Applications and word problems

The AP machinery is small , just two formulas , but its applications are wide. Below are the classical templates that appear repeatedly in board exams.

Common templates

Salary / cost progressions. A starting salary increases by a fixed amount each year. Identify $a$ (starting), $d$ (annual increment), and use $a_n$ to find a specific year's salary, or $S_n$ to find total earnings over $n$ years.

Seats / objects in rows. Each row has a fixed number of additional seats compared to the previous one. Total seats $= S_n$.

Brick / log stacking. Bottom row has many; each higher row has fewer (negative $d$). Total bricks $= S_n$. Sometimes the question is "how many rows can be stacked" , a quadratic in $n$.

Debt instalments. Total to be paid is fixed; instalments form an AP. From the sum and the number of instalments, find $a$ and $d$.

Physical sequences. Pendulum arcs, falling distances, etc. , verify carefully whether they actually form an AP (some problems are GPs in disguise).

A general strategy

1. Identify the quantity (salary, seats, distance) that progresses arithmetically.
2. Write down what is $a$, what is $d$, and what is $n$.
3. Decide whether the question wants a specific term (use $a_n$) or a cumulative sum (use $S_n$).
4. Solve and check the answer is physically meaningful.

Worked examples

Example 1. A man saves ₹$1000$ in the first month of his job and increases his savings by ₹$200$ each month. How much will he save in $24$ months?

AP: $1000, 1200, 1400, \ldots$. $a = 1000, d = 200, n = 24$.

$S_{24} = \dfrac{24}{2}[2 \cdot 1000 + 23 \cdot 200] = 12 \cdot [2000 + 4600] = 12 \cdot 6600 = 79200$.

He saves ₹$79{,}200$.

Example 2. An auditorium has $20$ seats in the first row. Each successive row has $2$ more seats than the previous. How many seats are in the $15$th row, and how many in the first $15$ rows?

$a = 20, d = 2, n = 15$. $a_{15} = 20 + 14 \cdot 2 = 48$.

$S_{15} = \dfrac{15}{2}(20 + 48) = \dfrac{15}{2} \cdot 68 = 510$.

Example 3. A pile of logs has $20$ in the bottom row, $19$ in the next, $18$ above that, and so on. If there are $200$ logs in all, how many rows are there?

$a = 20, d = -1$. $S_n = \dfrac{n}{2}[40 + (n - 1)(-1)] = \dfrac{n}{2}(41 - n)$. Set $= 200$: $n(41 - n) = 400 \Rightarrow n^2 - 41 n + 400 = 0$.

Discriminant $= 1681 - 1600 = 81 = 9^2$. $n = (41 \pm 9)/2 = 25$ or $16$.

$n = 25$ would mean the top row has $20 - 24 = -4$ logs , impossible. So $n = 16$. Top row has $20 - 15 = 5$ logs.

Example 4. A man arranges to pay a debt of ₹$3600$ in $40$ monthly instalments which form an AP. When $30$ instalments are paid, he discovers that he still owes one-third of the debt (₹$1200$). Find the value of the first instalment.

In $30$ instalments he has paid $3600 - 1200 = 2400$.

Using $S_{30} = 2400$ and $S_{40} = 3600$:

$S_{30} = \dfrac{30}{2}(2 a + 29 d) = 15(2 a + 29 d) = 2400 \Rightarrow 2 a + 29 d = 160$. $S_{40} = \dfrac{40}{2}(2 a + 39 d) = 20 (2 a + 39 d) = 3600 \Rightarrow 2 a + 39 d = 180$.

Subtract: $10 d = 20 \Rightarrow d = 2$. Then $2 a + 58 = 160 \Rightarrow a = 51$.

First instalment: ₹$51$.

Example 5. A spiral is drawn starting with two semicircles of radii $0.5$ cm and $1.0$ cm, then $1.5$ cm and $2.0$ cm, and so on. Find the total length of the spiral after $13$ semicircles. (Take $\pi = 22/7$.)

Length of a semicircle of radius $r$ is $\pi r$. Radii: $0.5, 1.0, 1.5, \ldots$ , an AP with $a = 0.5, d = 0.5$.

Length total $= \pi (r_1 + r_2 + \ldots + r_{13}) = \pi \cdot S_{13}$.

$S_{13} = \dfrac{13}{2}[1 + 12 \cdot 0.5] = \dfrac{13}{2} \cdot 7 = 45.5$.

Total length $= (22/7) \cdot 45.5 = 22 \cdot 6.5 = 143$ cm.

Try it yourself

1. ₹$1000$ is invested at simple interest of $10\%$ per annum. Show that the interest at the end of each successive year forms an AP and find the interest after $20$ years.
2. A spiral consists of $20$ semicircles with radii $0.5, 1, 1.5, \ldots$ cm. Find its total length.
3. $200$ logs are stacked: $20$ in the bottom row, $19$ in the next, etc. How many rows and how many in top row?
4. A man saves ₹$32$ in the first week and increases by ₹$4$ each week. How many weeks until total savings exceed ₹$700$?
5. A theatre has $30$ rows; the first row has $25$ seats and each row $4$ more. Total seats?
6. A debt of ₹$1500$ is paid in $10$ AP instalments. If the first is ₹$100$, find the common difference.
7. ₹$2000$ debt in $15$ AP instalments where the last is ₹$200$. Find the first.
8. A worker is given a starting wage of ₹$5000$, raised by ₹$300$ each year. Compute total over $15$ years.
9. Bricks in a stack form an AP with bottom $30$, top $5$, with rows decreasing by $5$. Total bricks?
10. Pendulum's arc reduces by a constant amount each swing. After $50$ swings, total distance travelled forms an AP , set up the formula given $a = 100$ cm, $d = -2$ cm.

Pitfalls / Insight

• Decide carefully if a sequence is truly an AP. Some "physical" sequences (e.g., compound interest) are GP, not AP.
• Reject impossible roots of quadratics in $n$ (negative or fractional terms).
• For instalments, the sum is the total debt , set up $S_n =$ total.

Insight. Almost every AP word problem is built from two facts: a first term and an increment. Spot those, then everything else is plugging into $a_n$ or $S_n$.