Arithmetic Progressions studies lists of numbers in which every term after the first is made by adding the same fixed number, the common difference d. The chapter shows how to test whether a list is an AP, how to find any term with aₙ = a + (n − 1)d, and how to add the first n terms with Sₙ = n/2 [2a + (n − 1)d], the idea Gauss used to add 1 to 100. Board questions use these two formulas on terms, counts of multiples, sums and daily-life problems on salaries, savings, prizes and stacks.
Key Concepts
1. Patterns in Lists of Numbers
Many lists of numbers in daily life follow a rule. The chapter opens with examples such as Reena’s monthly salary of ₹8000 with an annual increment of ₹500 (8000, 8500, 9000, …), the rungs of a ladder that decrease uniformly by 2 cm from a 45 cm bottom rung (45, 43, 41, …, 31), and Shakila’s money box, which gets ₹100 at age one and ₹50 more every year (100, 150, 200, 250, …). In each of these, the next term comes from adding a fixed number.
Other lists follow other rules. In a savings scheme where the amount becomes 5/4 times itself every 3 years, ₹8000 grows to 10000, 12500, 15625, 19531.25: each term is multiplied by a fixed number. The numbers of unit squares in squares of side 1, 2, 3, … are 12, 22, 32, …, and the rabbit pairs 1, 1, 2, 3, 5, 8 follow yet another rule. This chapter studies only the first kind of list.
2. Arithmetic Progression and Common Difference
Look at the lists 1, 2, 3, 4, … (add 1), 100, 70, 40, 10, … (add −30), 3, 3, 3, 3, … (add 0) and −1.0, −1.5, −2.0, −2.5, … (add −0.5).
Each number in a list is called a term. In each list, every term after the first is the previous term plus the same fixed number.
Definition: An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. The fixed number is called the common difference of the AP, written d. It can be positive, negative or zero.
For terms a1, a2, a3, …, an: a2 − a1 = a3 − a2 = … = an − an−1 = d.
In general, d = ak+1 − ak, where ak+1 and ak are the (k + 1)th and kth terms. Always subtract a term from the term that follows it, even when the following term is smaller. In 6, 3, 0, −3, …, d = 3 − 6 = −3, not 6 − 3 = 3.
| Sign of d | Behaviour of the AP | Example |
|---|---|---|
| d > 0 | Terms increase | 6, 9, 12, 15, … (d = 3) |
| d < 0 | Terms decrease | 6, 3, 0, −3, … (d = −3) |
| d = 0 | All terms are equal | 2, 2, 2, 2, … (d = 0) |
Illustration (NCERT Example 1): For the AP 3/2, 1/2, −1/2, −3/2, …, the first term a = 3/2 and d = 1/2 − 3/2 = −1.
3. General Form, Finite and Infinite APs
If the first term is a and the common difference is d, the AP is
a, a + d, a + 2d, a + 3d, …
This is the general form of an AP. To write an AP you need both the first term a and the common difference d. For example, a = 6, d = 3 gives 6, 9, 12, 15, …; a = 6, d = −3 gives 6, 3, 0, −3, …; a = −7, d = −2 gives −7, −9, −11, …; and a = 2, d = 0 gives 2, 2, 2, ….
A finite AP has a limited number of terms and so has a last term. Example: cash prizes for toppers of Classes I to XII, 200, 250, 300, …, 750. An infinite AP goes on without end and has no last term, like 1, 2, 3, 4, ….
The last term of a finite AP is written l. If the AP has m terms, then am = l.
4. Testing Whether a List Is an AP
A list is an AP only if the difference between each term and the one before it is the same all the way. The method:
- Find a2 − a1, a3 − a2, a4 − a3, and so on for the terms given.
- If all these differences are equal, the list is an AP and the common value is d.
- If even one difference is different, the list is not an AP.
Illustration (NCERT Example 2):
- 4, 10, 16, 22, …: differences 6, 6, 6. It is an AP with d = 6. Next two terms: 28 and 34.
- 1, −1, −3, −5, …: differences −1 − 1 = −2, −3 − (−1) = −2, −5 − (−3) = −2. It is an AP with d = −2. Next two terms: −7 and −9.
- −2, 2, −2, 2, −2, …: a2 − a1 = 4 but a3 − a2 = −4. Not an AP.
- 1, 1, 1, 2, 2, 2, 3, 3, 3, …: differences 0, 0, 1. Not an AP.
Lists like 2, 4, 8, 16, … (differences 2, 4, 8) and 1, 1, 2, 3, 5, … (differences 0, 1, 1, 2) are not APs.
5. The nth Term of an AP
Reena’s salary shows the idea. Her salary in the 3rd year is 8000 + 2 × 500, in the 4th year 8000 + 3 × 500, and in the 5th year 8000 + 4 × 500 = ₹10000. So the 15th year salary is 8000 + (15 − 1) × 500 = ₹15000, and the 25th year salary is 8000 + (25 − 1) × 500 = ₹20000.
In general, a2 = a + d, a3 = a + 2d, a4 = a + 3d, and the nth term has n − 1 steps of d added to a:
an = a + (n − 1)d
an is also called the general term of the AP. The formula links four quantities: an, a, n and d. If any three are known, the fourth can be found.
