Curriculum
Mathematics Grade 10
Grade 10 Mathematics
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Chapter 1. Real Numbers
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Chapter 2. Polynomials
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Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
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Chapter 4: QUADRATIC EQUATIONS
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Chapter 5: ARITHMETIC PROGRESSIONS
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Chapter 6: TRIANGLES
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Chapter 7: COORDINATE GEOMETRY
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Chapter 8: INTRODUCTION TO TRIGONOMETRY
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Chapter 9: SOME APPLICATIONS OF TRIGONOMETRY
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Chapter 10: CIRCLES
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Chapter 11: AREAS RELATED TO CIRCLES
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Chapter 12: SURFACE AREAS AND VOLUMES
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Chapter 13: STATISTICS
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Chapter 14: PROBABILITY
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Interactive Worksheets
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Chapter Wise Worksheets
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Solved Problems and solutions
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Chapter 5: ARITHMETIC PROGRESSIONS
Grade 10 · Mathematics · Chapter 5
Arithmetic Progressions
Understand patterns in sequences — find any term, compute any sum, and solve real-life problems using the power of APs.
| What Is an AP? | nth Term Formula | Sum of n Terms | Word Problems |
What You Will Learn
| ✦ Patterns in nature and daily life | ✦ Definition of an Arithmetic Progression |
| ✦ Common difference — positive, negative, zero | ✦ General form: a, a+d, a+2d, a+3d, … |
| ✦ nth term formula: aₙ = a + (n−1)d | ✦ Worked Examples 1–16 (fully solved) |
| ✦ Sum of first n terms: Sₙ = n/2 [2a + (n−1)d] | ✦ Real-life applications and word problems |
| ✦ Practice Sets A–D with answers | ✦ Chapter summary and exam quick-check |
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Section 1 Patterns Around Us — Introduction |
In nature, many things follow a certain pattern — the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple. We also see patterns in daily life:
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Reena’s Salary: Starting ₹8000/month with ₹500 annual increment. Salaries for years 1, 2, 3, …: 8000, 8500, 9000, … Each term is ₹500 more than the previous. |
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ii
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Ladder Rungs: Decrease by 2 cm from bottom to top. Lengths (cm): 45, 43, 41, 39, 37, 35, 33, 31 |
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iii
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Shakila’s Savings: ₹100 on daughter’s 1st birthday, ₹50 more each year: 100, 150, 200, 250, … |
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iv
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Savings Scheme (NOT an AP): Amount becomes 5/4 times every 3 years: 10000, 12500, 15625, 19531.25 — each term is multiplied, not added by a fixed number. |
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v
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Fibonacci (NOT an AP): Rabbit pairs: 1, 1, 2, 3, 5, 8 — each term is the sum of the two before it, not a fixed addition. |
Key Observation: In examples (i), (ii), (iii), each new term is obtained by adding a fixed number to the previous term. This is the defining property of an Arithmetic Progression.
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Section 2 Arithmetic Progressions — Definition |
Definition
An Arithmetic Progression (AP) is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term. This fixed number is the common difference (d) — it can be positive, negative, or zero.
Examples of APs:
| Terms | Common Difference d | Notes |
|---|---|---|
| 1, 2, 3, 4, … | d = +1 | Infinite AP |
| 100, 70, 40, 10, … | d = −30 | Decreasing AP |
| −3, −2, −1, 0, … | d = +1 | Negative terms |
| 3, 3, 3, 3, … | d = 0 | Constant AP |
| −1.0, −1.5, −2.0, −2.5, … | d = −0.5 | Decimal AP |
General Form of an AP
a, a + d, a + 2d, a + 3d, …
where a = first term and d = common difference
WORKED EXAMPLE 1 — Find a and d
For the AP: 3/2, 1/2, −1/2, −3/2, …, write the first term a and common difference d.
Common difference: d = 1/2 − 3/2 = −2/2 = d = −1
We can use any two consecutive terms to find d, once we know the list is an AP.
WORKED EXAMPLE 2 — Identify APs
Which of these form an AP? If they do, write the next two terms.
Differences: 6, 6, 6 — constant. Yes, AP with d = 6. Next: 28, 34
(ii) 1, −1, −3, −5, …
Differences: −2, −2, −2 — constant. Yes, AP with d = −2. Next: −7, −9
(iii) −2, 2, −2, 2, −2, …
Differences: +4, −4 — not equal. Not an AP. ✗
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, …
a₂−a₁=0, a₃−a₂=0, a₄−a₃=1 — not equal. Not an AP. ✗
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Section 3 nth Term of an AP |
Instead of adding d step by step, we can jump directly to any term using one formula.
How the Formula is Derived
| a₁ = a + (1−1)d = a |
| a₂ = a + (2−1)d = a + d |
| a₃ = a + (3−1)d = a + 2d |
| a₄ = a + (4−1)d = a + 3d |
| aₙ = a + (n−1)d ← General Formula |
nth Term Formula
aₙ = a + (n − 1) × d
a = first term d = common difference n = position of term
WORKED EXAMPLE 3 — Find a specific term
Find the 10th term of the AP: 2, 7, 12, …
aₙ = a + (n−1)d → a₁₀ = 2 + (10−1)×5 = 2 + 45 = 47
WORKED EXAMPLE 4 — Which term equals a value?
