MathsCBSE Class 10 · Arithmetic Progressions

Word Problems and Daily Life Applications of AP

Updated 14 October 2026 · Fact-checked

AP word problems turn a real situation into an arithmetic progression. Savings, instalments, salary increments, stacked pipes or daily practice that change by a fixed amount are examples. You find the first term and common difference, then use the nth term or sum formula. The board asks these as short answers, long answers and Section E case studies.

Find the fixed step first (here +1 pipe per row), then ask whether the question wants one row, the nth term, or all rows, the sum.

Illustration drawn for QuestionBuddy.

Understand Word Problems and Daily Life Applications of AP

A word problem on AP is an ordinary AP question in disguise. Somewhere in the story, a quantity changes by the same amount every step. A loan instalment might go up by ₹100 each month, a stack might have one pipe fewer in each row, or a runner might add 0.5 km each day. Once you spot that fixed change, the story becomes an AP with first term $a$ and common difference $d$ .

Why does this work? The formula $a_{n} = a + (n - 1) d$ only needs a starting value and a steady step. The formula $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ only needs those two plus the number of steps. Real patterns with a fixed increase or decrease satisfy exactly these conditions. If the change is not fixed, for example if it doubles each time, the situation is not an AP and these formulas do not apply.

The key decision in every word problem is this: is the question about one value or a total? The amount paid in the 12th month, the pipes in the top row, or the penalty on the 30th day is a single term $a_{n}$ . Total amount repaid, total pipes, or total penalty is a sum $S_{n}$ . Most lost marks come from mixing these two.

The unknown can be any of $a$ , $d$ , $n$ , $a_{n}$ or $S_{n}$ . If a total is given and time is asked, you get a **quadratic in $n$ (Chapter 4). Reject a root that is negative or a fraction, and give a reason from the context, such as "months cannot be negative". If two facts about different terms are given, you get a pair of linear equations** in $a$ and $d$ (Chapter 3).

Two conditions must hold before you apply the formulas. The steps must be equally spaced, such as each day, each row or each year. Also, $n$ must be a natural number. If $d$ is negative, check that the terms you count stay sensible: sales or pipes cannot be negative. Finish with a sentence answer and units, because the examiner reads the final line.

Word Problems and Daily Life Applications of AP: formulas to remember

Value at the nth step: $a_{n} = a + (n - 1) d$; Use it for a single value, such as the instalment in month $n$ or the pipes in row $n$ .
Total of the first n steps: $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$; Use it for totals when $a$ , $d$ and $n$ are known. It gives a quadratic in $n$ when the total is given.
Total using first and last values: $S_{n} = \frac{n}{2} (a + l)$; Faster when the first and last values are known, for example the bottom and top rows of a stack.
Number of steps from first to last value: $n = \frac{l - a}{d} + 1$; Counts rows, days or months between two values. Do not forget the $+ 1$ .
Total of a block of steps: $S_{m} - S_{k} = a_{k + 1} + a_{k + 2} + \dots + a_{m}$; Use it for totals such as the 3rd week only, which is days 15 to 21.
Difference of two values: $a_{m} - a_{k} = (m - k) d$; Gives $d$ quickly when values at two different steps are given, for example production in the 3rd and 7th years.

How to solve Word Problems and Daily Life Applications of AP questions

Use this method on any AP word problem, from a 2-mark question to a Section E case.

1Find what changes by a fixed amount each step (rupees, rows, km, minutes) and check that the step really is constant.
2Write $a$ and $d$ in words with units, for example $a =$ ₹1000 (first instalment), $d =$ ₹100 (increase per month). Make $d$ negative if the quantity decreases.
3Read each part and label it: one value means $a_{n}$ , a total means $S_{n}$ , a count of steps means $n$ .
4If two facts about terms are given, form two equations in $a$ and $d$ and solve the pair.
5Substitute into the right formula. If $n$ is unknown in a sum, form the quadratic, put it in standard form and factorise.
6Reject any root that is not a natural number, or that makes a term impossible (such as negative sales), and state the reason.
7Check one answer by substitution, then write a sentence answer with units.

Word Problems and Daily Life Applications of AP in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Word Problems and Daily Life Applications of AP

Suppose a family takes an interest-free loan of ₹1,18,000 from a relative to repair their house. They agree to repay it in monthly instalments. The first instalment is ₹1,000, and each instalment is ₹100 more than the one before it.

