1. Introduction to Arithmetic Progressions
A sequence is just an ordered list of numbers, each number being a term. When the difference between one term and the next is always the same number, the sequence is called an arithmetic progression (AP).
Why do they matter? Almost anything that changes by a fixed amount each step, such as the seats in a row of an auditorium, the value of a monthly saving, the distance covered in extra daily practice, or the candles arranged in a staircase, produces an arithmetic progression. If you can spot the pattern once, you can predict the 20th or even the 100th term without listing all the terms in between.
Sequence
An ordered list of numbers called terms.
Common difference
The fixed jump \(d\) from one term to the next.
\(n^{th}\) term
A formula \(a + (n-1)d\) gives any position directly.
Sum of \(n\) terms
\[S_n = \frac{n}{2}\left[2a + (n-1)d\right]\]
2. What You Will Learn
- Tell whether a given list of numbers is an arithmetic progression.
- Write the first term \(a\) and common difference \(d\), including negative differences.
- Express an AP in the general form \(a,\; a+d,\; a+2d,\; \dots\).
- Find any term using \(a_n = a + (n-1)d\), and find \(n\) the other way round.
- Find unknown terms and missing middle terms of an AP.
- Add a long run of terms with the sum formula.
- Frame and solve word problems from seating, savings, staircase and production situations.
- Answer competency and case-based questions with reasoning, not just calculation.
3. Sequences
A sequence is written as \(a_1, a_2, a_3, \dots\) where \(a_1\) is the first term, \(a_2\) the second term, and so on. The position is called the index. The table shows a few sequences and the rule that produces them.
| Sequence | Rule for the terms | Type |
|---|---|---|
| 2, 5, 8, 11, 14 | Add 3 each time | Arithmetic |
| 10, 7, 4, 1, -2 | Subtract 3 each time | Arithmetic |
| 2, 4, 8, 16, 32 | Multiply by 2 each time | Not arithmetic (geometric) |
| 1, 4, 9, 16, 25 | Square numbers | Not arithmetic |
| 3, 3, 3, 3, 3 | Always the same | Arithmetic (constant) |
Not every sequence is arithmetic. Only a fixed addition (or subtraction) at every step makes a sequence an AP. Sequences that multiply by a constant, square, or follow any other rule are not APs.
4. Arithmetic Progressions
A list like 3, 7, 11, 15, 19 shows the key feature of an AP: every next term is obtained by adding the same number to the preceding term.
An arithmetic progression is a sequence of numbers \(a_1, a_2, a_3, \dots\) in which each term (except the first) is obtained by adding a fixed number \(d\) to the preceding term. The fixed number \(d\) is called the common difference.
Depending on the sign of \(d\), an AP is:
Increasing
\(d > 0\). Example: 3, 7, 11, 15 (add 4).
Decreasing
\(d < 0\). Example: 10, 7, 4, 1 (add -3).
Constant
\(d = 0\). Example: 6, 6, 6, 6.
2, 4, 8, 16 is not an AP. The jumps are \(4 - 2 = 2\), \(8 - 4 = 4\), \(16 - 8 = 8\): they are not equal. A fixed multiplier gives a different kind of sequence.
5. Common Difference
The common difference is the number you add to any term to reach the next term. It can be positive, negative or zero, and it is found by subtracting any term from the next one.
\[d = a_2 - a_1 = a_3 - a_2 = \dots = a_n - a_{n-1}\]
Notice the order: \(d\) is always the later term minus the earlier term, not the other way round.
| Arithmetic progression | \(a\) | \(d\) | Type |
|---|---|---|---|
| 3, 7, 11, 15, 19 | 3 | 4 | Increasing |
| 10, 7, 4, 1, -2 | 10 | -3 | Decreasing |
| -1, -4, -7, -10 | -1 | -3 | Decreasing |
| 5, 5, 5, 5 | 5 | 0 | Constant |
| 1.5, 2, 2.5, 3 | 1.5 | 0.5 | Increasing |
Find the common difference of the AP \(-\dfrac{1}{3},-\dfrac{2}{3},-1,-\dfrac{4}{3}\).
\[d = a_2 - a_1 = -\frac{2}{3} - \left(-\frac{1}{3}\right) = -\frac{1}{3}\]
Check: \(a_3 - a_2 = -1 + \frac{2}{3} = -\frac{1}{3}\) ✓. The common difference is \(-\dfrac{1}{3}\) (a decreasing AP).
