers ($n$): Use the formula $a_n=a_1+(n-1)d$.

   • $6=50+(n-1)(-1)$
   • Subtract 50: $-44=(n-1)(-1)$
   • Divide by -1: $44=n-1$
   • Add 1: $n=45$. There are 45 layers of poles.

3. Find the total number of poles ($S_{45}$):

   • $S_{45}=\frac{45}{2}(50+6)$
   • $S_{45}=\frac{45}{2}(56)$
   • $S_{45}=45(28)=1260$. There are 1,260 poles in the pile.

Problem 25: Family Savings

The goal is to find the total savings over 20 years ($S_{20}$).

1. Identify the knowns:

   • First term ($a_1$): 6,000.
   • Common difference ($d$): $7,000-6,000=1,000$.
   • Number of terms ($n$): 20 years.

2. Calculate the total savings ($S_{20}$): Since we don't have the last term ($a_{20}$), it's most direct to use the formula $S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$.

   • $S_{20}=\frac{20}{2}(2(6000)+(20-1)(1000))$
   • $S_{20}=10(12000+(19)(1000))$
   • $S_{20}=10(12000+19000)$
   • $S_{20}=10(31000)=310,000$. They save Rs. 310,000 in total.

Problem 26: Mr. Saleem's Savings

We need to find the total savings for the month of October ($S_{31}$).

1. Identify the knowns:

   • First term ($a_1$): 500.
   • Common difference ($d$): $550-500=50$.
   • Number of terms ($n$): 31 (days in October).

2. Calculate the total savings ($S_{31}$): We use the formula $S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$.

   • $S_{31}=\frac{31}{2}(2(500)+(31-1)(50))$
   • $S_{31}=\frac{31}{2}(1000+(30)(50))$
   • $S_{31}=\frac{31}{2}(1000+1500)$
   • $S_{31}=\frac{31}{2}(2500)$
   • $S_{31}=31(1250)=38,750$. Mr. Saleem saves Rs. 38,750 during October.

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Key Formulas or Methods Used

1. Finding the n-th Term: Used to find a specific term in a sequence (like the number of marchers in the last row) or to solve for the number of terms ($n$). $a_n=a_1+(n-1)d$

2. Sum of an Arithmetic Series (when first and last terms are known): $S_n=\frac{n}{2}\left(a_1+a_n\right)$

3. Sum of an Arithmetic Series (when first term and common difference are known): $S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$

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Summary of Steps

1. Translate the Problem: Read the word problem and identify the first term ($a_1$), the common difference ($d$), the number of terms ($n$), and/or the last term ($a_n$).
2. Find Any Missing Information: If a key variable (like $n$ or $a_n$) needed for the sum is missing, use the n-th term formula, $a_n=a_1+(n-1)d$, to find it first.
3. Calculate the Total Sum: Choose the appropriate sum formula based on the information you have and substitute the values to find the total ($S_n$).