This MCQ module is based on: Sum of an Arithmetic Progression
TOPIC 24 OF 25
Sum of an Arithmetic Progression
🎓 Class 9
Mathematics
CBSE
Theory
Ch 8 — Predicting What Comes Next: Exploring Sequences and Progressions
⏱ ~40 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Sum of an Arithmetic Progression
Targeting Class 9 level in Sequences Series, with Intermediate difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
8.6 The Sum of an AP — A Beautiful Trick
Suppose we want \(1 + 2 + 3 + \cdots + 100\). Adding term by term takes a long time. There is a clever shortcut, attributed to a young Carl Friedrich Gauss, that works for any AP.
Story — Gauss at School
Legend has it that when a teacher asked his class to add 1 to 100, an 8-year-old Gauss wrote the answer in seconds: 5050. He had spotted that the first and last terms add to 101, the second and second-last to 101, and so on — fifty pairs of 101. So \(50 \times 101 = 5050\). The same trick gives the formula for the sum of any AP.
Pairing Method — Derivation
Let the AP have first term a, common difference d, last term \(l = a_n = a + (n-1)d\), and let \(S_n\) be the sum of the first n terms.
Write \(S_n\) forwards and then backwards:
\(S_n = a + (a+d) + (a+2d) + \cdots + (l-d) + l\)
\(S_n = l + (l-d) + (l-2d) + \cdots + (a+d) + a\)
Add the two equations term by term. Every pair adds to \(a + l\), and there are n such pairs:
\(2 S_n = n\,(a + l)\)
Sum Formulas — Two Equivalent Forms
\[ S_n = \frac{n}{2}\,(a + l) \quad \text{(when first and last are known)}\]
Substituting \(l = a + (n-1)d\):\[ S_n = \frac{n}{2}\,\big[\,2a + (n-1)d\,\big]\quad\text{(when } a, d, n\text{ are known)}\]
Fig. 8.3 — Pairing forwards and backwards: each column sums to (a + l).
Special Case — Sum of First n Natural Numbers
For 1, 2, 3, …, n we have \(a = 1,\; d = 1,\; l = n\). Then
\[ 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}. \]
8.7 Worked Examples — Sum Formula
Example 1 — Sum of natural numbers up to 100
Find \(1 + 2 + 3 + \cdots + 100\).
Solution
\(S_{100} = \dfrac{100(101)}{2} = 5050\).
Example 2 — Sum of first 30 terms
Find the sum of the first 30 terms of the AP: 4, 9, 14, 19, …
Solution
\(a = 4,\; d = 5,\; n = 30\).
\(S_{30} = \dfrac{30}{2}\,[2(4) + 29(5)] = 15\,[8 + 145] = 15 \times 153 = 2295\).
\(S_{30} = \dfrac{30}{2}\,[2(4) + 29(5)] = 15\,[8 + 145] = 15 \times 153 = 2295\).
Example 3 — Using first and last term
The first term of an AP is 7 and its 25th term is 79. Find the sum of the first 25 terms.
Solution
\(a = 7,\; l = 79,\; n = 25\). \(S_{25} = \dfrac{25}{2}(7 + 79) = \dfrac{25}{2} \times 86 = 25 \times 43 = 1075\).
Example 4 — Decreasing AP sum
Find \(40 + 37 + 34 + \cdots + 1\).
Solution
\(a = 40,\; d = -3,\; l = 1\). First find n: \(1 = 40 + (n-1)(-3) \Rightarrow -39 = -3(n-1) \Rightarrow n-1 = 13 \Rightarrow n = 14\).
\(S_{14} = \dfrac{14}{2}(40 + 1) = 7 \times 41 = 287\).
\(S_{14} = \dfrac{14}{2}(40 + 1) = 7 \times 41 = 287\).
Example 5 — Sum of even numbers
Find the sum of all even numbers from 2 to 200.
Solution
AP: 2, 4, 6, …, 200. \(a = 2,\; d = 2,\; l = 200\). n = (200 − 2)/2 + 1 = 100. \(S = \dfrac{100}{2}(2 + 200) = 50 \times 202 = 10{,}100\).
Example 6 — Sum given two terms
The 4th term of an AP is 9, and the 10th term is 21. Find the sum of the first 12 terms.
Solution
\(a + 3d = 9\) … (i); \(a + 9d = 21\) … (ii). Subtract: \(6d = 12 \Rightarrow d = 2\). Then \(a = 9 - 6 = 3\).
\(S_{12} = \dfrac{12}{2}\,[2(3) + 11(2)] = 6\,[6 + 22] = 6 \times 28 = 168\).
\(S_{12} = \dfrac{12}{2}\,[2(3) + 11(2)] = 6\,[6 + 22] = 6 \times 28 = 168\).
Example 7 — Number of terms from sum
How many terms of the AP 9, 17, 25, … must be added to get 636?