Which term of the AP: 21, 18, 15, … is −81? Is any term 0?
−81 = 21 + (n−1)(−3) → −81 = 24 − 3n → 3n = 105 → n = 35
For aₙ = 0: 0 = 21 + (n−1)(−3) → 3(n−1) = 21 → n = 8. The 8th term is 0.
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35th term = −81
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8th term = 0
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WORKED EXAMPLE 5 — Build AP from two terms
The 3rd term of an AP is 5 and the 7th term is 9. Find the AP.
Subtract (1) from (2): 4d = 4 → d = 1. From (1): a = 3
AP: 3, 4, 5, 6, 7, …
WORKED EXAMPLE 6 — Is a number in the AP?
Check whether 301 is a term of the AP: 5, 11, 17, 23, …
n is not a whole number. 301 is NOT a term. ✗
WORKED EXAMPLE 7 — Count terms meeting a condition
How many two-digit numbers are divisible by 3?
99 = 12 + (n−1)×3 → 87 = 3(n−1) → n−1 = 29 → n = 30
WORKED EXAMPLE 8 — Term counting from the last
Find the 11th term from the last of the AP: 10, 7, 4, …, −62.
Total terms: −62 = 10 + (n−1)(−3) → n = 25
11th from last = (25−11+1)th = 15th term
a₁₅ = 10 + 14×(−3) = 10 − 42 = −32
Alternate method: Reverse AP (a = −62, d = +3) → a₁₁ = −62 + 10×3 = −32 ✓
WORKED EXAMPLE 9 — Simple Interest as AP
₹1000 at 8% per year simple interest. Find the interest at the end of 30 years.
a₃₀ = 80 + 29×80 = 80 + 2320 = ₹2400
WORKED EXAMPLE 10 — Rose Garden Rows
Rose plants per row: 23, 21, 19, …, 5. How many rows?
5 = 23 + (n−1)(−2) → −18 = −2(n−1) → n = 10 rows
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Section 4 Sum of First n Terms of an AP |
Adding terms one by one is slow. Gauss, at age 10, solved 1+2+…+100 instantly by pairing first and last terms. The same trick works for any AP.
Derivation (Gauss Method)
Write forward: S = a + (a+d) + … + [a+(n−1)d] …(1)
Write backward: S = [a+(n−1)d] + … + (a+d) + a …(2)
Add (1)+(2): 2S = n × [2a+(n−1)d]
S = n/2 × [2a + (n−1)d]
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Formula 1 — d is known Sₙ = n/2 × [2a + (n−1)d] |
Formula 2 — last term l is known Sₙ = n/2 × (a + l) |
WORKED EXAMPLE 11 — Sum of AP
Find the sum of first 22 terms of the AP: 8, 3, −2, …
S = 22/2 × [2(8) + 21(−5)] = 11 × [16 − 105] = 11 × (−89) = −979
WORKED EXAMPLE 12 — Find 20th term from sum
S₁₄ = 1050, first term = 10. Find the 20th term.
910 = 91d → d = 10
a₂₀ = 10 + 19×10 = 200
WORKED EXAMPLE 13 — How many terms for a given sum?
How many terms of AP: 24, 21, 18, … must be taken so their sum is 78?
78 = n/2 [48 + (n−1)(−3)] = n/2 [51 − 3n]
156 = 51n − 3n² → 3n² − 51n + 156 = 0 → n² − 17n + 52 = 0
(n−4)(n−13) = 0 → n = 4 or n = 13 (both valid)
Both answers are correct because terms 5 to 13 sum to zero (d < 0, so some terms are negative).
WORKED EXAMPLE 14 — Classic sums
Find (i) sum of first 1000 positive integers (ii) sum of first n positive integers
(ii) Sₙ = n/2 × (1 + n) = n(n+1)/2
WORKED EXAMPLE 15 — Sum from nth term formula
Find the sum of first 24 terms where aₙ = 3 + 2n.
S₂₄ = 24/2 × [2(5) + 23(2)] = 12 × [10+46] = 12 × 56 = 672
WORKED EXAMPLE 16 — TV Set Manufacturing
Year 3 output: 600 sets. Year 7: 700 sets. Find (i) Year 1, (ii) Year 10, (iii) 7-year total.
Subtract: 4d = 100 → d = 25. From (1): a = 550
(i) Year 1: 550 sets
(ii) a₁₀ = 550 + 9×25 = 775 sets
(iii) S₇ = 7/2 [2(550) + 6(25)] = 7/2 [1250] = 4375 sets
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Year 1 = 550
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Year 10 = 775
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7-Year Total = 4375
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Section 5 Practice Exercises with Answers |
Set A — Identify and Build APs
State if it is an AP. If yes, find a, d and the next two terms.