Students often write \(d = a_1 - a_2 = 3\). Because the sequence mostly grows, the sign of a negative \(d\) is easy to lose. Always compute \(d = a_2 - a_1\).
6. General Form of an AP
If the first term is \(a\) and the common difference is \(d\), the AP grows like this:
\[a,\; a + d,\; a + 2d,\; a + 3d,\; \dots\]
where:
- \(a\) is the first term;
- \(d\) is the common difference;
- \(n\) is the position (or number of terms);
- \(a_n\) is the \(n^{th}\) term.
Write the first five terms of the AP with \(a = 3\) and \(d = 4\).
\[3,\; 3+4,\; 3+2(4),\; 3+3(4),\; 3+4(4)\]
So the AP is 3, 7, 11, 15, 19.
Write the first five terms of the AP with \(a = 10\) and \(d = -3\).
\[10,\; 10 - 3,\; 10 - 6,\; 10 - 9,\; 10 - 12\]
So the AP is 10, 7, 4, 1, -2.
7. The \(n^{th}\) Term of an AP
Writing out a full list is fine for the first few terms, but to reach the 20th term we need a formula. The pattern in the general form is easy to read:
\[ \begin{aligned} a_1 &= a = a + (1-1)d\\ a_2 &= a + d = a + (2-1)d\\ a_3 &= a + 2d = a + (3-1)d\\ a_4 &= a + 3d = a + (4-1)d\\ a_n &= a + (n-1)d \end{aligned} \]
Each term carries one fewer copy of \(d\) than its position number: the 4th term has \(3d\), so the \(n^{th}\) term has \((n-1)d\).
\[ \boxed{a_n = a + (n-1)d} \]
Find the 10th term of the AP 2, 7, 12, 17, ...
Here \(a = 2\), \(d = 5\), \(n = 10\). Using \(a_n = a + (n-1)d\):
\[a_{10} = 2 + (10-1)(5) = 2 + 45 = 47\]
So the 10th term is 47. Check: 2, 7, 12, 17, 22, 27, 32, 37, 42, 47 ✓.
Find the 15th term of the AP 20, 17, 14, 11, ...
Here \(a = 20\), \(d = 17 - 20 = -3\), \(n = 15\).
\[a_{15} = 20 + (15-1)(-3) = 20 - 42 = -22\]
So the 15th term is -22.
Which term of the AP 21, 18, 15, ... is -81?
Here \(a = 21\), \(d = -3\), \(a_n = -81\). Put these into the formula:
\[-81 = 21 + (n-1)(-3) \implies (n-1)(-3) = -102 \implies n - 1 = 34 \implies n = 35\]
So -81 is the 35th term.
Find the 20th term from the end of the AP 3, 8, 13, ..., 253.
Let the last term \(l = 253\) and \(d = 5\). Terms counted from the end form the same AP reversed, so the 20th term from the end is the last term minus \((20-1)d\):
\[253 - (20-1)(5) = 253 - 95 = 158\]
So the 20th term from the end is 158.
8. Finding the Number of Terms
Whenever a question asks "how many terms" or "which term", solve the \(n^{th}\) term formula for \(n\):
\[n = \frac{a_n - a}{d} + 1\]
How many terms are there in the AP 7, 13, 19, ..., 205?
Here \(a = 7\), \(d = 6\), and \(a_n = 205\).
\[n = \frac{205 - 7}{6} + 1 = \frac{198}{6} + 1 = 33 + 1 = 34\]
So there are 34 terms.
Is 68 a term of the AP 7, 10, 13, ...?
Here \(a = 7\), \(d = 3\), \(a_n = 68\). Then
\[n = \frac{68 - 7}{3} + 1 = \frac{61}{3} + 1\]
Since \(\dfrac{61}{3}\) is not an integer, \(n\) is not a whole number, so 68 is not a term of this AP.
9. Unknown Terms and Missing Terms
The formula \(a_n = a + (n-1)d\) links four quantities. If three of them are known, the fourth can be found.
The 7th term of an AP is 20 and the 13th term is 32. Find \(a\) and \(d\).
The 7th term: \(a + 6d = 20\). The 13th term: \(a + 12d = 32\).
