Applications and Word Problems

Spotting an AP in a Word Problem

Many real situations grow by a constant amount each step — exactly an AP. The signal words are 'increases/decreases by a fixed amount each …', 'each row has 2 more than the previous', and so on.

The strategy

1. Identify $a$ (the first term) and $d$ (the constant change).
2. Decide whether you need the nth term ($a_n$, for a single value at a position) or the sum ($S_n$, for a total).
3. Apply the right formula and interpret the answer with units.

Key Point: Use $a_n = a + (n-1)d$ for 'the value at step $n$' and $S_n = \dfrac{n}{2}[2a + (n-1)d]$ for 'the total of $n$ steps'.

[Board Important] Read the question to decide: 'how much in the 10th year?' → nth term; 'total over 10 years?' → sum.

Savings, Salary and Instalments

These are classic AP applications.

Savings example

'₹100 in month 1, increasing ₹20 each month.' Here $a = 100$, $d = 20$. The 12th month's saving is $a_{12}$; the total for a year is $S_{12}$.

Loan instalments

When a loan is repaid in instalments that decrease (or increase) by a fixed amount, the instalments form an AP, and the total repaid is $S_n$.

Key Point: 'In the $n$th period' asks for $a_n$; 'total over $n$ periods' asks for $S_n$.

[Board Important] For 'how much does he save in total in $n$ months', always use the sum formula $S_n$, not the nth term.

Rows, Seats and Stacks

Arrangements that grow by a fixed number per row are APs.

Seating example

'A theatre has 20 seats in row 1, 22 in row 2, 24 in row 3, …' Here $a = 20$, $d = 2$. Seats in the $n$th row $= a_n$; total seats in $n$ rows $= S_n$.

Stacking / logs

Logs stacked with one fewer in each higher row form a (decreasing) AP; the total number of logs is $S_n$.

Key Point: Count rows/positions as $n$; a fixed step per row makes it an AP.

[Board Important] A common board problem: 'In how many rows are 200 logs stacked if the rows have 20, 19, 18, …?' — set $S_n = 200$ and solve the quadratic.

Distance, Production and Other Patterns

Production

'A factory made 600 sets in year 3 and 700 in year 7; production increases uniformly.' The yearly production is an AP — use two terms to find $a$ and $d$, then the required year or total.

Distance / potatoes-race type

Some problems involve repeatedly covering distances that increase by a fixed amount; the total distance is a sum $S_n$.

Key Point: If two data points are given (e.g. value in year 3 and year 7), form two equations $a + 2d = \dots$ and $a + 6d = \dots$ to find $a$ and $d$ first.

[Board Important] Always translate the worded positions correctly: 'year 3' means $a_3 = a + 2d$ (not $a + 3d$).

Solved Examples

Example 1: Savings (nth term)

A man saves ₹100 in the first month and ₹20 more each month. How much does he save in the 12th month?

Solution:

1. $a = 100$, $d = 20$, $n = 12$.
2. $a_{12} = 100 + 11(20) = 320$.

Final Answer: ₹320.

Takeaway: 'In the $n$th month' → use $a_n$.

Example 2: Savings (total)

For the same saver (₹100, then +₹20/month), find the total saved in 12 months.

Solution:

1. $S_{12} = \dfrac{12}{2}[2(100) + 11(20)] = 6[200 + 220] = 6(420) = 2520$.

Final Answer: ₹2520.

Takeaway: 'Total over $n$ months' → use $S_n$.

Example 3: Seats in rows

A hall has 20 seats in the first row and 2 more in each successive row. How many seats are in the 15th row, and how many in total in 15 rows?

Solution:

1. $a = 20$, $d = 2$. $a_{15} = 20 + 14(2) = 48$.
2. $S_{15} = \dfrac{15}{2}[40 + 14(2)] = \dfrac{15}{2}(68) = 510$.

Final Answer: 48 seats in row 15; 510 seats in total.