Illustration (NCERT Example 3): 10th term of 2, 7, 12, …: a = 2, d = 5, n = 10. a10 = 2 + 9 × 5 = 47.
Illustration (NCERT Example 9): ₹1000 invested at 8% simple interest per year. The interest at the end of years 1, 2, 3, … is 1000 × 8 × 1/100 = 80, then 160, then 240, …. This is an AP with a = 80 and d = 80. Interest at the end of 30 years = a30 = 80 + 29 × 80 = ₹2400.
6. Problems on the nth Term
Most nth-term questions are one of five types.
Type A: Which term is a given number? Put an equal to the number and solve for n. NCERT Example 4: in 21, 18, 15, …, a = 21 and d = −3. For an = −81: −81 = 21 + (n − 1)(−3) = 24 − 3n, so 3n = 105 and n = 35. For an = 0: 21 + (n − 1)(−3) = 0 gives 3(n − 1) = 21, so n = 8. The 8th term is 0.
Type B: Is a number a term at all? n must be a positive integer, because it counts the position of a term. If solving gives a fraction or a negative value, the number is not a term. NCERT Example 6: for 5, 11, 17, 23, …, 301 = 5 + (n − 1) × 6 = 6n − 1 gives n = 302/6 = 151/3. This is not a whole number, so 301 is not a term.
Type C: Find the AP from two conditions. Write each condition in terms of a and d and solve the pair of linear equations. NCERT Example 5: a3 = 5 and a7 = 9 give a + 2d = 5 and a + 6d = 9. Subtracting, 4d = 4, so d = 1 and a = 3. The AP is 3, 4, 5, 6, 7, …
Type D: Counting terms. Use the last term: l = a + (n − 1)d. NCERT Example 7: the two-digit numbers divisible by 3 are 12, 15, 18, …, 99. So 99 = 12 + (n − 1) × 3, giving n − 1 = 29 and n = 30.
Type E: A term counted from the end. In a finite AP of n terms, the kth term from the end is the (n − k + 1)th term from the start. Or reverse the AP: the first term becomes l and the common difference becomes −d, so the kth term from the end is l − (k − 1)d. NCERT Example 8: in 10, 7, 4, …, −62, there are 25 terms. The 11th term from the end is the (25 − 11 + 1) = 15th term: 10 + 14 × (−3) = −32.
A useful fact: am − an = (m − n)d. For example, if the 17th term exceeds the 10th term by 7, then 7d = 7 and d = 1.
7. Sum of the First n Terms
Gauss’s method. Asked at the age of 10 to add the positive integers from 1 to 100, Gauss wrote the sum forwards and backwards:
S = 1 + 2 + 3 + … + 99 + 100
S = 100 + 99 + … + 3 + 2 + 1
Adding the two lines pair by pair gives 2S = 101 + 101 + … + 101 (100 times) = 100 × 101, so S = 5050.
The general formula. Apply the same idea to the AP a, a + d, a + 2d, …, a + (n − 1)d:
S = a + (a + d) + … + [a + (n − 1)d] (1)
S = [a + (n − 1)d] + [a + (n − 2)d] + … + a (2)
Adding (1) and (2) term by term, every pair adds up to 2a + (n − 1)d, and there are n pairs. So 2S = n[2a + (n − 1)d], which gives
Sn = n/2 [2a + (n − 1)d]
Since a + (n − 1)d = an, this can also be written Sn = n/2 (a + an). When the AP has only n terms, an is the last term l, so
Sn = n/2 (a + l)
Illustration (Shakila’s money box): a = 100, d = 50, n = 21. S = 21/2 [200 + 20 × 50] = 21/2 × 1200 = ₹12600 in the box on her daughter’s 21st birthday.
Sum of the first n positive integers (NCERT Example 14): Here a = 1 and l = n, so Sn = n(n + 1)/2. For the first 1000 positive integers, S = 1000/2 × 1001 = 500500.
8. Problems on the Sum and the Link an = Sn − Sn−1
Finding n from a sum. This gives a quadratic equation in n. Keep only positive integer values. NCERT Example 13: how many terms of 24, 21, 18, … give a sum of 78? 78 = n/2 [48 + (n − 1)(−3)] = n/2 (51 − 3n). So 3n2 − 51n + 156 = 0, that is n2 − 17n + 52 = 0, or (n − 4)(n − 13) = 0. Both n = 4 and n = 13 work. Two answers are possible because the terms from the 5th to the 13th add up to zero: a is positive and d is negative, so the positive and negative terms cancel.
Finding d or a from a sum. NCERT Example 12: S14 = 1050 and a = 10. 1050 = 14/2 [20 + 13d] = 140 + 91d, so d = 10. Then a20 = 10 + 19 × 10 = 200.
Sum of a list given by its nth term. NCERT Example 15: an = 3 + 2n gives 5, 7, 9, 11, …, an AP with a = 5 and d = 2. S24 = 12 [10 + 46] = 672. Any an that is linear in n (of the form pn + q) gives an AP with d = p.
The term from the sum. The nth term is the sum of the first n terms minus the sum of the first (n − 1) terms:
an = Sn − Sn−1, and a1 = S1.
For Sn = n2 + n: a1 = S1 = 2 and a2 = S2 − S1 = 6 − 2 = 4.