Subtracting the first equation from the second: \(6d = 12 \implies d = 2\). Then \(a + 12 = 20 \implies a = 8\).
So \(a = 8\) and \(d = 2\). The 10th term would be \(8 + 9(2) = 26\).
In an AP, \(a = 4\) and the 10th term is 40. Find \(d\).
\[40 = 4 + (10-1)d \implies 9d = 36 \implies d = 4\]
So the common difference is 4.
Find the missing term in the AP 2, __, 26.
The middle term must be the arithmetic mean of its two neighbours (equal gaps each side):
\[\text{middle term} = \frac{2 + 26}{2} = 14\]
So the AP is 2, 14, 26, with \(d = 12\). Check: \(14 - 2 = 12\) and \(26 - 14 = 12\) ✓.
Write the first four terms of the AP with \(a = 3\) and \(d = -2\).
\[3,\; 3 - 2,\; 3 - 4,\; 3 - 6 \implies \text{3, 1, -1, -3}\]
The AP is 3, 1, -1, -3.
The arithmetic mean rule generalises: if \(x\), \(y\), \(z\) are three consecutive terms of an AP, then \(y = \dfrac{x + z}{2}\). Use it to verify quickly when a missing middle term is hidden in larger APs.
10. Sum of the First \(n\) Terms
When a question asks for a total, such as total seats or total savings, add the AP up to a certain term. Counting term by term is slow for 100 terms, so we derive a formula.
Write the sum forward, then backward, and add the two lines. Each column adds to the same value:
\[ \begin{aligned} S_n &= a + (a+d) + \dots + (a + (n-1)d)\\ S_n &= (a + (n-1)d) + \dots + (a+d) + a\\[4pt] 2S_n &= n\left[2a + (n-1)d\right] \end{aligned} \]
\[ \boxed{S_n = \frac{n}{2}\left[2a + (n-1)d\right]} \]
If the last term \(l = a + (n-1)d\) is known, a shorter equivalent form is
\[ S_n = \frac{n}{2}\left(a + l\right) \]
Find the sum of the first 20 terms of the AP 3, 5, 7, ...
Here \(a = 3\), \(d = 2\), \(n = 20\).
\[S_{20} = \frac{20}{2}\left[2(3) + (20-1)(2)\right] = 10(6 + 38) = 10(44) = 440\]
So the sum is 440.
Find the sum of the AP 2, 5, 8, ..., 26.
First count the terms: \(a = 2\), \(d = 3\), \(a_n = 26\).
\[n = \frac{26 - 2}{3} + 1 = 9\]
Last term \(l = 26\), so \(S_9 = \dfrac{9}{2}(2 + 26) = \dfrac{9}{2}(28) = 126\).
So the sum is 126. Check: \(2+5+8+11+14+17+20+23+26 = 126\) ✓.
The two versions of the formula are equal only when \(l = a + (n-1)d\). Do not use \(\dfrac{n}{2}(a+l)\) with a guessed last term, and do not forget the factor \(\dfrac{n}{2}\) in either form.
11. Application Word Problems
Word problems ask you to decide whether you need a single term or a total. The checklist below leads to the right formula every time.
A cinema hall has 40 seats in the first row and 8 more seats in each successive row. How many seats are in the 25th row? Here \(a = 40\), \(d = 8\), \(n = 25\). \(a_{25} = 40 + 24(8) = 40 + 192 = 232\). So the 25th row has 232 seats.
An auditorium has 30 rows. The first row has 20 seats and each successive row has 2 more seats. What is the total number of seats? Each row count is an AP with \(a = 20\), \(d = 2\), \(n = 30\):
\[S_{30} = \frac{30}{2}\left[2(20) + (30-1)(2)\right] = 15(40 + 58) = 15(98) = 1470\]
So there are 1470 seats in the auditorium.
A student deposits ₹300 in the first month and ₹100 more in each successive month. What is the total deposited after 12 months? Here \(a = 300\), \(d = 100\), \(n = 12\):
\[S_{12} = \frac{12}{2}\left[2(300) + (12-1)(100)\right] = 6(600 + 1100) = 6(1700) = 10200\]
The total deposit is ₹10,200.