In how many months will the loan be fully repaid? How much is the last instalment?

Try it first, then show the solutionSolution

Instalments form an AP with $a = 1000$ , $d = 100$ , and $S_{n} = 118000$ .
$\frac{n}{2} [2 (1000) + (n - 1) 100] = 118000$
$\frac{n}{2} (1900 + 100 n) = 118000$
$n (1900 + 100 n) = 236000$
$100 n^{2} + 1900 n - 236000 = 0$ , so $n^{2} + 19 n - 2360 = 0$
$(n + 59) (n - 40) = 0$ , so $n = 40$ or $n = - 59$
Reject $n = - 59$ because the number of months cannot be negative, so $n = 40$ .
Last instalment: $a_{40} = 1000 + 39 \times 100 = 4900$
Check: $S_{40} = \frac{40}{2} (1000 + 4900) = 20 \times 5900 = 118000$ . This is correct.

Answer: The loan is repaid in 40 months, and the last instalment is ₹4,900.

Memory hook: Rising instalments: the total given means use S_n, and the months come from a quadratic.

Word Problems and Daily Life Applications of AP

Imagine a girl in your school's cricket team practises batting in the nets. On the first day she faces 30 balls, and each day after that she faces 5 more balls than the day before. Her coach sets a cap of 100 balls a day, so once she reaches 100 balls she stays at 100 balls every day after that.

How many balls does she face in total over the first 25 days of practice?

Try it first, then show the solutionSolution

Until the cap, the balls per day form an AP with $a = 30$ , $d = 5$ .
Day she reaches the cap: $30 + (n - 1) 5 = 100$ gives $n - 1 = 14$ , so $n = 15$ .
Balls in days 1 to 15: $S_{15} = \frac{15}{2} (30 + 100) = 15 \times 65 = 975$ .
Days 16 to 25 are $25 - 15 = 10$ days at 100 balls each, giving $10 \times 100 = 1000$ balls.
Total in 25 days $= 975 + 1000 = 1975$ .
Check: without the cap, day 16 would have $30 + 15 \times 5 = 105$ balls, which is more than 100, so the cap really applies from day 16.

Answer: She faces 1,975 balls in the first 25 days.

Memory hook: Capped nets practice: the AP runs only until the cap, then add the flat days separately.

Tricks to remember Word Problems and Daily Life Applications of AP

Short memory aids. Say each one aloud once, then use it on a question the same day.

One or All?
Ask "one or all?" before every part. If the question asks about one month, row or day, use $a_{n}$ . If it asks about everything so far, use $S_{n}$ .
Use it when: Use it at the start of every part of a word problem or case study.
Fence-post rule
Ten gaps need eleven posts. When you count rows or days from a first value to a last value, the count is $\frac{l - a}{d} + 1$ .
Use it when: Use it whenever you count terms between two given values.
Context kills roots
A quadratic gives two roots, but real life accepts only natural numbers that keep every term sensible.
Use it when: Use it after factorising a quadratic in $n$ , and always write the reason for rejecting a root.

Word Problems and Daily Life Applications of AP: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Pipes stacked in rows · 3 marks · Short answerBoard level

Some pipes are stacked in rows. The bottom row has 25 pipes, each row above has one pipe fewer than the row below it, and the top row has 6 pipes. How many rows are there, and how many pipes are there in all?

Try it first, then show the solutionSolution

Counting from the bottom, the rows form an AP with $a = 25$ , $d = - 1$ , $l = 6$ . (½ mark)
$6 = 25 + (n - 1) (- 1)$ gives $n - 1 = 19$ . (½ mark)
So $n = 20$ rows. (½ mark)
Total pipes: $S_{20} = \frac{20}{2} (25 + 6)$ . (1 mark)
$S_{20} = 10 \times 31 = 310$ . (½ mark)

Answer: There are 20 rows and 310 pipes in all.

Example 2: Prizes in decreasing AP · 3 marks · Short answerBoard level

A school uses ₹3,600 to give 8 prizes for overall performance. Each prize is ₹50 less than the prize just above it. Find the value of each prize.