Arithmetic mean. If a, b, c are in AP, then b − a = c − b, so b = (a + c)/2. b is called the arithmetic mean of a and c. This gives a quick way to find an unknown: if 2k + 1, 13, 5k − 3 are in AP, then 2 × 13 = (2k + 1) + (5k − 3), so 26 = 7k − 2 and k = 4.
Formula and Theorem Sheet
| Result | Formula | Notes |
|---|---|---|
| Common difference | d = ak+1 − ak | Later term minus earlier term; can be positive, negative or zero |
| nth term | an = a + (n − 1)d | n must be a positive integer |
| Last term of n terms | l = an = a + (n − 1)d | Used to count terms |
| Difference of two terms | am − an = (m − n)d | Quick route to d |
| kth term from the end | l − (k − 1)d, or the (n − k + 1)th term | Reverse the AP: first term l, difference −d |
| Sum of first n terms | Sn = n/2 [2a + (n − 1)d] | Four quantities: S, a, d, n |
| Sum with last term | Sn = n/2 (a + l) | When d is not given |
| Term from sums | an = Sn − Sn−1; a1 = S1 | When Sn is given in terms of n |
| First n positive integers | 1 + 2 + … + n = n(n + 1)/2 | From Sn = n/2 (a + l) with a = 1, l = n |
| Arithmetic mean | b = (a + c)/2 when a, b, c are in AP | Equivalent to 2b = a + c |
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Important Definitions
- Arithmetic progression (AP): a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term.
- Common difference (d): the fixed number added to each term to get the next one. d = ak+1 − ak. It can be positive, negative or zero.
- General form of an AP: a, a + d, a + 2d, a + 3d, …, where a is the first term and d the common difference.
- nth term or general term (an): the term in position n, given by an = a + (n − 1)d.
- Finite AP: an AP with a limited number of terms. It has a last term.
- Infinite AP: an AP whose terms go on without end. It has no last term.
- Last term (l): the final term of a finite AP. If the AP has m terms, l = am.
- Sum of first n terms (Sn): a1 + a2 + … + an = n/2 [2a + (n − 1)d].
- Arithmetic mean: if a, b, c are in AP, then b = (a + c)/2 is called the arithmetic mean of a and c.
Solved Examples (NCERT-Based)
Example 1: Which situations make an AP? (NCERT Exercise 5.1)
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. (iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre. (iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.
Solution:
(i) Fares: 15, 23, 31, 39, …. Each difference is 8. AP, with a = 15 and d = 8.
(ii) Let the air be x at the start. After each stroke 3/4 of the remaining air is left: x, 3x/4, 9x/16, 27x/64, …. Differences: 3x/4 − x = −x/4 and 9x/16 − 3x/4 = −3x/16. These are unequal. Not an AP.
(iii) Costs: 150, 200, 250, 300, …. Each difference is 50. AP, with a = 150 and d = 50.
(iv) Amounts: 10000(1.08), 10000(1.08)2, 10000(1.08)3, …, that is 10800, 11664, 12597.12, …. Differences 864 and 933.12 are unequal. Not an AP.
Example 2: Which term is 78? (NCERT Exercise 5.2)
Which term of the AP : 3, 8, 13, 18, … ,is 78?
Solution: a = 3, d = 8 − 3 = 5, an = 78.
78 = 3 + (n − 1) × 5
75 = (n − 1) × 5
n − 1 = 15, so n = 16.
78 is the 16th term.
Example 3: Is −150 a term? (NCERT Exercise 5.2)
Check whether −150 is a term of the AP : 11, 8, 5, 2 …
Solution: a = 11, d = 8 − 11 = −3. Suppose −150 is the nth term.
−150 = 11 + (n − 1)(−3)
−161 = −3(n − 1)
n − 1 = 161/3, so n = 164/3.
n is not a positive integer. So −150 is not a term of the AP.
Example 4: The 31st term from two given terms (NCERT Exercise 5.2)
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
a11 = a + 10d = 38 (1)
a16 = a + 15d = 73 (2)
(2) − (1): 5d = 35, so d = 7. From (1): a = 38 − 70 = −32.
a31 = a + 30d = −32 + 210 = 178.
Example 5: Three-digit numbers divisible by 7 (NCERT Exercise 5.2)
How many three-digit numbers are divisible by 7?
Solution: The first three-digit multiple of 7 is 105 (7 × 15) and the last is 994 (7 × 142). The list 105, 112, 119, …, 994 is an AP with a = 105, d = 7, l = 994.
994 = 105 + (n − 1) × 7
889 = 7(n − 1)
n − 1 = 127, so n = 128.
There are 128 three-digit numbers divisible by 7.
Example 6: 20th term from the last (NCERT Exercise 5.2)
Find the 20th term from the last term of the AP : 3, 8, 13, …, 253.
Solution: Reverse the AP: 253, 248, 243, …, 3. Now the first term is 253 and d = −5.
20th term = 253 + (20 − 1)(−5) = 253 − 95 = 158.
Example 7: First three terms from two sums (NCERT Exercise 5.2)
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
a4 + a8 = (a + 3d) + (a + 7d) = 2a + 10d = 24, so a + 5d = 12 (1)
a6 + a10 = (a + 5d) + (a + 9d) = 2a + 14d = 44, so a + 7d = 22 (2)
(2) − (1): 2d = 10, so d = 5. From (1): a = 12 − 25 = −13.