A staircase is built with 1 brick in the top step, 2 bricks in the next, then 3, 4, and so on. How many bricks are needed for 12 steps? The counts 1, 2, 3, ..., 12 form an AP with \(a = 1\), \(d = 1\), \(n = 12\), \(l = 12\):
\[S_{12} = \frac{12}{2}(1 + 12) = 6(13) = 78\]
So 78 bricks are needed.
A factory makes 800 units in the first week and increases production by 25 units every week. In which week will production reach 1050 units? Here \(a = 800\), \(d = 25\), \(a_n = 1050\):
\[1050 = 800 + (n-1)(25) \implies 25(n-1) = 250 \implies n = 11\]
Production reaches 1050 units in the 11th week.
How many multiples of 4 are there between 10 and 250? The first multiple of 4 from 10 is 12 and the last is 248. They form an AP with \(a = 12\), \(d = 4\), \(a_n = 248\):
\[n = \frac{248 - 12}{4} + 1 = 59 + 1 = 60\]
So there are 60 multiples of 4 between 10 and 250.
A runner covers 2 km on the first day and increases the daily distance by 0.5 km each day. Distance on the 21st day: \(a_{21} = 2 + 20(0.5) = 12\) km, so the runner covers 12 km on day 21.
12. Important CBSE Question Types
Almost every board question on AP is one of eight forms. Learning to recognise the type tells you instantly which formula to open.
| Type | Typical wording | Approach |
|---|---|---|
| 1. Find the \(n^{th}\) term | "Find the 12th term of ..." | Use \(a_n = a + (n-1)d\) |
| 2. Which term / find \(n\) | "Which term of ... is 96?" | Set \(a_n\) equal and solve for \(n\) |
| 3. Find \(a\) and \(d\) | "The 4th term is 14 and the 9th term is 29" | Form two equations, subtract |
| 4. Count the terms | "How many terms are there in ...?" | Use \(n = \dfrac{a_n - a}{d} + 1\) |
| 5. Is it a term? | "Is 72 a term of ...?" | Check \(n\) comes out a positive integer |
| 6. Missing terms | "Find the middle term of ..." | Arithmetic mean: \(\dfrac{x + z}{2}\) or solve for \(d\) |
| 7. Application | Seats, savings, production, staircase | Pick \(a_n\) or \(S_n\) from the story |
| 8. Competency / case-based | A situation with 2-3 sub-questions | Show reasoning with the data given |
(1) The 10th term of 3, 8, 13, ... is \(3 + 9(5) = 48\). (2) Which term of 6, 11, 16, ... is 96? \(n = \dfrac{96-6}{5} + 1 = 19\). (3) If the 4th term is 14 and the 9th term is 29, subtract the equations: \(5d = 15 \implies d = 3\), and \(a + 3(3) = 14 \implies a = 5\). (4) Multiples of 5 between 20 and 100: \(n = \dfrac{95 - 25}{5} + 1 = 15\). (5) 72 in 9, 13, 17, ...? \(\dfrac{72 - 9}{4}\) is not a whole number, so no. (6) Missing term in 8, __, __, 26: \(8 + 3d = 26 \implies d = 6\), so the terms are 14 and 20.
13. Competency and Reasoning Questions
Competency questions check whether you understand why the AP tools work, not just the plug-and-play formula. They often hide a sign error or ask you to interpret an equation rather than compute.
From the following, pick the sequence that is not an AP: 2, 5, 8, 11; 3, 6, 12, 24; 20, 16, 12; 7, 7, 7.
Reasoning. The second: 3, 6, 12, 24. Its differences are \(3, 6, 12\), which are not equal. The others have differences \(3, -4, 0\) respectively, so each is an AP.
If the \(n^{th}\) term of a sequence is \(a_n = 4n + 3\), prove that it is an AP and find \(a\) and \(d\).
Reasoning. \(a_{n+1} - a_n = [4(n+1)+3] - [4n+3] = 4\), a constant, so it is an AP with \(d = 4\). The first term \(a_1 = 4(1) + 3 = 7\), so \(a = 7\).
A student finds the 10th term of the AP 2, 5, 8, ... as \(2 + 10 \times 3 = 32\). Where is the error?
Reasoning. The multiplier must be \((n-1)d = 9 \times 3 = 27\), not \(10 \times 3\). The correct 10th term is \(2 + 27 = 29\).