Solution
\(a = 9,\; d = 8,\; S_n = 636\).
\(\dfrac{n}{2}\,[2(9) + (n-1)(8)] = 636 \Rightarrow n[18 + 8n - 8] = 1272 \Rightarrow n[10 + 8n] = 1272\).
\(8n^2 + 10n - 1272 = 0 \Rightarrow 4n^2 + 5n - 636 = 0\). Solve: \(n = \dfrac{-5 \pm \sqrt{25 + 10176}}{8} = \dfrac{-5 \pm \sqrt{10201}}{8} = \dfrac{-5 \pm 101}{8}\).
n = 96/8 = 12 (rejecting negative). So we need 12 terms.
\(\dfrac{n}{2}\,[2(9) + (n-1)(8)] = 636 \Rightarrow n[18 + 8n - 8] = 1272 \Rightarrow n[10 + 8n] = 1272\).
\(8n^2 + 10n - 1272 = 0 \Rightarrow 4n^2 + 5n - 636 = 0\). Solve: \(n = \dfrac{-5 \pm \sqrt{25 + 10176}}{8} = \dfrac{-5 \pm \sqrt{10201}}{8} = \dfrac{-5 \pm 101}{8}\).
n = 96/8 = 12 (rejecting negative). So we need 12 terms.
8.8 Real-World Applications
Example 8 — Stadium total seats
Recall the stadium with 20, 24, 28, … seats per row for 30 rows. How many seats in total?
Solution
\(a = 20,\; d = 4,\; n = 30\). From Part 1 we found \(a_{30} = 136\).
\(S_{30} = \dfrac{30}{2}(20 + 136) = 15 \times 156 = 2340\) seats.
\(S_{30} = \dfrac{30}{2}(20 + 136) = 15 \times 156 = 2340\) seats.
Example 9 — Total savings
A boy saves ₹50 in the first week and increases his savings by ₹10 every week. How much will he have saved at the end of one year (52 weeks)?
Solution
\(a = 50,\; d = 10,\; n = 52\). \(S_{52} = \dfrac{52}{2}\,[2(50) + 51(10)] = 26\,[100 + 510] = 26 \times 610 = ₹15{,}860\).
Example 10 — Loan repayment
A loan is repaid in monthly installments forming an AP. The first installment is ₹6,000 and each subsequent installment increases by ₹200. If the loan is fully repaid in 24 months, what is the total amount repaid?
Solution
\(a = 6000,\; d = 200,\; n = 24\). \(S_{24} = \dfrac{24}{2}\,[2(6000) + 23(200)] = 12\,[12000 + 4600] = 12 \times 16600 = ₹1{,}99{,}200\).
Example 11 — Brick stacking
A pile of bricks has 25 bricks in the bottom row, 23 in the next, 21 in the next, and so on, with one brick at the top. Find the total number of bricks.
Solution
AP from 25 down to 1, d = −2. Number of terms: \(1 = 25 + (n-1)(-2) \Rightarrow n = 13\).
\(S_{13} = \dfrac{13}{2}(25 + 1) = \dfrac{13 \times 26}{2} = 169\) bricks.
\(S_{13} = \dfrac{13}{2}(25 + 1) = \dfrac{13 \times 26}{2} = 169\) bricks.
Fig. 8.4 — Brick stack: each higher row has 2 fewer bricks than the row below.
Example 12 — Telephone pole spacing
200 logs are stacked in a triangular pile: 20 in the bottom row, 19 in the next, 18 in the next, and so on. How many rows are there, and how many logs are in the top row?
Solution
a = 20, d = −1, S = 200. \(\dfrac{n}{2}\,[40 - (n-1)] = 200 \Rightarrow n(41 - n) = 400 \Rightarrow n^2 - 41n + 400 = 0\).
\(n = \dfrac{41 \pm \sqrt{1681 - 1600}}{2} = \dfrac{41 \pm 9}{2}\) → n = 25 or n = 16.
Check n = 25: top row would have \(20 - 24 = -4\) logs — impossible. So n = 16 rows; top row has \(20 - 15 = 5\) logs.
\(n = \dfrac{41 \pm \sqrt{1681 - 1600}}{2} = \dfrac{41 \pm 9}{2}\) → n = 25 or n = 16.
Check n = 25: top row would have \(20 - 24 = -4\) logs — impossible. So n = 16 rows; top row has \(20 - 15 = 5\) logs.
Activity 8.2 — Visualising the Sum
L4 AnalyseMaterials: Squared graph paper, two coloured pencils, ruler.
Predict: The sum 1 + 2 + 3 + ... + 10 = ? Compute first; we'll verify with a picture.
- Draw a "staircase" with one square in row 1, two in row 2, three in row 3, …, ten in row 10. Colour it RED.
- Now draw a second identical staircase rotated 180° (flipped) on top so that together the two form a 10 × 11 rectangle. Colour the new one BLUE.