First three terms: −13, −8, −3.
Example 8: Subba Rao’s salary (NCERT Exercise 5.2)
Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
Solution: Salaries in 1995, 1996, 1997, …: 5000, 5200, 5400, …, an AP with a = 5000 and d = 200.
7000 = 5000 + (n − 1) × 200
2000 = 200(n − 1)
n − 1 = 10, so n = 11.
The 11th year counted from 1995 is 1995 + 10 = 2005. His income reached ₹7000 in 2005.
Example 9: How many terms give 636? (NCERT Exercise 5.3)
How many terms of the AP : 9, 17, 25, … must be taken to give a sum of 636?
Solution: a = 9, d = 8, Sn = 636.
636 = n/2 [18 + (n − 1) × 8] = n/2 (8n + 10) = n(4n + 5)
4n2 + 5n − 636 = 0
Discriminant = 25 + 4 × 4 × 636 = 25 + 10176 = 10201 = 1012.
n = (−5 ± 101)/8 gives n = 12 or n = −106/8, which is rejected. 12 terms must be taken.
Example 10: Sum of first n terms from two sums (NCERT Exercise 5.3)
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
S7 = 7/2 (2a + 6d) = 7(a + 3d) = 49, so a + 3d = 7 (1)
S17 = 17/2 (2a + 16d) = 17(a + 8d) = 289, so a + 8d = 17 (2)
(2) − (1): 5d = 10, so d = 2 and a = 7 − 6 = 1.
Sn = n/2 [2 × 1 + (n − 1) × 2] = n/2 × 2n = n2.
Example 11: Terms from a given Sn (NCERT Exercise 5.3)
If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
- S1 = 4 − 1 = 3, so the first term is 3.
- S2 = 8 − 4 = 4 (sum of first two terms).
- a2 = S2 − S1 = 4 − 3 = 1.
- S3 = 12 − 9 = 3, so a3 = S3 − S2 = 3 − 4 = −1.
- S10 = 40 − 100 = −60 and S9 = 36 − 81 = −45, so a10 = −60 − (−45) = −15.
- an = Sn − Sn−1 = (4n − n2) − [4(n − 1) − (n − 1)2] = 4n − n2 − 4n + 4 + n2 − 2n + 1 = 5 − 2n.
Example 12: Seven cash prizes (NCERT Exercise 5.3)
A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
Solution: Let the first prize be ₹a. Then d = −20, n = 7, S7 = 700.
700 = 7/2 [2a + 6(−20)] = 7(a − 60)
a − 60 = 100, so a = 160.
The prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40. Check: 160 + 140 + 120 + 100 + 80 + 60 + 40 = 700.
Example 13: Stack of 200 logs (NCERT Exercise 5.3)
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?
Solution: Logs per row from the bottom: 20, 19, 18, …, an AP with a = 20, d = −1. Let there be n rows, so Sn = 200.
200 = n/2 [40 + (n − 1)(−1)] = n(41 − n)/2
400 = 41n − n2
n2 − 41n + 400 = 0
(n − 16)(n − 25) = 0, so n = 16 or n = 25.
n = 25 would make the top row a25 = 20 − 24 = −4 logs, which is impossible, so it is rejected. For n = 16, the top row has a16 = 20 − 15 = 5 logs.
The logs are placed in 16 rows, with 5 logs in the top row.
Competency-Based Questions (with answers)
1. Case-based: The school auditorium
A school auditorium has 20 seats in the first row, 22 in the second row, 24 in the third row, and so on, with each row having 2 more seats than the row in front of it. There are 30 rows.
(i) How many seats are in the 15th row? (ii) Which row has 50 seats? (iii) How many seats are there in the auditorium?
Answer: The seats form an AP with a = 20, d = 2, n = 30.
(i) a15 = 20 + 14 × 2 = 48 seats.
(ii) 50 = 20 + (n − 1) × 2 gives n − 1 = 15, so the 16th row.
(iii) Last row: a30 = 20 + 29 × 2 = 78. S30 = 30/2 (20 + 78) = 15 × 98 = 1470 seats.
2. Case-based: The ladder
The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top. The bottom rung is 45 cm long and the ladder has 8 rungs.
(i) Write the lengths as an AP and state a and d. (ii) Find the length of the top rung using the nth term formula. (iii) Find the total length of wood needed for the rungs.
Answer: (i) 45, 43, 41, 39, 37, 35, 33, 31 with a = 45 and d = −2.
(ii) a8 = 45 + 7(−2) = 31 cm.
(iii) S8 = 8/2 (45 + 31) = 4 × 76 = 304 cm.
3. Source-based: Gauss and the sum 1 to 100
Read the source: “He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050.” Gauss wrote the sum S = 1 + 2 + 3 + … + 99 + 100, then wrote it again in reverse order, S = 100 + 99 + … + 3 + 2 + 1, and added the two lines.
(i) What does each pair add up to when the two lines are added? (ii) Use the same method to find 1 + 2 + … + 50. (iii) Hence find 51 + 52 + … + 100.
Answer: (i) Each pair adds to 101, and there are 100 pairs, so 2S = 10100 and S = 5050.