A list of numbers decreases by the same amount each step. Is the common difference positive or negative?
Reasoning. Negative. If the list falls, each term is less than the previous one, so \(d = a_{k+1} - a_k < 0\). The size of the drop is \(|d|\).
Two APs have first terms 12 and -12, and the same common difference 4. Which has the larger 20th term?
Reasoning. The common part \((20-1)d = 76\) is identical, so the larger first term wins: \(12 + 76 = 88\) against \(-12 + 76 = 64\). The first AP.
For an AP with first term \(p\) and third term \(q\), write the middle term.
Reasoning. The middle term is the arithmetic mean of the two neighbours: \(\dfrac{p + q}{2}\). The same result follows from \(a_2 = p + d\) and \(q = a_2 + d\).
14. Case-Based Examples
Case-based questions describe a situation and then ask 2-3 related sub-questions. Answer the sub-questions using the case data only, and show all working.
Case Study 1 — Community Hall Seating
A community hall has 24 rows. The first row has 20 seats and each successive row has 3 more seats.
Question 1
Find the number of seats in the 15th row.
Question 2
Find the total number of seats in the hall.
Question 3
Verify, using the formula, that the 15th row has exactly 62 seats.
Solution. Here \(a = 20\), \(d = 3\).
(1) \(a_{15} = 20 + (15-1)(3) = 20 + 42 = 62\). The 15th row has 62 seats.
(2) \(S_{24} = \dfrac{24}{2}\left[2(20) + (24-1)(3)\right] = 12(40 + 69) = 12(109) = 1308\). The hall holds 1308 seats.
(3) Substituting \(n = 15\) into \(a_n = 20 + (n-1)(3)\) gives \(20 + 42 = 62\), confirming the 15th row has 62 seats.
Case Study 2 — Monthly Investment Plan
A client invests ₹1000 in the first month and increases the investment by ₹500 every month.
Question 1
What is the investment in the 10th month?
Question 2
What is the total invested over 12 months?
Question 3
In which month does the investment first reach ₹4500?
Solution. Here \(a = 1000\), \(d = 500\).
(1) \(a_{10} = 1000 + 9(500) = 1000 + 4500 = 5500\). The amount in month 10 is ₹5500.
(2) \(S_{12} = \dfrac{12}{2}\left[2(1000) + 11(500)\right] = 6(2000 + 5500) = 6(7500) = 45000\). The total invested is ₹45,000.
(3) \(1000 + (n-1)(500) = 4500 \implies n - 1 = 7 \implies n = 8\). It is reached in the 8th month.
Case Study 3 — Marathon Training Plan
A trainee runs 4 km on the first day and adds 2 km to the daily run on every subsequent day.
Question 1
Find the distance run on the 10th day.
Question 2
Find the total distance covered in the first 10 days.
Question 3
On which day does the distance first reach 20 km?
Solution. Here \(a = 4\), \(d = 2\).
(1) \(a_{10} = 4 + 9(2) = 22\). The distance on day 10 is 22 km.
(2) \(S_{10} = \dfrac{10}{2}\left[2(4) + 9(2)\right] = 5(8 + 18) = 5(26) = 130\). The total covered is 130 km.
(3) \(4 + (n-1)(2) = 20 \implies n - 1 = 8 \implies n = 9\). The distance reaches 20 km on the 9th day.
Case Study 4 — Trees Along a Road
Trees are planted along both sides of a straight road exactly 1 km long. The first tree is at the start, the last at the finish, with 40 m between successive trees.
Question 1
How many trees are there on one side of the road?
Question 2
How many trees are there in total on both sides?
Question 3
How far from the start is the 10th tree on a side?
Solution. Positions form an AP with \(a = 0\), \(d = 40\), \(l = 1000\) m.
(1) \(n = \dfrac{1000 - 0}{40} + 1 = 25 + 1 = 26\). There are 26 trees on one side.
(2) Both sides: \(26 \times 2 = 52\) trees in total.
(3) 10th tree: \(a_{10} = 0 + 9(40) = 360\) m from the start.
15. CBSE Board-Style Questions
These original questions are written in the pattern and difficulty of CBSE board questions on this chapter.
Q-B1.
The \(n^{th}\) term of an AP is \(3 + 4n\). Find the first term and the common difference.
Q-B2.
