Applications and Word Problems on Arithmetic Progressions for CBSE Class 10

Mathematics is at its most engaging when it solves tangible real-world challenges. In everyday life, countless processes proceed with constant incremental increases or decreases: saving a fixed extra amount every week, paying off a debt with escalating monthly installments, stacking pipes in warehouse racks, or calculating production outputs in manufacturing plants.

In CBSE Class 10 Mathematics, word problems on Arithmetic Progressions appear regularly in Section C and Section D, carrying 3, 4, or 5 marks. The key to mastering these questions lies in identifying whether the problem asks for an isolated single value (the $n$-th term, $a_n$) or a cumulative total (the sum of terms, $S_n$).

What You Will Learn

• How to translate narrative word problems into arithmetic progressions
• Distinguishing between $n$-th term queries ($a_n$) and cumulative sum queries ($S_n$)
• Problem Type 1: Financial loan repayments and savings schedules
• Problem Type 2: Stacking logs of wood and warehouse pipes
• Problem Type 3: The classic Potato Race problem (NCERT high-yield question)
• Problem Type 4: Uniform annual production in manufacturing industries
• Board exam presentation templates and error-prevention tips

1. The Decision Matrix: $a_n$ vs. $S_n$

Before picking up your pen to calculate, analyze what the question is asking:

2. Problem Type 1: Financial Installments and Loans

Solved Example: Repaying a Cash Loan

Problem: A man repays a loan of $\text{₹}3250$ by paying $\text{₹}20$ in the first month and then increases the payment by $\text{₹}15$ every month. How long will it take him to clear the loan?

Solution:

1. Identify the AP Parameters:
   • First installment: $a = 20$.
   • Monthly increase: $d = 15$.
   • Total loan amount to be cleared (sum of all installments): $S_n = 3250$.
   • Let the number of months required be $n$.
2. Apply the Sum Formula: $S_n = \frac{n}{2}[2a + (n - 1)d] = 3250$ $\frac{n}{2}[2(20) + (n - 1)(15)] = 3250$ $n[40 + 15n - 15] = 6500$ $n[15n + 25] = 6500$ $15n^2 + 25n - 6500 = 0$
3. Divide the entire quadratic equation by 5: $3n^2 + 5n - 1300 = 0$
4. Factorize (product $= 3 \times (-1300) = -3900$, sum $= 5$ $\implies +65$ and $-60$): $3n^2 - 60n + 65n - 1300 = 0$ $3n(n - 20) + 65(n - 20) = 0$ $(n - 20)(3n + 65) = 0$ $n = 20 \quad \text{or} \quad n = -\frac{65}{3}$
5. Since time (number of months) cannot be negative or fractional, $n = -\frac{65}{3}$ is rejected.
6. Therefore, <u>it will take him $20$ months to clear the loan</u>.

3. Problem Type 2: Stacking Logs of Wood (NCERT Classic)

Problem: 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it, and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:

1. Identify the AP:
   • Bottom row ($a_1$): $a = 20$.
   • Next row ($a_2$): $19$.
   • Common difference: $d = 19 - 20 = -1$.
   • Total number of logs: $S_n = 200$.
2. Find Number of Rows ($n$): $S_n = \frac{n}{2}[2a + (n - 1)d] = 200$ $\frac{n}{2}[2(20) + (n - 1)(-1)] = 200$ $n[40 - n + 1] = 400$ $n[41 - n] = 400 \implies 41n - n^2 = 400$ $n^2 - 41n + 400 = 0$
3. Factorize (product $= 400$, sum $= -41$ $\implies -25$ and $-16$): $(n - 16)(n - 25) = 0 \implies n = 16 \quad \text{or} \quad n = 25$
4. Evaluate Both Values of $n$ by Finding Logs in Top Row ($a_n$):
   • If $n = 25$: $a_{25} = a + (25 - 1)d = 20 + 24(-1) = 20 - 24 = -4$ A row cannot contain a negative number of logs ($-4$). Hence, $n = 25$ is physically impossible and must be rejected!
   • If $n = 16$: $a_{16} = a + (16 - 1)d = 20 + 15(-1) = 20 - 15 = 5$ This is a valid positive count (5 logs in the 16th row).
5. Conclusion: <u>The 200 logs are placed in $16$ rows, and there are $5$ logs in the top row</u>.

4. Problem Type 3: The Potato Race (NCERT Classic)

Problem: In a potato race, a bucket is placed at the starting point, which is $5\text{ m}$ from the first potato, and the other potatoes are placed $3\text{ m}$ apart in a straight line. There are $10$ potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it into the bucket, and runs back to pick up the next potato until all the potatoes are in the bucket. What is the total distance the competitor has to run?

    Bucket       Potato 1       Potato 2       Potato 3       ... Potato 10
      [B] -------- (5m) --------- (3m) --------- (3m) -----

Solution:

1. Analyze the Distance for Each Potato:
   • To pick up the 1st potato: competitor runs $5\text{ m}$ to potato and $5\text{ m}$ back to bucket $\implies 2 \times 5 = 10\text{ m}$.
   • To pick up the 2nd potato: distance from bucket is $5 + 3 = 8\text{ m}$. Round trip $\implies 2 \times 8 = 16\text{ m}$.
   • To pick up the 3rd potato: distance from bucket is $5 + 3 + 3 = 11\text{ m}$. Round trip $\implies 2 \times 11 = 22\text{ m}$.
2. The Resulting Sequence of Distances: $10, 16, 22, 28, \dots$
   • This forms an AP with first term $a = 10$, common difference $d = 16 - 10 = 6$, and total potatoes $n = 10$.
3. Calculate Total Distance Run ($S_{10}$): $S_{10} = \frac{10}{2}[2a + (10 - 1)d]$ $S_{10} = 5[2(10) + 9(6)] = 5[20 + 54] = 5[74] = 370\text{ m}$
4. Therefore, <u>the total distance the competitor has to run is $370\text{ metres}$</u>.

5. Summary and Examination Tips

Exam Tip: In log stacking or row problems where solving for $n$ yields two positive integers, ALWAYS compute $a_n$ for both values. The larger value will inevitably give a negative number of logs, providing the formal reason to discard it.

Common Mistake: Forgetting to write the physical units in the final conclusion (e.g., ₹, months, metres, rows). Always state the final answer in a complete sentence with correct units!