Arithmetic Progressions: Word Problems and Real-World Applications Class 10

When an automotive factory plans to ramp up vehicle assembly from $1000$ cars in its first month by adding $50$ additional cars every subsequent month, when an apple orchard worker calculates how many total steps they walk picking up dropped fruit placed at fixed intervals along a line, or when an architect calculates the decreasing lengths of ladder rungs, the mathematics governing their calculations is an Arithmetic Progression (AP).

In CBSE Class 10 Mathematics, Chapter 5 (Arithmetic Progressions), Section D and Section E board exam questions frequently feature intricate real-world word problems. Students must extract the first term ($a$), the common difference ($d$), and determine whether the question requires finding a specific term ($a_n$) or the cumulative total ($S_n$).

In this guide, we master the four most famous real-world AP word problems in the NCERT curriculum.

What You Will Learn

• How to translate descriptive narratives into mathematical AP sequences ($a, d, n$)
• Distinguishing when to use $a_n = a + (n - 1)d$ vs. $S_n = \frac{n}{2}[2a + (n - 1)d]$
• Problem 1: The Spiral of Semicircles (Lengths of 13 consecutive spirals)
• Problem 2: The Potato Race (Cumulative running distance)
• Problem 3: The Ladder Rungs (Uniformly decreasing lengths)
• Problem 4: 200 Logs Stacked in Rows (Finding row count and top-row logs)
• Step-by-step algebraic quadratic factorization and rejecting extraneous roots

1. Problem 1: The Spiral of Consecutive Semicircles (NCERT Classic)

Problem Statement:

A spiral is made up of successive semicircles, with centers alternately at $A$ and $B$, starting with center at $A$, of radii $0.5\text{ cm}, 1.0\text{ cm}, 1.5\text{ cm}, 2.0\text{ cm}, \dots$. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi = 22/7$).

                         Center A    Center B
                            *           *
                   Spiral 1: Radius r1 = 0.5 cm, Length l1 = π(0.5)
                   Spiral 2: Radius r2 = 1.0 cm, Length l2 = π(1.0)
                   Spiral 3: Radius r3 = 1.5 cm, Length l3 = π(1.5)

Step-by-Step Solution:

1. Length of a Semicircle of Radius $r$: $l = \pi r$
2. List the Lengths of Successive Semicircles:
   • $l_1 = \pi(0.5) = \frac{\pi}{2}$
   • $l_2 = \pi(1.0) = \pi$
   • $l_3 = \pi(1.5) = \frac{3\pi}{2}$
   • $l_4 = \pi(2.0) = 2\pi$
3. Identify the AP Parameters:
   • First term: $a = l_1 = 0.5\pi = \mathbf{\frac{\pi}{2}}$
   • Common difference: $d = l_2 - l_1 = \pi - 0.5\pi = 0.5\pi = \mathbf{\frac{\pi}{2}}$
   • Number of semicircles: $n = \mathbf{13}$
4. Calculate Total Length of the Spiral ($S_{13}$): $S_n = \frac{n}{2} [2a + (n - 1)d]$ $S_{13} = \frac{13}{2} \left[ 2\left(\frac{\pi}{2}\right) + (13 - 1)\left(\frac{\pi}{2}\right) \right]$ $S_{13} = \frac{13}{2} \left[ \pi + 12\left(\frac{\pi}{2}\right) \right] = \frac{13}{2} [\pi + 6\pi] = \frac{13}{2} [7\pi]$
5. Substitute $\pi = \frac{22}{7}$: $S_{13} = \frac{13}{2} \times 7 \times \frac{22}{7}$ Notice that $7$ cancels completely, and $22 / 2 = 11$: $S_{13} = 13 \times 11 = \mathbf{143\text{ cm}}$
6. Therefore, <u>the total length of the spiral is $143\text{ cm}$</u>.

2. Problem 2: The Potato Race Problem (NCERT Classic)

Problem Statement:

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, runs to pick up the nearest potato, runs back with it to drop it in the bucket, and continues until all potatoes are in the bucket. What is the total distance the competitor has to run?

    Bucket       Potato 1       Potato 2       Potato 3            Potato 10
     [ \___/ ] ──── 5 m ──── ( • ) ─── 3 m ─── ( • ) ─── 3 m ─── ( • ) ... ( • )

Step-by-Step Solution:

1. Analyze Distance Run for Each Potato: A competitor runs to the potato and returns to the bucket, covering twice the distance!
   • For $1^{\text{st}}$ potato: Distance $= 2 \times 5 = \mathbf{10\text{ m}}$.
   • For $2^{\text{nd}}$ potato: Distance $= 2 \times (5 + 3) = 2 \times 8 = \mathbf{16\text{ m}}$.
   • For $3^{\text{rd}}$ potato: Distance $= 2 \times (5 + 3 + 3) = 2 \times 11 = \mathbf{22\text{ m}}$.
2. Identify the AP Parameters: The distances run form an Arithmetic Progression: $10, 16, 22, 28, \dots$
   • First term: $a = \mathbf{10\text{ m}}$
   • Common difference: $d = 16 - 10 = \mathbf{6\text{ m}}$
   • Total number of potatoes: $n = \mathbf{10}$
3. Calculate Total Distance Run ($S_{10}$): $S_n = \frac{n}{2} [2a + (n - 1)d]$ $S_{10} = \frac{10}{2} [2(10) + (10 - 1) \times 6]$ $S_{10} = 5 [20 + (9 \times 6)] = 5 [20 + 54] = 5 \times 74 = \mathbf{370\text{ metres}}$
4. Therefore, <u>the competitor runs a total distance of $370\text{ metres}$</u>.

3. Problem 3: Stacking 200 Logs (Quadratic Extraneous Roots)

Problem Statement:

$200$ 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?

Step-by-Step Solution:

1. Identify the AP: $20, 19, 18, \dots$
   • First term $a = 20$
   • Common difference $d = 19 - 20 = \mathbf{-1}$
   • Total logs $S_n = 200$
2. Apply Sum Formula: $S_n = \frac{n}{2} [2a + (n - 1)d]$ $200 = \frac{n}{2} [2(20) + (n - 1)(-1)]$ $400 = n [40 - n + 1] = n [41 - n] = 41n - n^2$ $\mathbf{n^2 - 41n + 400 = 0}$
3. Factorize: Product $= 400$, Sum $= -41$ ($-16$ and $-25$): $(n - 16)(n - 25) = 0 \implies \mathbf{n = 16} \quad \text{or} \quad \mathbf{n = 25}$
4. Determine the Valid Root: Calculate the number of logs in the top row ($a_n$) for both values:
   • For $n = 25$: $a_{25} = a + (25 - 1)d = 20 + 24(-1) = 20 - 24 = \mathbf{-4}$ The number of physical logs cannot be negative! Therefore, $n = 25$ is mathematically extraneous and rejected.
   • For $n = 16$: $a_{16} = a + (16 - 1)d = 20 + 15(-1) = 20 - 15 = \mathbf{5\text{ logs}}$
5. Therefore, <u>the logs are stacked in $16$ rows, with $5$ logs in the top row</u>.

4. Summary and Examination Tips

Exam Tip: Whenever an AP problem yields two positive integer values for $n$ from a quadratic equation (like $n = 16$ and $n = 25$), ALWAYS compute $a_n$ for both! One of the roots will inevitably produce a physically impossible negative count.

Common Mistake: In the potato race, taking the first term as $5$. The competitor runs to the potato ($5\text{ m}$) AND runs back to the bucket ($5\text{ m}$), making the first term $a = 10\text{ m}$!