(ii) Pairs add to 51, and there are 50 of them: 2S = 50 × 51, so S = 1275.
(iii) 5050 − 1275 = 3775.
4. Assertion-Reason
Assertion (A): 301 is not a term of the list 5, 11, 17, 23, ….
Reason (R): The position n of a term in an AP must be a positive integer.
Choose: (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.
Answer: (a) 301 = 5 + (n − 1) × 6 gives n = 151/3, which is not a positive integer. R is exactly why A holds.
5. Assertion-Reason
Assertion (A): The list 1, 1, 2, 3, 5, 8 is an arithmetic progression.
Reason (R): In an AP, ak+1 − ak is the same for all values of k.
Choose: (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.
Answer: (d) The differences are 0, 1, 1, 2, 3, which are not equal, so A is false. R is the correct test for an AP.
6. Assertion-Reason
Assertion (A): If the sum of the first n terms of an AP is Sn = 3n2 + 2n, then its second term is 11.
Reason (R): The nth term of an AP is an = Sn − Sn−1.
Choose: (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.
Answer: (a) S2 = 12 + 4 = 16 and S1 = 3 + 2 = 5, so a2 = 16 − 5 = 11. A follows directly from R.
7. Assertion-Reason
Assertion (A): 0 is a term of the AP 21, 18, 15, ….
Reason (R): Every AP with a negative common difference has 0 as one of its terms.
Choose: (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.
Answer: (c) 21 + (n − 1)(−3) = 0 gives n = 8, so A is true. R is false: in 10, 7, 4, 1, −2, …, solving 10 − 3(n − 1) = 0 gives n = 13/3, so 0 is not a term.
8. Error analysis: The 11th term from the end
For the AP 10, 7, 4, …, −62, a student found 25 terms and wrote: “The 11th term from the end is the 25 − 11 = 14th term, so it is 10 + 13(−3) = −29.” Find the error and give the correct answer.
Answer: Counting back from the 25th term, the 1st term from the end is the 25th, the 2nd is the 24th, and so on. The kth term from the end is the (n − k + 1)th term. The 11th from the end is the 25 − 11 + 1 = 15th term: 10 + 14(−3) = −32. The student forgot the “+ 1”.
Important Questions for Board Exams
1-Mark Questions
- 30th term of the AP: 10, 7, 4, …, is (A) 97 (B) 77 (C) −77 (D) −87
a = 10, d = −3. a30 = 10 + 29(−3) = 10 − 87 = −77. (C) - Find the common difference of the AP 1/p, (1 − p)/p, (1 − 2p)/p, …
d = (1 − p)/p − 1/p = −p/p = −1. - If 2k + 1, 13, 5k − 3 are three consecutive terms of an AP, find k.
2 × 13 = (2k + 1) + (5k − 3), so 26 = 7k − 2 and k = 4. - Find the sum of the first 20 positive integers.
n(n + 1)/2 = 20 × 21/2 = 210.
2-Mark Questions
- The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
a17 − a10 = (a + 16d) − (a + 9d) = 7d = 7, so d = 1. - How many multiples of 4 lie between 10 and 250?
First multiple after 10 is 12; last before 250 is 248. 248 = 12 + (n − 1) × 4 gives n − 1 = 59, so n = 60. - For what value of n, are the nth terms of two APs: 63, 65, 67, … and 3, 10, 17, … equal?
63 + (n − 1) × 2 = 3 + (n − 1) × 7 gives 60 = 5(n − 1), so n − 1 = 12 and n = 13. (Both 13th terms equal 87.) - Find the sum of the odd numbers between 0 and 50.
1, 3, 5, …, 49: 49 = 1 + (n − 1) × 2 gives n = 25. S = 25/2 (1 + 49) = 625.
3-Mark Questions
- If the 3rd and the 9th terms of an AP are 4 and −8 respectively, which term of this AP is zero?
a + 2d = 4 and a + 8d = −8. Subtracting, 6d = −12, so d = −2 and a = 8. 0 = 8 + (n − 1)(−2) gives n − 1 = 4, so the 5th term is zero. - The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
350 = 17 + (n − 1) × 9 gives n − 1 = 37, so n = 38. S = 38/2 (17 + 350) = 19 × 367 = 6973. - Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
d = 18 − 14 = 4 and a = 14 − 4 = 10. S51 = 51/2 [20 + 50 × 4] = 51/2 × 220 = 51 × 110 = 5610. - A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
a = 200, d = 50, n = 30. S30 = 15 [400 + 29 × 50] = 15 × 1850 = ₹27750.
5-Mark Questions
- A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
The length of a semicircle of radius r is πr. So the lengths are π(0.5), π(1.0), π(1.5), …, π(6.5), an AP with first term 0.5π, d = 0.5π, n = 13.
Total length = 13/2 [0.5π + 6.5π] = 13/2 × 7π = 45.5π = 45.5 × 22/7 = 143 cm. - In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
The potatoes are 5, 8, 11, … m from the bucket. For each potato she runs there and back, so the distances are 2 × 5, 2 × 8, 2 × 11, …, that is 10, 16, 22, …, an AP with a = 10, d = 6, n = 10.
S10 = 10/2 [20 + 9 × 6] = 5 × 74 = 370 m. - In an AP: (i) given a12 = 37, d = 3, find a and S12. (ii) given a3 = 15, S10 = 125, find d and a10.