Try it first, then show the solutionSolution

The prizes form an AP with first term $a$ (the first prize), $d = - 50$ , $n = 8$ and $S_{8} = 3600$ . (½ mark)
$\frac{8}{2} [2 a + 7 (- 50)] = 3600$ (½ mark)
$4 (2 a - 350) = 3600$ , so $2 a - 350 = 900$ . (½ mark)
$2 a = 1250$ , so $a = 625$ . (½ mark)
The prizes are ₹625, ₹575, ₹525, ₹475, ₹425, ₹375, ₹325 and ₹275. (½ mark)
Check: $\frac{8}{2} (625 + 275) = 4 \times 900 = 3600$ . (½ mark)

Answer: The prizes are ₹625, ₹575, ₹525, ₹475, ₹425, ₹375, ₹325 and ₹275.

Example 3: A block of days in a training plan · 3 marks · Short answerBoard level

A runner runs 2.5 km on day 1 of a training plan and increases her distance by 0.5 km every day. (i) On which day does she run exactly 10 km? (ii) Find the total distance she runs in the 3rd week, that is, from day 15 to day 21.

Try it first, then show the solutionSolution

The daily distances form an AP with $a = 2.5$ km and $d = 0.5$ km. (½ mark)
(i) $2.5 + (n - 1) (0.5) = 10$ gives $(n - 1) (0.5) = 7.5$ . (½ mark)
So $n - 1 = 15$ and $n = 16$ . She runs exactly 10 km on day 16. (½ mark)
(ii) $a_{15} = 2.5 + 14 (0.5) = 9.5$ km and $a_{21} = 2.5 + 20 (0.5) = 12.5$ km. (½ mark)
Days 15 to 21 form a block of 7 terms, so the sum is $\frac{7}{2} (9.5 + 12.5)$ . (½ mark)
$= \frac{7}{2} \times 22 = 77$ km. (½ mark)

Answer: (i) She runs exactly 10 km on day 16. (ii) She runs 77 km in the 3rd week.

Example 4: Penalty for a delayed project · 5 marks · Long answerHarder board level

A contract says that if a building is delayed, the contractor pays a penalty of ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, and so on, with each day's penalty ₹50 more than the day before. (i) Find the penalty for the 30th day. (ii) Find the total penalty if the work is delayed by 30 days. (iii) The contractor paid a total penalty of ₹9,200. For how many days was the work delayed?

Try it first, then show the solutionSolution

The daily penalties form an AP with $a = 200$ and $d = 50$ . (½ mark)
(i) $a_{30} = 200 + 29 \times 50 = 200 + 1450 = 1650$ , so the penalty for the 30th day is ₹1,650. (½ mark)
(ii) $S_{30} = \frac{30}{2} (200 + 1650)$ (1 mark)
$= 15 \times 1850 = 27750$ , so the total penalty for 30 days is ₹27,750. (½ mark)
(iii) $\frac{n}{2} [400 + (n - 1) 50] = 9200$ , so $n (350 + 50 n) = 18400$ . (1 mark)
$50 n^{2} + 350 n - 18400 = 0$ , so $n^{2} + 7 n - 368 = 0$ . (½ mark)
$(n + 23) (n - 16) = 0$ , so $n = 16$ or $n = - 23$ . Reject $- 23$ because days cannot be negative. (½ mark)
Check: $S_{16} = 8 [400 + 750] = 8 \times 1150 = 9200$ . This is correct. (½ mark)

Answer: (i) ₹1,650 (ii) ₹27,750 (iii) The work was delayed by 16 days.

Example 5: Case study: comparing two fan factories · 4 marks · Case-basedHOTS / competency

Suppose two factories that make ceiling fans start in the same year. Factory P's production rises by a fixed number of fans every year; it made 600 fans in its 3rd year and 700 fans in its 7th year. Factory Q made 820 fans in its 1st year, and its production falls by 20 fans every year.
(i) How many fans did Factory P make in its 1st year? (1 mark)
(ii) In which year do the two factories make the same number of fans, and how many fans does each make that year? (1 mark)
(iii) Which factory made more fans in total in the first 7 years, and by how many? Explain why, even though both make the same number in the 7th year. (2 marks)