Takeaway: One question, two formulas — read what's asked.

Example 4: Logs stacked

200 logs are stacked so that the bottom row has 20 logs, the next 19, the next 18, and so on. In how many rows are they placed?

Solution:

1. $a = 20$, $d = -1$. Set $S_n = 200$: $\dfrac{n}{2}[40 + (n-1)(-1)] = 200$.
2. $n(41 - n) = 400 \Rightarrow n^2 - 41n + 400 = 0 \Rightarrow (n - 16)(n - 25) = 0$.
3. $n = 16$ or $25$. But $n = 25$ would need a 25th row with $20 + 24(-1) = -4$ logs (impossible), so $n = 16$.

Final Answer: 16 rows.

Takeaway: Reject the root that makes a term negative (impossible logs).

Example 5: Production from two years

A manufacturer made 600 TV sets in the 3rd year and 700 in the 7th year. Production rises uniformly. Find the production in the 1st year.

Solution:

1. $a_3 = a + 2d = 600$; $a_7 = a + 6d = 700$.
2. Subtract: $4d = 100 \Rightarrow d = 25$; then $a = 600 - 50 = 550$.

Final Answer: 550 sets in the 1st year.

Takeaway: Two given terms give $a$ and $d$; 'year 3' = $a + 2d$.

Example 6: Total production

For the manufacturer above ($a = 550$, $d = 25$), find the total production in the first 7 years.

Solution:

1. $S_7 = \dfrac{7}{2}[2(550) + 6(25)] = \dfrac{7}{2}[1100 + 150] = \dfrac{7}{2}(1250) = 4375$.

Final Answer: 4375 sets.

Takeaway: Once $a$ and $d$ are known, the sum follows directly.

Example 7: Simple interest as an AP

₹1000 is invested at 8% simple interest per year. The amounts at the end of each year form an AP. Find the amount after 30 years.

Solution:

1. SI per year $= 8\%$ of 1000 $= ₹80$. End of year 1: $1080$, year 2: $1160$, … an AP with $a = 1080$, $d = 80$.
2. After 30 years: $a_{30} = 1080 + 29(80) = 1080 + 2320 = 3400$.

Final Answer: ₹3400.

Takeaway: Simple interest adds a fixed amount each year, forming an AP.

Example 8: Salary increments

A person's starting salary is ₹25000 with an annual increment of ₹2000. In which year does the salary become ₹39000?

Solution:

1. $a = 25000$, $d = 2000$. Set $a_n = 39000$: $25000 + (n-1)2000 = 39000$.
2. $(n-1)2000 = 14000 \Rightarrow n - 1 = 7 \Rightarrow n = 8$.

Final Answer: The 8th year.

Takeaway: 'When does it reach a value?' → set $a_n$ to that value.

Example 9: Trees planted

A school plants trees so that the number planted increases by 5 each grade: Grade 1 plants 10, Grade 2 plants 15, and so on up to Grade 12. How many trees in total?

Solution:

1. $a = 10$, $d = 5$, $n = 12$.
2. $S_{12} = \dfrac{12}{2}[20 + 11(5)] = 6[20 + 55] = 6(75) = 450$.

Final Answer: 450 trees.

Takeaway: 'Total over all grades' → sum formula.

Example 10: Potato race (distances)

In a race, a bucket is 5 m from the first potato, and potatoes are 3 m apart. A competitor runs from the bucket to each potato and back. For 10 potatoes, find the total distance run (the to-and-fro distances form an AP).

Solution:

1. To the 1st potato and back: $2 \times 5 = 10$ m. To the 2nd: $2 \times 8 = 16$ m. So $a = 10$, $d = 6$.
2. $S_{10} = \dfrac{10}{2}[20 + 9(6)] = 5[20 + 54] = 5(74) = 370$.

Final Answer: 370 m.

Takeaway: Set up the first two round-trip distances to find $a$ and $d$, then sum.