SL 1.2 Question Bank

ARITHMETIC SEQUENCES AND SERIES

This question bank contains 11 questions covering arithmetic sequences and series, distributed across different paper types according to IB AAHL curriculum standards.

📌 Multiple Choice Questions (4 Questions)

MCQ 1. The 7th term of an arithmetic sequence is 23 and the 15th term is 47. What is the first term?

A) 2 B) 5 C) 8 D) 11

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Solution:

Let first term = $a$, common difference = $d$

$u_7 = a + 6d = 23$ … (1)

$u_{15} = a + 14d = 47$ … (2)

Subtract (1) from (2): $8d = 24$, so $d = 3$

Substitute into (1): $a + 6(3) = 23$, so $a = 5$

✅ Answer: B) 5

MCQ 2. An arithmetic sequence has first term 12 and common difference -3. Which term is the first negative term?

A) 4th term B) 5th term C) 6th term D) 7th term

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Solution:

General term: $u_n = 12 + (n-1)(-3) = 12 – 3n + 3 = 15 – 3n$

For negative term: $15 – 3n < 0$

$15 < 3n$, so $n > 5$

First negative term is when $n = 6$

Check: $u_5 = 15 – 3(5) = 0$, $u_6 = 15 – 3(6) = -3$

✅ Answer: C) 6th term

MCQ 3. The sum of the first 10 terms of an arithmetic sequence is 185. If the first term is 4, what is the 10th term?

A) 31 B) 33 C) 35 D) 37

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Solution:

Using $S_n = \frac{n}{2}(u_1 + u_n)$

$185 = \frac{10}{2}(4 + u_{10})$

$185 = 5(4 + u_{10})$

$37 = 4 + u_{10}$

$u_{10} = 33$

✅ Answer: B) 33

MCQ 4. How many terms of the arithmetic sequence 8, 11, 14, 17, … are needed for the sum to equal 275?

A) 8 B) 9 C) 10 D) 11

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Solution:

First term $a = 8$, common difference $d = 3$

Using $S_n = \frac{n}{2}[2a + (n-1)d]$

$275 = \frac{n}{2}[2(8) + (n-1)(3)]$

$550 = n[16 + 3n – 3]$

$550 = n[13 + 3n]$

$550 = 13n + 3n^2$

$3n^2 + 13n – 550 = 0$

Using quadratic formula or factoring: $n = 10$ or $n = -18.33$

Since $n$ must be positive: $n = 10$

✅ Answer: C) 10

📌 Paper 1 Questions (No Calculator) – 2 Questions

Paper 1 – Q1. The first three terms of an arithmetic sequence are $2k + 3$, $5k – 2$, and $8k – 7$.

(a) Find the value of $k$. [4 marks]

(b) Hence find the 20th term of the sequence. [3 marks]

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Solution:

(a) Finding k:

For arithmetic sequence, common difference must be constant:

$d_1 = (5k – 2) – (2k + 3) = 3k – 5$

$d_2 = (8k – 7) – (5k – 2) = 3k – 5$

Since $d_1 = d_2$:

$3k – 5 = 3k – 5$ ✓ (This is always true)

Alternative approach using middle term property:

$2(5k – 2) = (2k + 3) + (8k – 7)$

$10k – 4 = 10k – 4$ ✓

This means any value of k works. However, let’s verify the common difference exists:

Common difference: $d = 3k – 5$

(b) Finding 20th term:

First term: $u_1 = 2k + 3$

Common difference: $d = 3k – 5$

$u_{20} = u_1 + 19d$

$u_{20} = (2k + 3) + 19(3k – 5)$

$= 2k + 3 + 57k – 95$

$= 59k – 92$

✅ Answer: (a) Any value of k (sequence is arithmetic for all k); (b) $u_{20} = 59k – 92$

Paper 1 – Q2. The sum of the first $n$ terms of an arithmetic sequence is given by $S_n = 2n^2 + 5n$.