(i) 37 = a + 11 × 3, so a = 4. S12 = 12/2 (4 + 37) = 6 × 41 = 246.
(ii) a + 2d = 15, so 2a + 4d = 30. S10 = 5(2a + 9d) = 125, so 2a + 9d = 25. Subtracting, 5d = −5, so d = −1 and a = 17. a10 = 17 + 9(−1) = 8.
Common Mistakes and Examiner Tips
- Subtracting in the wrong order to find d. In 6, 3, 0, −3, …, d = 3 − 6 = −3. Always take the later term minus the earlier term.
- Checking only one difference. 1, 1, 1, 2, 2, 2, … has a2 − a1 = a3 − a2 = 0, yet a4 − a3 = 1. Check every consecutive difference you can form from the given terms before calling a list an AP, and show them.
- Writing a + nd instead of a + (n − 1)d. The first term has no d added. The 10th term of 2, 7, 12, … is 2 + 9 × 5 = 47, not 2 + 10 × 5 = 52.
- Accepting a fractional or negative n. If n comes out as 151/3 or −2, the number is not a term of the AP. Say so in words.
- Off-by-one error for terms from the end. The kth term from the end of an n-term AP is the (n − k + 1)th term. The safer route is to reverse the AP: first term l, common difference −d.
- Wrong first or last term when counting multiples. For “multiples of 4 between 10 and 250”, start at 12 and stop at 248. For “three-digit numbers divisible by 7”, start at 105 and stop at 994. Divide the end points by the number to find them.
- Confusing an and Sn. “How much is saved in the 12th week” asks for a12. “How much is saved in 12 weeks” asks for S12. Underline the words in the question.
- Treating a given Sn expression as an. If Sn = 4n − n2, the second term is S2 − S1 = 1, not 4(2) − 22 = 4.
- Keeping both roots of the quadratic in n without checking. In the log stack, n = 25 gives a top row of −4 logs, so it is rejected. For 24, 21, 18, … with sum 78 (the textbook’s Example 13), both n = 4 and n = 13 are valid. Test each root against the situation and give a reason.
- Stopping at a number in word problems. Finish with the unit and a sentence, such as “His income reached ₹7000 in 2005”, and convert the term number back to a year or row when asked.
Quick Revision Points
- An AP is a list in which each term after the first is the previous term plus a fixed number d.
- d = ak+1 − ak, the later term minus the earlier term.
- d can be positive (increasing AP), negative (decreasing AP) or zero (constant AP).
- General form: a, a + d, a + 2d, a + 3d, ….
- A list is an AP only if all consecutive differences are equal.
- Lists made by multiplying by a fixed number, such as 2, 4, 8, 16, …, are not APs.
- nth term: an = a + (n − 1)d.
- If solving for n gives a non-integer or a negative value, the number is not a term.
- am − an = (m − n)d.
- kth term from the end = l − (k − 1)d = the (n − k + 1)th term.
- Sn = n/2 [2a + (n − 1)d] = n/2 (a + l).
- 1 + 2 + 3 + … + n = n(n + 1)/2; for n = 100 the sum is 5050.
- an = Sn − Sn−1, and a1 = S1.
- If an is linear in n, like 3 + 2n, the list is an AP whose d is the coefficient of n.
- Finding n from a sum gives a quadratic; keep only positive integer roots that fit the situation.
- If a, b, c are in AP, 2b = a + c and b is the arithmetic mean of a and c.
Weightage in Board Exams
Arithmetic Progressions is Chapter 5 of the Class 10 Maths textbook. Check the current CBSE course structure for the marks given to the unit it belongs to; the chapter itself is assessed in a predictable way.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Finding d, a given term, whether a list is an AP, the value of k for three terms in AP, an from Sn |
| Short answers | Which term is a number, is a number a term, counting multiples in a range, term from the end, sum of a simple AP |
| Long answers | Finding a and d from two conditions, then a term or a sum; finding n from a sum by solving a quadratic; word problems on salaries, savings, prizes, penalties and stacks |
| Case-based | A daily-life pattern (seats, savings, rungs) split into parts on an, n and Sn |
The textbook marks Exercise 5.4 as optional and states that its questions are not from the examination point of view. For every word problem, write the AP, state a, d and n, pick the formula, and finish with a sentence answer.
Class 10 Maths · Chapter 5 – swipe through all 10 cards to understand the whole chapter.
What is an AP?
A list in which each term, except the first, is the previous term plus a fixed number.
You need both the first term a and the common difference d to write an AP.
- Salary 8000, 8500, 9000, … is an AP with d = 500
- A finite AP has a last term l; an infinite AP does not
- 2, 4, 8, 16, … is not an AP: it multiplies by 2
Finding d
Subtract any term from the term that follows it.
In 6, 3, 0, −3, …, d = 3 − 6 = −3, even though the next term is smaller.
- Positive d: terms increase
- Negative d: terms decrease
- d = 0: all terms are equal, like 3, 3, 3, …
Is the list an AP?
Every consecutive difference must be the same.
One unequal difference is enough to say the list is not an AP.
- 4, 10, 16, 22: differences 6, 6, 6, so AP
- −2, 2, −2, 2: differences 4, −4, so not AP
- 1, 1, 2, 3, 5: differences 0, 1, 1, 2, so not AP
The general term
The nth term is the first term plus (n − 1) steps of d.