How many two-digit numbers are divisible by 7?
Q-B3.
Find the sum of the first 20 terms of the AP 5, 8, 11, ...
Q-B4.
Which term of the AP 9, 13, 17, ... is 217?
Q-B5.
For the AP 100, 96, 92, ..., find the 12th term.
Q-B6.
Find the sum of the first 100 positive odd integers.
Q-B7.
A hall has 26 rows. The first row has 18 seats and each successive row has 2 more seats. Find the total number of seats in the hall.
Q-B8.
Is the sequence whose \(n^{th}\) term is \(2n^2 - 1\) an arithmetic progression? Justify.
Answers. (B1) \(a_1 = 7\), so \(a = 7\); \(d = 4\). (B2) Two-digit multiples of 7 run from 14 to 98: \(n = \dfrac{98 - 14}{7} + 1 = 13\). (B3) \(S_{20} = 10(10 + 57) = 670\). (B4) \(n = \dfrac{217 - 9}{4} + 1 = 53\). (B5) \(a_{12} = 100 + 11(-4) = 56\). (B6) \(S_{100} = \dfrac{100}{2}(1 + 199) = 10000\). (B7) \(S_{26} = 13(36 + 50) = 1118\) seats. (B8) No. \(a_1 = 1, a_2 = 7, a_3 = 17\) have differences 6 and 10, not equal; an AP's \(n^{th}\) term must be linear in \(n\).
16. Important Formulae
\[a,\; a + d,\; a + 2d,\; a + 3d,\; \dots\]
\[ \boxed{a_n = a + (n-1)d} \]
\[d = a_2 - a_1 = a_n - a_{n-1}\]
\[S_n = \frac{n}{2}\left[2a + (n-1)d\right] = \frac{n}{2}(a + l)\]
Match the formula to the question. For "how many seats in a row" use \(a_n\); for "total seats in all rows" use \(S_n\). Keep the sign of \(d\), and remember \((n-1)\) not \(n\).
17. Common Examination Mistakes
| Common mistake | Correct method |
|---|---|
| Confusing the first term \(a\) with the \(n^{th}\) term \(a_n\) | \(a\) is the term itself; \(a_n = a + (n-1)d\) is what you are usually asked to find |
| Using \(n\) instead of \(n-1\) in \(a_n = a + (n-1)d\) | The 1st term is reached with \(n = 1\) giving \(a + 0 \cdot d\); always \((n-1)\) copies of \(d\) |
| Writing \(d = a_1 - a_2\) for a decreasing AP | Always \(d = a_2 - a_1\): later term minus earlier term |
| Forgetting the sign of a negative \(d\) when substituting | Substitute \(d\) with its sign, e.g. \(20 + 14(-3)\), and simplify carefully |
| Assuming every sequence is an AP | Check that the difference between consecutive terms is constant |
| Solving "which term" but stopping at \((n-1)\) | Isolate \(n\) fully: \(n = \dfrac{a_n - a}{d} + 1\), then check it is a positive integer |
| Applying \(\dfrac{n}{2}(a+l)\) without the actual last term | Use the exact count and last term, or use \(\dfrac{n}{2}\left[2a + (n-1)d\right]\) |
| Mixing up a term question with a total question | "In row x" means \(a_n\); "total in x rows" means \(S_n\) |
| Using the arithmetic mean rule outside an AP | \(y = \dfrac{x+z}{2}\) holds only when \(x, y, z\) are consecutive terms of an AP |
| Not verifying the final answer | With \(a_n\) and \(d\), extend the sequence a few terms to check the pattern |
For the AP 21, 18, 15, ..., a student finds the \(n^{th}\) term as \(21 + (n-1)(-3)\) but then writes \(a_{10} = 21 - 27 = -6\) while claiming \(d = 3\). The sign error changes the answer: the correct 10th term is \(21 + 9(-3) = 21 - 27 = -6\), same value, but the reasoning that "subtract anyway" must use \(d = -3\) formally.
18. CBSE Exam Tips
- Start by naming \(a\), \(d\) and what is asked (\(a_n\) or \(S_n\)) before touching the formula; this organises the whole answer.
- Write the formula first and then substitute: \(a_n = a + (n-1)d\) with values. Substitution steps carry marks even if a later arithmetic step is wrong.