Try it first, then show the solutionSolution

(i) For P, $a_{7} - a_{3} = 4 d$ , so $4 d = 700 - 600 = 100$ and $d = 25$ . (½ mark)
$a = a_{3} - 2 d = 600 - 50 = 550$ , so Factory P made 550 fans in its 1st year. (½ mark)
(ii) In year $n$ , P makes $550 + 25 (n - 1)$ and Q makes $820 - 20 (n - 1)$ . Equating them gives $45 (n - 1) = 270$ . (½ mark)
$n - 1 = 6$ , so $n = 7$ . Each makes $550 + 6 \times 25 = 700$ fans in the 7th year. (½ mark)
(iii) P: $S_{7} = \frac{7}{2} (550 + 700) = \frac{7}{2} \times 1250 = 4375$ . (½ mark)
Q: $S_{7} = \frac{7}{2} (820 + 700) = \frac{7}{2} \times 1520 = 5320$ . (½ mark)
Q made $5320 - 4375 = 945$ more fans. (½ mark)
Reason: in each of years 1 to 6, Q made more fans than P (for example 820 against 550 in year 1). They become equal only in year 7, so Q's total stays ahead. (½ mark)

Answer: (i) 550 fans (ii) In the 7th year, when each factory makes 700 fans (iii) Factory Q, by 945 fans, because it made more than P in every one of the first 6 years.

Example 6: Logs stacked in rows: how many rows? · 3 marks · Short answerHarder board level

200 logs are available, and some of them are stacked so that the top row has 3 logs and each row below has 2 logs more than the row above it. When the stack is finished, it holds exactly 168 logs. How many rows are there, how many logs are in the bottom row, and how many logs are left over?

Try it first, then show the solutionSolution

Counting from the top, the rows form an AP with $a = 3$ , $d = 2$ . For $n$ rows, $S_{n} = \frac{n}{2} [6 + 2 (n - 1)] = n (n + 2)$ . (1 mark)
$n (n + 2) = 168$ gives $n^{2} + 2 n - 168 = 0$ , so $(n + 14) (n - 12) = 0$ . The number of rows cannot be negative, so $n = 12$ . (1 mark)
Bottom row: $a_{12} = 3 + 11 \times 2 = 25$ logs. Logs left over: $200 - 168 = 32$ . (1 mark)

Answer: 12 rows; 25 logs in the bottom row; 32 logs are left over.

Example 7: Equal sums on either side of a plot · 5 marks · Long answerHarder board level

The plots in a housing colony are numbered 1, 2, 3, ..., 288 in a row. Show that there is a plot number $x$ such that the sum of the plot numbers before $x$ equals the sum of the plot numbers after $x$ , and find $x$ .

Try it first, then show the solutionSolution

The plot numbers form an AP with $a = 1$ , $d = 1$ , so $S_{n} = \frac{n}{2} [2 + (n - 1)] = \frac{n ( n + 1 )}{2}$ . (½ mark)
Sum of plot numbers before $x$ : $S_{x - 1} = \frac{( x - 1 ) x}{2}$ . (½ mark)
Sum of all plot numbers: $S_{288} = \frac{288 \times 289}{2} = 144 \times 289 = 41616$ . (1 mark)
Sum of plot numbers after $x$ : $S_{288} - S_{x} = 41616 - \frac{x ( x + 1 )}{2}$ . (½ mark)
Equate the two sums: $\frac{x ( x - 1 )}{2} = 41616 - \frac{x ( x + 1 )}{2}$ . (½ mark)
$\frac{x ^{2} - x + x ^{2} + x}{2} = 41616$ , so $x^{2} = 41616$ . (½ mark)
$x = 204$ , since $20 4^{2} = 41616$ . Reject $x = - 204$ because a plot number is a natural number. As 204 lies between 1 and 288, such a plot exists. (1 mark)
Check: before, $\frac{203 \times 204}{2} = 20706$ ; after, $41616 - \frac{204 \times 205}{2} = 41616 - 20910 = 20706$ . (½ mark)

Answer: Such a plot exists, and it is plot number 204. The sum on each side is 20,706.

Where students lose marks in Word Problems and Daily Life Applications of AP

The slips examiners see most often, and what to write instead.