Links four quantities aₙ, a, n and d: know three, find the fourth.
- 10th term of 2, 7, 12, … is 2 + 9 × 5 = 47
- In 21, 18, 15, …, −81 is the 35th term
- n must be a positive integer, or the number is not a term
Finding a and d
Write each given term as a + (n − 1)d and solve the pair of equations.
a3 = 5 and a7 = 9 give 4d = 4, so d = 1, a = 3 and the AP 3, 4, 5, 6, …
- a11 = 38, a16 = 73 give d = 7, a = −32, a31 = 178
- If a17 exceeds a10 by 7, then 7d = 7 and d = 1
- Subtract the two equations first to get d quickly
How many terms?
Put the last term equal to a + (n − 1)d and solve for n.
kth term from the end = l − (k − 1)d, the (n − k + 1)th term.
- Two-digit multiples of 3: 12 to 99, so 30 terms
- Three-digit multiples of 7: 105 to 994, so 128 terms
- 11th from the end of 10, 7, …, −62 is the 15th term, −32
Sum of first n terms
Add the sum to itself written backwards: n equal pairs appear.
Gauss added 1 to 100 this way: 100 pairs of 101, halved, gives 5050.
- Use n/2 (a + l) when d is not given
- 1 + 2 + … + n = n(n + 1)/2
- Money box 100, 150, 200, … for 21 years totals ₹12600
aₙ = Sₙ − Sₙ₋1
The nth term is the sum of n terms minus the sum of (n − 1) terms.
If Sₙ = 4n − n2, then a1 = 3, a2 = 1 and aₙ = 5 − 2n.
- Find S1 first: it is the first term
- Subtract consecutive sums to get each term
- Never read the Sₙ expression as the nth term
When the sum is given
Putting Sₙ equal to the given sum gives a quadratic in n.
Keep only positive integer roots that make sense in the situation.
- 9, 17, 25, … needs 12 terms to reach 636
- 24, 21, 18, … sums to 78 for both n = 4 and n = 13
- 200 logs from 20, 19, 18, …: 16 rows, 5 logs on top
The middle term
If a, b, c are in AP, the middle term is the average of the other two.
Same as 2b = a + c, the quickest route to an unknown k.
- Arithmetic mean of 7 and 15 is 11
- 2k + 1, 13, 5k − 3 in AP gives k = 4
- a, b, c are in AP exactly when b − a = c − b
📝 Practice Arithmetic Progressions - 10 board questions
CBSE previous-year and competency-based · with answers & explanations
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Why A: Step 1 (d = a₂ – a₁): d = 2√2 – √2 = √2. Step 2 (check the next pair): 3√2 – 2√2 = √2, the same, so d = √2.
Why not B: 1 is the jump in the whole-number coefficients 1, 2, 3, 4, forgetting that each is multiplied by √2.
Why not C: 2√2 is the second term itself, not the gap between terms.
Why not D: -√2 is a₁ – a₂, the subtraction done the wrong way round; the terms are increasing, so d is positive.
Remember: d = later term minus earlier term; √2 rides along like a unit.
Why A: Step 1 (write terms from aₙ): a₁ = √2 + 1, a₂ = 2√2 + 1. Step 2 (d = a₂ – a₁): (2√2 + 1) – (√2 + 1) = √2, the coefficient of n.
Why not B: √2 n depends on n, but d must be the same fixed number for every pair.
Why not C: 1 is the constant term; it cancels when two terms are subtracted.
Why not D: √2 + 1 is the first term a₁, not the gap between terms.
Remember: If aₙ = pn + q, then d = p, the coefficient of n.
Why C: Step 1 (d = a₂ – a₁): d = -10/4 – (-15/4) = 5/4. Step 2 (aₘ – aₙ = (m – n)d): a₁₆ – a₁₂ = 4 × 5/4 = 5.
Why not A: 4 is the number of steps, m – n, without multiplying by d.
Why not B: 5/4 is d itself, one step instead of four.
Why not D: 25/4 = 5 × 5/4 uses 5 steps; from the 12th to the 16th term there are only 4.
Remember: aₘ – aₙ = (m – n)d; you never need a itself.
Why B: Step 1 (put m = 1 and m = 2 in Sₘ): S₁ = 2 + 3 = 5 and S₂ = 8 + 6 = 14. Step 2 (aₙ = Sₙ – Sₙ₋₁): a₂ = S₂ – S₁ = 14 – 5 = 9. Check: a₁ = 5, d = 4, so the AP is 5, 9, 13, …
Why not A: 10 would need S₂ = 15, but S₂ = 2(2²) + 3(2) = 14.
Why not C: 12 would need S₁ = 2, but S₁ = 2(1²) + 3(1) = 5.
Why not D: 4 is the common difference d = a₂ – a₁, not the second term.
Remember: aₙ = Sₙ – Sₙ₋₁; the sum formula is not the term formula.
Why D: Using R, d = a₂ – a₁ = 1 – 5 = -4 (and -3 – 1 = -4). So d is -4, not 4, and A is false. R is the correct definition of d, so R is true.
Why not A: A is false, so ‘both true’ cannot be right.