- Treat a negative common difference with extra care, especially when solving "which term" equations.
- For "which term" or counts, check that \(n\) is a positive integer; a fraction means the number is not in the AP.
- For a term "from the end", reverse the AP mentally and use \(l - (n-1)d\), where \(l\) is the last term.
- For missing middle terms, use the arithmetic mean \(\dfrac{x+z}{2}\) of the two neighbours.
- In word problems, write the units in the final answer (seats, ₹, km, m).
- For case-based questions, read the whole case once before answering any sub-question, then reuse the same \(a\) and \(d\) throughout.
19. Quick Revision
| Concept | Key point |
|---|---|
| Sequence | An ordered list of numbers; each number is a term |
| Arithmetic progression | A sequence with a constant difference between consecutive terms |
| Common difference | \(d = a_2 - a_1 = a_n - a_{n-1}\) |
| General form | \(a,\; a + d,\; a + 2d,\; a + 3d,\; \dots\) |
| \(n^{th}\) term | \(a_n = a + (n-1)d\) |
| Number of terms | \(n = \dfrac{a_n - a}{d} + 1\), must be a positive integer |
| Sum of \(n\) terms | \(S_n = \dfrac{n}{2}\left[2a + (n-1)d\right] = \dfrac{n}{2}(a + l)\) |
| Middle term rule | \(x,\; y,\; z\) consecutive \(\implies y = \dfrac{x+z}{2}\) |
| Types by \(d\) | \(d > 0\) increasing, \(d < 0\) decreasing, \(d = 0\) constant |
| Not an AP | Differences not equal, e.g. 2, 4, 8, 16, ... |
20. Self-Practice Questions
Thirty original questions written in the CBSE exam style. Answers and solutions are in Section 21.
Conceptual
Q1.
For the AP 5, 10, 15, 20, state the first term \(a\) and the common difference \(d\).
Q2.
Which of the following are arithmetic progressions? \(1, 3, 5, 7\); \(5, 5, 5, 5\); \(10, 7, 4, 1\); \(2, 4, 8, 16\).
Q3.
Find the common difference of the AP 10, 7, 4, 1.
Q4.
Write the general form of an AP with first term \(a\) and common difference \(d\).
Q5.
Why is 2, 4, 8, 16 not an arithmetic progression?
Q6.
In the AP 3, 7, 11, 15, ..., identify \(a\) and \(d\).
Standard — nth term
Q7.
Find the 12th term of the AP 5, 8, 11, 14, ...
Q8.
Find the 9th term of the AP 20, 17, 14, 11, ...
Q9.
Find the 18th term of the AP 2, 4, 6, 8, ...
Q10.
Which term of the AP 5, 10, 15, ... is 100?
Q11.
Which term of the AP 21, 18, 15, ... is -30?
Q12.
Find the 10th term of the AP -7, -4, -1, 2, ...
Q13.
How many terms are there in the AP 10, 13, 16, ..., 94?
Q14.
Is 95 a term of the AP 4, 11, 18, ...? If yes, which term?
Using the sum formula
Q15.
Find the sum of the first 50 even numbers: 2, 4, 6, ..., 100.
Q16.
Find the sum of the first 12 terms of the AP 1, 6, 11, 16, ...
Q17.
Find the sum of the first 15 terms of the AP 4, 8, 12, ...
Q18.
Find the sum of the AP 2, 5, 8, ..., 26.
Application based
Q19.
An auditorium has 30 rows. The first row has 25 seats and each successive row has 4 more seats. Find the total number of seats.
Q20.
A student saves ₹800 in the first month and ₹200 more each following month. Find the total saved in 10 months.
Q21.
A hall has 35 seats in the first row and 6 more in each successive row. How many seats are in the 25th row?
Q22.
How many multiples of 3 are there between 20 and 200?
Q23.
A plant is 10 cm tall on day 1 and grows 2 cm every day. What is its height on day 15?
Q24.
A cyclist rides 3 km on day 1 and 1 km more each day. Find the distance on day 30 and the total distance in 30 days.
Q25.
At 8:00 am the temperature is 30°C and it falls by 2°C each hour. What is the temperature at 11:00 am?
Q26.
A factory makes 500 items in week 1 and 40 more each week. In which week does production reach 900 items?
Q27.