Mistake: Using the nth term when the question asks for a total
The student sees the word "month" and finds the value for that month instead of adding all the months.
Wrong working
Savings of ₹10, ₹20, ₹30, ... Total in 12 months $= a_{12} = 10 + 11 \times 10 = 120$
Right working
Total $= S_{12} = \frac{12}{2} (10 + 120) = 780$ , so the total is ₹780.
Do this: Underline "total", "in all" or "altogether" and use $S_{n}$ for those parts.
Usually costs: Most of the marks for that part.
Mistake: Not rejecting the extra root, or rejecting it without a reason
The student stops after factorising and writes both roots, or crosses one out silently.
Wrong working
$(n + 23) (n - 16) = 0$ , so $n = 16, - 23$ days.
Right working
$n = 16$ or $n = - 23$ . Since the number of days cannot be negative, $n = 16$ .
Do this: Always write one line giving the reason from the context.
Usually costs: ½ to 1 mark.
Mistake: Missing the +1 when counting rows or days
The student counts the gaps instead of the terms.
Wrong working
Rows from 25 down to 6: $n = \frac{25 - 6}{1} = 19$
Right working
$n = \frac{6 - 25}{- 1} + 1 = 20$ rows.
Do this: Use $l = a + (n - 1) d$ and solve it fully instead of a shortcut.
Usually costs: The answer and every later step that uses $n$.
Mistake: Taking d as positive for a decreasing quantity
The student reads "₹50 less" but writes $d = 50$ .
Wrong working
$4 (2 a + 350) = 3600$ gives $a = 275$ , so the first prize is ₹275.
Right working
$d = - 50$ : $4 (2 a - 350) = 3600$ gives $a = 625$ , so the first prize is ₹625.
Do this: Write the sign of $d$ in words first: decreasing means negative $d$ .
Usually costs: The whole question.

Exam tips

Begin every word problem with a line such as "Here $a =$ ₹200, $d =$ ₹50". Examiners give method marks for correct modelling.
In Section E cases, the 1-mark parts usually need a single term or $d$ , and the 2-mark part needs a total or the number of steps. Show working even for the 1-mark parts.
Keep the units the same throughout, for example all in minutes or all in rupees, and convert only in the final answer.
End with a sentence answer that contains units, such as "The loan is repaid in 40 months".
When the numbers allow, check your answer by adding the first few terms or by substituting into $S_{n}$ .