Why not B: A is false, so neither ‘both true’ option can be right.
Why not C: This swaps the truth values; the reason is the true statement and the assertion is the false one.
Remember: A decreasing AP always has a negative d.
Why A: Step 1 (d = a₂ – a₁): d = 19/4 – 20/4 = -1/4. Step 2 (aₙ = a + (n – 1)d): a₁₀ = 5 + 9 × (-1/4) = 20/4 – 9/4 = 11/4.
Why not B: 4/11 is the reciprocal of the correct value.
Why not C: 13/4 = 5 – 7/4 comes from using 7 steps of d instead of 9.
Why not D: 4/13 inverts a wrongly counted term.
Remember: The 10th term is 9 steps after the first, so use (n – 1)d.
Why A: Step 1 (aₙ = a + (n – 1)d): a₁₅ = a + 14d and a₁₁ = a + 10d. Step 2 (subtract, a cancels): a₁₅ – a₁₁ = 4d = 48, so d = 12.
Why not B: 16 divides 48 by 3, miscounting the steps from the 11th to the 15th term.
Why not C: -12 has the wrong sign; a later term minus an earlier one is positive here, so d is positive.
Why not D: -16 has both the wrong sign and the wrong step count.
Remember: aₘ – aₙ = (m – n)d, so a₁₅ – a₁₁ = 4d.
Why B: Step 1 (symmetric terms): let the numbers be a – d, a, a + d. Step 2 (add, d cancels): 3a = 30, so the middle term a = 10.
Why not A: 4 does not satisfy 3a = 30; it would need a sum of 12.
Why not C: 16 would make the other two numbers add to 14, which is not twice 16 as the AP requires.
Why not D: 8 would need a sum of 24, not 30.
Remember: In an AP of three (or any odd number of) terms, the middle term is the average: sum ÷ number of terms.
Why B: Step 1 (d = a₂ – a₁, same denominator): d = (1 – 4x)/(2x) – 1/(2x) = (1 – 4x – 1)/(2x) = -4x/(2x). Step 2 (cancel x): -4x/(2x) = -2. Check: (1 – 8x)/(2x) – (1 – 4x)/(2x) = -4x/(2x) = -2 as well.
Why not A: -2x keeps the x that should cancel between -4x and 2x.
Why not C: 2 is a₁ – a₂, the subtraction done the wrong way round.
Why not D: 2x has both the wrong sign and the uncancelled x.
Remember: With a common denominator, subtract the numerators and simplify fully before choosing.
Why B: Step 1 (d = a₂ – a₁): d = -26 – (-29) = 3. Step 2 (aₙ = a + (n – 1)d): -29 + 3(n – 1) = 16, so 3(n – 1) = 45, n – 1 = 15 and n = 16.
Why not A: The 11th term is -29 + 10 × 3 = 1, not 16.
Why not C: 10th gives -29 + 27 = -2, not 16.
Why not D: 31st is the position of the last term 61, not of 16.
Remember: Set aₙ equal to the given value and solve for n; n must come out a positive whole number.
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- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Quadratic Equations Class 10 Notes
- Triangles Class 10 Notes
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Frequently Asked Questions
An arithmetic progression (AP) is a list of numbers in which each term, except the first, is obtained by adding a fixed number to the term before it. That fixed number is the common difference d. For example, 8000, 8500, 9000, … is an AP with d = 500. The common difference can be positive, negative or zero.
Find the differences a₂ − a₁, a₃ − a₂, a₄ − a₃ and so on. If all of them are equal, the list is an AP and that value is d. If even one difference is different, it is not an AP. For example, 1, 1, 2, 3, 5 has differences 0, 1, 1, 2, so it is not an AP.
The nth term, also called the general term, is aₙ = a + (n − 1)d, where a is the first term and d the common difference. The first term has no d added, the second has one d, and so on. For 2, 7, 12, …, the 10th term is 2 + 9 × 5 = 47.
Put the number equal to a + (n − 1)d and solve for n. If n comes out as a positive integer, the number is that term. If n is a fraction or negative, the number is not a term. For 5, 11, 17, …, putting 301 gives n = 151/3, so 301 is not a term.
Sₙ = n/2 [2a + (n − 1)d] uses the first term, the common difference and the number of terms. Sₙ = n/2 (a + l) uses the first and last terms and is handy when d is not given. Both come from writing the sum forwards and backwards and adding, which gives n pairs of equal value.
Use aₙ = Sₙ − Sₙ₋₁, and a₁ = S₁. If Sₙ = 4n − n², then S₁ = 3 and S₂ = 4, so the first term is 3 and the second term is 4 − 3 = 1. Working in general gives aₙ = 5 − 2n. Do not substitute n into the Sₙ expression and call it the nth term.
Finding n from a given sum gives a quadratic equation, which can have two roots. If a is positive and d is negative, some terms are positive and some negative, and they can cancel. For 24, 21, 18, …, both 4 terms and 13 terms add up to 78. In a word problem, reject a root that makes a quantity impossible, like a negative number of logs.
If a, b, c are in AP, then b − a = c − b, so b = (a + c)/2. This b is called the arithmetic mean of a and c. It is useful for finding an unknown: if 2k + 1, 13, 5k − 3 are in AP, then 26 = 7k − 2, so k = 4.