- Count the squares in the rectangle: 10 × 11 = 110. The two staircases are equal, so each = 110/2 = 55.
- Verify: 1 + 2 + 3 + ... + 10 = 55. ✓ This visualises why \(S = n(n+1)/2\).
Competency-Based Questions
Scenario: An NGO plants trees on a reclaimed plot. They plant 10 trees in the first week, 13 in the second, 16 in the third, and so on, increasing by 3 every week. Their target is to plant a total of 1000 trees.
Q1. How many trees will they plant in the 12th week?
L3 ApplyAnswer: (b) 43. a = 10, d = 3, n = 12. \(a_{12} = 10 + 11(3) = 43\).
Q2. After how many weeks will the running total cross 200 trees? Analyse with a step-by-step computation.
L4 AnalyseAnswer: \(S_n = \dfrac{n}{2}\,[20 + (n-1)(3)] = \dfrac{n(3n + 17)}{2}\). Test n = 10: \(\dfrac{10 \times 47}{2} = 235\). Test n = 9: \(\dfrac{9 \times 44}{2} = 198\). So total first crosses 200 in week 10.
Q3. The NGO director claims, "If we keep this pattern, we'll reach 1000 trees in about 20 weeks." Evaluate this claim.
L5 EvaluateAnswer: \(S_{20} = \dfrac{20}{2}\,[20 + 19 \times 3] = 10 \times 77 = 770\). At 20 weeks the total is only 770, not 1000. So the claim is incorrect. To reach 1000: solve \(\dfrac{n(3n+17)}{2} \ge 1000 \Rightarrow 3n^2 + 17n - 2000 \ge 0 \Rightarrow n \approx 23.1\). So they need about 24 weeks.
Q4. Design a faster plantation plan that still uses an AP, plants 10 trees in week 1, and reaches a total of 1000 trees in exactly 20 weeks. What common difference is needed?
L6 CreateSolution: a = 10, n = 20, S = 1000. \(1000 = \dfrac{20}{2}\,[20 + 19d] = 10[20 + 19d] \Rightarrow 100 = 20 + 19d \Rightarrow d = 80/19 \approx 4.21\). Round up to d = 5 trees/week; then \(S_{20} = 10[20 + 95] = 1150\) — comfortably above target.
Assertion–Reason Questions
Assertion (A): The sum of the first n natural numbers is n(n+1)/2.
Reason (R): The natural numbers form an AP with a = 1 and d = 1.
Reason (R): The natural numbers form an AP with a = 1 and d = 1.
Answer: (a) — Substituting a = 1, l = n in S = n(a+l)/2 gives n(n+1)/2. R explains A.
Assertion (A): If S_n = n² for all n ≥ 1, the sequence is an AP.
Reason (R): Then \(a_n = S_n - S_{n-1} = n^2 - (n-1)^2 = 2n - 1\), which is linear in n.
Reason (R): Then \(a_n = S_n - S_{n-1} = n^2 - (n-1)^2 = 2n - 1\), which is linear in n.
Answer: (a) — Both correct: a_n = 2n − 1 is linear → AP with d = 2 (the odd numbers). R explains A.
Frequently Asked Questions
What is the formula for the sum of n terms of an AP in Class 9?
S_n = (n/2)(2a + (n - 1) d), where a is the first term, d is the common difference, and n is the number of terms. Equivalently, S_n = (n/2)(a + l), where l is the n-th (last) term being summed.
How was the sum 1 + 2 + 3 + ... + 100 computed by Gauss?
Gauss paired the terms: (1 + 100) + (2 + 99) + (3 + 98) + ... = 50 pairs each summing to 101 = 50 x 101 = 5050. The same trick generalises to S_n = (n/2)(first + last) for any AP.
What is the sum of the first n natural numbers?
Apply the AP sum with a = 1, d = 1, last term l = n: S = (n/2)(1 + n) = n(n+1)/2. For example, the sum of the first 50 natural numbers is 50 x 51 / 2 = 1275.
How do you find the number of terms n if you know S_n, a and d?
Substitute the known values into S_n = (n/2)(2a + (n - 1)d). This gives a quadratic in n; solve and take the positive integer root. For example, given S_n = 78, a = 3 and d = 4, you get n^2 + n*1 form quadratic and find n = 6.
What is the sum of the AP 5 + 11 + 17 + ... up to 20 terms?
Here a = 5, d = 6 and n = 20. S_20 = (20/2)(2 x 5 + (20 - 1) x 6) = 10 x (10 + 114) = 10 x 124 = 1240.
How can the AP sum formula be used in real-life problems?
It models any quantity that increases by a fixed amount each step: total seats in a stadium where each row has 2 more seats than the previous; total savings if you save a fixed extra amount per month; total bricks in a triangular stack; or total distance covered when speed grows by a constant per second.
AI Tutor
Mathematics Class 9 — Ganita Manjari
Ready