Practice questions on Word Problems and Daily Life Applications of AP

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 3 marks · Short answerHarder board level
Asha saves ₹5 on the first day of a month, ₹7 on the second day, ₹9 on the third day, and so on. Her brother saves ₹20 every day. (i) After how many days are their total savings equal? (ii) How much has each saved by then? (iii) Who has saved more after 20 days, and by how much?
Hint
Asha's total after n days is $\frac{n}{2} [10 + 2 (n - 1)] = n (n + 4)$ ; her brother's is 20n.
Answer
(i) $n (n + 4) = 20 n$ gives $n + 4 = 20$ (as $n \neq = 0$ ), so $n = 16$ days. (ii) Each has saved $20 \times 16 =$ ₹320 (check: Asha $\frac{16}{2} (5 + 35) = 320$ ). (iii) After 20 days Asha has $20 \times 24 =$ ₹480 and her brother ₹400, so Asha has ₹80 more.
Q2 · 3 marks · Short answerBoard level
A man repays a loan of ₹3,250 by paying ₹20 in the first month and increasing the payment by ₹15 every month. In how many months will the loan be cleared?
Hint
Set $\frac{n}{2} [40 + 15 (n - 1)] = 3250$ and simplify to a quadratic. Its discriminant is a perfect square.
Answer
$n (25 + 15 n) = 6500$ gives $3 n^{2} + 5 n - 1300 = 0$ . The discriminant is $25 + 15600 = 15625 = 12 5^{2}$ , so $n = \frac{- 5 + 125}{6} = 20$ . The negative root is rejected. Check: $10 (40 + 285) = 3250$ . The loan is cleared in 20 months.
Q3 · 5 marks · Long answerHarder board level
In Plan A, a person deposits ₹1,000 in the first month and ₹100 more each month. In Plan B, the person deposits ₹1,500 in the first month and ₹60 more each month. (i) From which month is the Plan A deposit larger than the Plan B deposit? (ii) Find the total deposited under each plan in the first 12 months. (iii) After how many months are the totals under the two plans equal, and from which month is the Plan A total larger?
Hint
For (iii), write $S_{A} (n) = n (950 + 50 n)$ and $S_{B} (n) = n (1470 + 30 n)$ .
Answer
(i) $1000 + 100 (n - 1) > 1500 + 60 (n - 1)$ gives $40 (n - 1) > 500$ , $n - 1 > 12.5$ , so the Plan A deposit is larger from the 14th month. (2)
(ii) Plan A: $\frac{12}{2} [2000 + 1100] = 18600$ , so ₹18,600. Plan B: $\frac{12}{2} [3000 + 660] = 21960$ , so ₹21,960. (1)
(iii) $S_{A} (n) = \frac{n}{2} [2000 + 100 (n - 1)] = n (950 + 50 n)$ and $S_{B} (n) = n (1470 + 30 n)$ . Equal when $950 + 50 n = 1470 + 30 n$ , so $n = 26$ (each total is $26 \times 2250 =$ ₹58,500). After that Plan A's total is larger, from the 27th month. (2)
Q4 · 5 marks · Long answerBoard level
Meera studies for 40 minutes on day 1 and increases her study time by 5 minutes each day. (i) On which day does she first study for 2 hours? (ii) Find her total study time up to and including that day, in hours and minutes. (iii) Find her total study time in the first 20 days, in hours and minutes.
Hint
Work in minutes, so 2 hours is 120 minutes. Convert to hours only at the end.
Answer
(i) $40 + 5 (n - 1) = 120$ gives $n = 17$ , so she first studies for 2 hours on day 17. (ii) $S_{17} = \frac{17}{2} (40 + 120) = 1360$ minutes, which is 22 hours 40 minutes. (iii) $S_{20} = 10 [80 + 95] = 1750$ minutes, which is 29 hours 10 minutes.
Q5 · 3 marks · Short answerHOTS / competency
Suppose a festival stall sells 24 lamps on day 1, and each day after that it sells 3 fewer lamps than the day before. A student is asked after how many days the total sales reach 78 lamps. The student gets two answers, 4 and 13. Which answer fits the situation? Justify your choice.
Hint
Solve $\frac{n}{2} [48 - 3 (n - 1)] = 78$ . Then find the sales on the 13th day.
Answer
$n (51 - 3 n) = 156$ gives $n^{2} - 17 n + 52 = 0$ , so $(n - 4) (n - 13) = 0$ . Mathematically both roots work, but $a_{13} = 24 - 36 = - 12$ , which means negative sales. Sales stop at day 9, when $a_{9} = 0$ . So the answer is 4 days. Check: $24 + 21 + 18 + 15 = 78$ .
Q6 · 1 mark · MCQBoard level
Suppose a school hall has 20 seats in the first row, and each row has 2 more seats than the row in front of it. If there are 25 rows, how many seats are there in all?
(a) 68
(b) 1000
(c) 1100
(d) 1200
Hint
Total means $S_{n}$ , not $a_{n}$ . The 25th row has $20 + 24 \times 2$ seats.
Answer
(c) 1100. The last row has $a_{25} = 20 + 48 = 68$ seats, so $S_{25} = \frac{25}{2} (20 + 68) = 25 \times 44 = 1100$ . Option (a) is only the last row.
Q7 · 3 marks · Short answerHarder board level
In a school garden, plants are set out in rows, and each row has the same number of plants more than the row before it. The 4th and 8th rows together have 52 plants, and the 6th and 11th rows together have 77 plants. Find the number of plants in the 1st row and the total number of plants in the first 15 rows.
Hint
Write both conditions in terms of a and d and subtract.
Answer
$a_{4} + a_{8} = 2 a + 10 d = 52$ and $a_{6} + a_{11} = 2 a + 15 d = 77$ . Subtracting, $5 d = 25$ , so $d = 5$ and $2 a = 2$ , $a = 1$ . The 1st row has 1 plant. $a_{15} = 1 + 14 \times 5 = 71$ , so $S_{15} = \frac{15}{2} (1 + 71) = 540$ plants.