Tickets sold on day 1 is 120 and sales fall by 5 each day. How many tickets are sold on day 12?
Competency / Reasoning
Q28.
If the \(n^{th}\) term of a sequence is \(4n + 3\), show that it is an AP and find \(a\) and \(d\).
Q29.
The 5th and 9th terms of an AP are 16 and 28. Find \(a\) and \(d\).
Q30 — Roadside trees.
Trees are planted every 50 m along a straight road of length 1.2 km, starting at the beginning. (i) How many trees are planted on one side? (ii) How far from the start is the 10th tree?
21. Answers and Solutions
A1.
\(a = 5\), \(d = 5\).
A2.
\(1, 3, 5, 7\); \(5, 5, 5, 5\); and \(10, 7, 4, 1\) are APs. \(2, 4, 8, 16\) is not (differences 2, 4, 8 are not equal).
A3.
\(d = 7 - 10 = -3\).
A4.
\(a,\; a + d,\; a + 2d,\; a + 3d,\; \dots\)
A5.
The differences are \(4 - 2 = 2\), \(8 - 4 = 4\), \(16 - 8 = 8\), which are not constant.
A6.
\(a = 3\), \(d = 4\).
A7.
\(a = 5\), \(d = 3\), \(n = 12\): \(5 + 11(3) = 38\).
A8.
\(a = 20\), \(d = -3\), \(n = 9\): \(20 + 8(-3) = -4\).
A9.
\(a = 2\), \(d = 2\), \(n = 18\): \(2 + 17(2) = 36\).
A10.
\(n = \dfrac{100 - 5}{5} + 1 = 20\), so 100 is the 20th term.
A11.
\(21 + (n-1)(-3) = -30 \implies n - 1 = 17 \implies n = 18\), the 18th term.
A12.
\(a = -7\), \(d = 3\), \(n = 10\): \(-7 + 9(3) = 20\).
A13.
\(n = \dfrac{94 - 10}{3} + 1 = 29\) terms.
A14.
\(n = \dfrac{95 - 4}{7} + 1 = 14\), so yes, it is the 14th term.
A15.
Even numbers form an AP with \(a = 2\), \(l = 100\), \(n = 50\): \(S = \dfrac{50}{2}(2 + 100) = 2550\).
A16.
\(a = 1\), \(d = 5\), \(n = 12\): \(S = 6(2 + 55) = 342\).
A17.
\(a = 4\), \(d = 4\), \(n = 15\): \(S = \dfrac{15}{2}(8 + 56) = 480\).
A18.
\(n = \dfrac{26 - 2}{3} + 1 = 9\), so \(S_9 = \dfrac{9}{2}(2 + 26) = 126\).
A19.
\(a = 25\), \(d = 4\), \(n = 30\): \(S = 15(50 + 116) = 2490\) seats.
A20.
\(a = 800\), \(d = 200\), \(n = 10\): \(S = 5(1600 + 1800) = 17000\). Total ₹17,000.
A21.
\(a = 35\), \(d = 6\), \(n = 25\): \(35 + 24(6) = 179\) seats.
A22.
Multiples from 21 to 198: \(n = \dfrac{198 - 21}{3} + 1 = 60\).
A23.
\(a = 10\), \(d = 2\), \(n = 15\): \(10 + 14(2) = 38\) cm.
A24.
Day 30: \(3 + 29(1) = 32\) km. Total: \(S = 15(6 + 29) = 525\) km.
A25.
From 8:00 to 11:00 is 3 hours: \(30 - 3(2) = 24\)°C at 11:00 am.
A26.
\(500 + 40(n-1) = 900 \implies n - 1 = 10 \implies n = 11\), week 11.
A27.
\(a = 120\), \(d = -5\), \(n = 12\): \(120 + 11(-5) = 65\) tickets.
A28.
\(a_{n+1} - a_n = [4(n+1)+3] - (4n+3) = 4\), constant, so it is an AP with \(a = 7\) and \(d = 4\).
A29.
\(a + 4d = 16\) and \(a + 8d = 28\) give \(4d = 12 \implies d = 3\), then \(a = 4\).
A30.
(i) Positions 0, 50, ..., 1200 m: \(n = \dfrac{1200 - 0}{50} + 1 = 25\) trees on one side. (ii) 10th tree: \(0 + 9(50) = 450\) m from the start.