(a) Find the first term. [2 marks]

(b) Find the common difference. [3 marks]

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Solution:

(a) Finding first term:

$u_1 = S_1 = 2(1)^2 + 5(1) = 2 + 5 = 7$

(b) Finding common difference:

Method 1: Find $u_2$

$S_2 = 2(2)^2 + 5(2) = 8 + 10 = 18$

$u_2 = S_2 – S_1 = 18 – 7 = 11$

$d = u_2 – u_1 = 11 – 7 = 4$

Method 2: Use general formula

For $n \geq 2$: $u_n = S_n – S_{n-1}$

$u_n = [2n^2 + 5n] – [2(n-1)^2 + 5(n-1)]$

$= 2n^2 + 5n – 2(n^2 – 2n + 1) – 5n + 5$

$= 2n^2 + 5n – 2n^2 + 4n – 2 – 5n + 5$

$= 4n + 3$

Common difference = coefficient of n = 4

Using $u_n = 4n + 3$:

$u_{15} = 4(15) + 3 = 60 + 3 = 63$

Or using $u_n = u_1 + (n-1)d$:

$u_{15} = 7 + (15-1)(4) = 7 + 56 = 63$

✅ Answer: (a) 7; (b) 4; (c) 63

📌 Paper 2 Questions (Calculator Allowed) – 4 Questions

Paper 2 – Q1. A theatre has 20 rows of seats. The first row has 15 seats, the second row has 18 seats, the third row has 21 seats, and so on in an arithmetic sequence.

(a) Find the number of seats in the 20th row. [3 marks]

(b) Find the total number of seats in the theatre. [3 marks]

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Solution:

First term $a = 15$, common difference $d = 18 – 15 = 3$

(a) Seats in 20th row:

$u_{20} = 15 + (20-1)(3) = 15 + 57 = 72$

(b) Total seats:

$S_{20} = \frac{20}{2}[2(15) + (20-1)(3)]$

$= 10[30 + 57] = 10 \times 87 = 870$

$u_n > 50$

$15 + (n-1)(3) > 50$

$15 + 3n – 3 > 50$

$3n + 12 > 50$

$3n > 38$

$n > 12.67$

Therefore, the 13th row is the first with more than 50 seats

Verification: $u_{13} = 15 + 12(3) = 51$ ✓

✅ Answer: (a) 72 seats; (b) 870 seats; (c) 13th row

Paper 2 – Q2. The sum of the first 8 terms of an arithmetic sequence is 156. The sum of the next 8 terms is 604.

(a) Find the common difference. [4 marks]

(b) Find the first term. [2 marks]

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Solution:

(a) Finding common difference:

Given: $S_8 = 156$

Sum of next 8 terms = $S_{16} – S_8 = 604$

Therefore: $S_{16} = 156 + 604 = 760$

Using $S_n = \frac{n}{2}[2a + (n-1)d]$:

$S_8 = \frac{8}{2}[2a + 7d] = 4[2a + 7d] = 156$

$2a + 7d = 39$ … (1)

$S_{16} = \frac{16}{2}[2a + 15d] = 8[2a + 15d] = 760$

$2a + 15d = 95$ … (2)

Subtract (1) from (2):

$8d = 56$, so $d = 7$

(b) Finding first term:

Substitute $d = 7$ into equation (1):

$2a + 7(7) = 39$

$2a + 49 = 39$

$2a = -10$, so $a = -5$

✅ Answer: (a) $d = 7$; (b) $a = -5$

Paper 2 – Q3. An arithmetic sequence has $u_3 = 22$ and $u_7 = 42$.

(a) Find the first term and common difference. [4 marks]

(b) Express $\sum_{k=1}^{20} u_k$ as a single numerical value. [3 marks]

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Solution:

(a) Finding first term and common difference:

$u_3 = a + 2d = 22$ … (1)

$u_7 = a + 6d = 42$ … (2)

Subtract (1) from (2):

$4d = 20$, so $d = 5$

Substitute into (1):

$a + 2(5) = 22$, so $a = 12$

(b) Finding sum:

$\sum_{k=1}^{20} u_k = S_{20}$

$S_{20} = \frac{20}{2}[2(12) + (20-1)(5)]$

$= 10[24 + 95]$

$= 10 \times 119 = 1190$

✅ Answer: (a) $a = 12$, $d = 5$; (b) 1190

Paper 2 – Q4. The terms $x$, $2x + 5$, and $4x + 3$ are consecutive terms of an arithmetic sequence.