Q8 · 1 mark · Assertion-reasonHOTS / competency
Logs are stacked with 30 logs in the bottom row, 29 in the next row, 28 in the row above that, and so on. A total of 360 logs is stacked this way.
Assertion (A): The stack has 16 rows.
Reason (R): When the quadratic in $n$ has two positive roots, the larger root must always be rejected.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Solve $\frac{n}{2} [60 - (n - 1)] = 360$ , then test each root against the number of logs in the top row.
Answer
(c). $n (61 - n) = 720$ gives $n^{2} - 61 n + 720 = 0$ , so $(n - 16) (n - 45) = 0$ . For $n = 45$ , the top row would have $30 - 44 = - 14$ logs, so that root is rejected; for $n = 16$ the top row has 15 logs. So A is true. R is false: a root is rejected only when the context makes it impossible, and sometimes both positive roots are valid.
Q9 · 3 marks · Short answerBoard level
A ladder has rungs 20 cm apart. The rungs get shorter uniformly from 50 cm at the bottom to 30 cm at the top. The top and bottom rungs are 3 m apart. (i) How many rungs are there? (ii) How much wood is needed for all the rungs, in metres?
Hint
3 m is 300 cm, which is 15 gaps of 20 cm. Count the rungs, not the gaps.
Answer
(i) Number of gaps $= \frac{300}{20} = 15$ , so there are $15 + 1 = 16$ rungs. (ii) The lengths form an AP from 50 cm to 30 cm with 16 terms, so the total is $\frac{16}{2} (50 + 30) = 8 \times 80 = 640$ cm, which is 6.4 m of wood.
Q10 · 4 marks · Short answerHOTS / competency
Imagine a gardener has to water 20 plants in a straight row using one bucket. The tap is in line with the row, 4 m before the first plant, and the plants are 3 m apart. She fills the bucket at the tap, walks to one plant, waters it and walks back to the tap, and repeats this for every plant in order. (i) Find the total distance she walks to water all 20 plants and return to the tap. (ii) If she can walk at most 700 m, what is the greatest number of plants she can water this way, starting from the first plant?
Hint
The one-way distances are 4, 7, 10, ... m, and each trip is twice the one-way distance. For (ii), show the total for $n$ plants is $n (3 n + 5)$ m and try values of $n$ .
Answer
(i) One-way distances: $a = 4$ , $d = 3$ , $a_{20} = 4 + 19 \times 3 = 61$ m. Their sum is $\frac{20}{2} (4 + 61) = 650$ m, so she walks $2 \times 650 = 1300$ m. (ii) For $n$ plants the total is $2 \times \frac{n}{2} [8 + 3 (n - 1)] = n (3 n + 5)$ m. For $n = 14$ this is $14 \times 47 = 658$ m, and for $n = 15$ it is $15 \times 50 = 750$ m, which is more than 700 m. So she can water at most 14 of the 20 plants.
Q11 · 3 marks · Short answerHarder board level
Ritu reads 27 pages of a novel on day 1 and 3 pages fewer on each following day. On which day does her total first reach 135 pages? The quadratic you get has two positive roots. Explain what each root means in the situation.
Hint
Solve $\frac{n}{2} [54 - 3 (n - 1)] = 135$ , then find $a_{10}$ .
Answer
$n (57 - 3 n) = 270$ gives $3 n^{2} - 57 n + 270 = 0$ , so $n^{2} - 19 n + 90 = 0$ and $(n - 9) (n - 10) = 0$ . Her total first reaches 135 pages on day 9 (check: $S_{9} = \frac{9}{2} (27 + 3) = 135$ ). The root $n = 10$ gives the same total only because $a_{10} = 27 - 9 \times 3 = 0$ : she reads no pages on day 10, so the total stays at 135. The answer is day 9.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which formula gives the amount in the 10th month only?Show answer
$a_{10} = a + 9 d$
2.Which formula gives the total for the first $n$ months when the last amount is known?Show answer
$S_{n} = \frac{n}{2} (a + l)$
3.A stack has one fewer log in each row above. What is $d$ when you count from the bottom?Show answer
$d = - 1$
4.How do you find the total for days 11 to 20 only?Show answer
$S_{20} - S_{10}$
5.Why do you reject $n = - 7$ in a savings problem?Show answer
The number of months or weeks cannot be negative. $n$ must be a natural number.

Word Problems and Daily Life Applications of AP: frequently asked questions

How do I know whether a word problem is an AP?

Check whether the quantity changes by the same fixed amount at each equal step, such as each day, row or month. If it does, it is an AP. If it multiplies or changes by different amounts, it is not an AP.

How do I decide between the nth term and the sum formula?

If the question asks about one particular month, row or day, use the nth term. If it asks for the total, the amount in all, or the amount so far, use the sum formula.

What if both roots of the quadratic are positive?

Check each root against the situation. If one root makes some term negative or impossible, such as negative sales, reject it and explain why. If both roots make sense, give both answers.

Are compound interest or doubling problems part of this chapter?

No. Compound interest and doubling do not change by a fixed amount each step, so they are not APs. Simple interest added each year does form an AP, because the interest added is the same every year.