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(1) 4, 10, 16, 22, …
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(2) 2, 4, 8, 16, …
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(3) −1.2, −3.2, −5.2, −7.2, …
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(4) 0.2, 0.22, 0.222, 0.2222, …
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(5) 0, −4, −8, −12, …
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(6) a, 2a, 3a, 4a, …
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Reveal Answers ▼
(4) Not AP (differences: 0.02, 0.002, …) | (5) AP, d=−4 → next: −16, −20 | (6) AP, d=a → next: 5a, 6a
Set B — Finding Specific Terms
Use aₙ = a + (n−1)d. Show all working.
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(1) Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
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(2) Which term of AP: 3, 8, 13, 18, … is 78?
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(3) Determine the AP whose 3rd term is 16 and 7th term exceeds 5th term by 12.
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(4) How many three-digit numbers are divisible by 7?
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(5) The 17th term of an AP exceeds the 10th term by 7. Find d.
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(6) Check whether −150 is a term of AP: 11, 8, 5, 2, …
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Reveal Answers ▼
(4) AP: 105 to 994; n = 128 | (5) d = 1 | (6) n = 54.33 — not whole; −150 is NOT a term
Set C — Sum of Terms
Use Sₙ = n/2 [2a + (n−1)d] or Sₙ = n/2 (a+l). Show all working.
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(1) Sum of first 10 terms of 2, 7, 12, …
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(2) Sum of first 12 terms of −37, −33, −29, …
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(3) Sum: 34 + 32 + 30 + … + 10
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(4) First term = 5, last term = 45, sum = 400. Find n and d.
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(5) Sum of first 40 positive integers divisible by 6.
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(6) Sum of odd numbers between 0 and 50.
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Reveal Answers ▼
(5) AP: 6, 12, …, 240; S = 4920 | (6) AP: 1, 3, …, 49; n=25; S = 625
Set D — Real-Life Word Problems
Identify the AP, write a and d, solve and verify.
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(1) Subba Rao joined in 1995 with annual salary ₹5000 and ₹200 increment each year. In which year did his income reach ₹7000?
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(2) A contractor pays ₹200 on day 1, ₹250 on day 2, ₹300 on day 3 (₹50 extra each day) for delay. He delayed by 30 days. Total penalty?
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(3) 200 logs stacked: 20 in bottom row, 19 next, 18 above, and so on. How many rows? How many logs in the top row?
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(4) In a school, each section of Class n plants n trees. There are 3 sections per class (Class I to XII). Total trees planted?
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(5) Ramkali saved ₹5 in week 1 and increased by ₹1.75 per week. In which week do her savings become ₹20.75?
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(6) ₹700 to be split into 7 prizes, each ₹20 less than the previous. Find the value of each prize.
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Reveal Answers ▼
(3) n²−39n+400=0 → 16 rows; top row has 20−15 = 5 logs | (4) 3(1+2+…+12) = 3×78 = 234 trees
(5) n = 10th week | (6) 7a−420=700 → a=160; Prizes: 160, 140, 120, 100, 80, 60, 40
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Section 6 Chapter Summary |
Everything on One Page
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Definition Each term = previous + fixed d. Form: a, a+d, a+2d, a+3d, … |
Common Difference d = aḱ₊₁ − aḱ Can be +ve, −ve or 0. Finite AP has a last term. |
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nth Term Formula aₙ = a + (n−1)d Last term = l. aₙ = Sₙ − Sₙ₋₁ |
Sum Formula 1 Sₙ = n/2 × [2a + (n−1)d] Use when a and d known. |
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Sum Formula 2 Sₙ = n/2 × (a + l) Use when first and last terms known. |
Arithmetic Mean If a, b, c in AP then b = (a+c)/2. b is the arithmetic mean of a and c. |
8-Point Quick-Check Before Your Exam
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1d must be constant throughout — check at least two consecutive differences.
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2You need both a and d to fully describe an AP — neither alone is enough.
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3If n from aₙ formula is not a positive integer, that value is NOT in the AP.
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4kth term from the last: reverse the AP (new a = l, new d = −d), then apply the formula.
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5Use Sₙ = n/2 (a+l) only when the last term is known; otherwise use the d-formula.
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6When Sₙ quadratic gives two values of n, both may be valid — always check each.
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7In word problems: state clearly what a and d represent physically before writing equations.
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8Use aₙ = Sₙ − Sₙ₋₁ whenever only the sum formula is given instead of individual terms.
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Grade 10 Maths Chapter 5 — Arithmetic Progressions
Topics covered: definition of AP, common difference, general form of AP, finite and infinite arithmetic progressions, nth term formula aₙ = a+(n−1)d, sum of first n terms Sₙ = n/2[2a+(n−1)d], sum using first and last term, arithmetic mean, worked examples 1–16, real-life word problems including salary, simple interest, logs, trees. CBSE Class 10 Maths board exam revision.