(a) Find the value of $x$. [3 marks]

(b) Write down the first three terms of the sequence. [1 mark]

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Solution:

(a) Finding x:

For arithmetic sequence: $d_1 = d_2$

$(2x + 5) – x = (4x + 3) – (2x + 5)$

$x + 5 = 2x – 2$

$5 + 2 = 2x – x$

$x = 7$

(b) First three terms:

When $x = 7$:

First term: $x = 7$

Second term: $2(7) + 5 = 19$

Third term: $4(7) + 3 = 31$

Sequence: 7, 19, 31 (common difference = 12)

$a = 7$, $d = 12$, $n = 25$

$S_{25} = \frac{25}{2}[2(7) + (25-1)(12)]$

$= 12.5[14 + 288]$

$= 12.5 \times 302 = 3775$

✅ Answer: (a) $x = 7$; (b) 7, 19, 31; (c) 3775

📌 Paper 3 Question (Extended Response) – 1 Question

Paper 3 – Extended Question. A construction company is building a pyramid structure using concrete blocks. The top layer has 1 block, the second layer has 4 blocks, the third layer has 7 blocks, and so on, forming an arithmetic sequence.

(a) Show that the number of blocks in the $n$th layer is given by $u_n = 3n – 2$. [2 marks]

(b) The pyramid has 20 layers. Find the total number of blocks needed to build the pyramid. [4 marks]

(i) The total weight of the pyramid in tonnes. [2 marks]

(ii) The total cost of all blocks. [2 marks]

(d) The company wants to build a larger pyramid where the total number of blocks is at least 2000. Find the minimum number of layers needed. [5 marks]

📖 Show Answer

Complete Solution:

(a) Showing the formula:

Layer 1: 1 block

Layer 2: 4 blocks

Layer 3: 7 blocks

First term: $a = 1$

Common difference: $d = 4 – 1 = 3$

General term: $u_n = a + (n-1)d$

$u_n = 1 + (n-1)(3)$

$u_n = 1 + 3n – 3$

$u_n = 3n – 2$ ✓

(b) Total blocks for 20 layers:

Using sum formula: $S_n = \frac{n}{2}[2a + (n-1)d]$

$S_{20} = \frac{20}{2}[2(1) + (20-1)(3)]$

$= 10[2 + 57]$

$= 10 \times 59 = 590 \text{ blocks}$

Alternative: Using $S_n = \frac{n}{2}(u_1 + u_n)$

$u_{20} = 3(20) – 2 = 58$

$S_{20} = \frac{20}{2}(1 + 58) = 10 \times 59 = 590$

(i) Total weight:

Total blocks = 590

Weight per block = 25 kg

Total weight = 590 × 25 = 14,750 kg

Converting to tonnes: 14,750 kg ÷ 1000 = 14.75 tonnes

(ii) Total cost:

Cost per block = $15

Total cost = 590 × $15 = $8,850

(d) Finding minimum layers for 2000+ blocks:

We need: $S_n \geq 2000$

Using $S_n = \frac{n}{2}[2(1) + (n-1)(3)]$

$\frac{n}{2}[2 + 3n – 3] \geq 2000$

$\frac{n}{2}[3n – 1] \geq 2000$

$n(3n – 1) \geq 4000$

$3n^2 – n \geq 4000$

$3n^2 – n – 4000 \geq 0$

Using quadratic formula:

$n = \frac{1 \pm \sqrt{1 + 48000}}{6} = \frac{1 \pm \sqrt{48001}}{6}$

$n = \frac{1 \pm 219.09}{6}$

$n = \frac{220.09}{6} = 36.68$ or $n = -36.35$ (rejected)

Since n must be a positive integer: $n = 37$

Verification: $S_{37} = \frac{37}{2}[2 + 36 \times 3] = \frac{37}{2}[110] = 37 \times 55 = 2035$ ✓

Check $S_{36} = \frac{36}{2}[2 + 35 \times 3] = 18[107] = 1926$ < 2000

Summary:
• 20-layer pyramid: 590 blocks, 14.75 tonnes, $8,850
• For ≥2000 blocks: minimum 37 layers needed

✅ Final Answers:
(a) Shown: $u_n = 3n – 2$
(b) 590 blocks
(c)(i) 14.75 tonnes; (ii) $8,850
(d) 37 layers