Mathematics — Secondary Examination
Arithmetic Progressions
Chapter 5 — Comprehensive Examination Paper
General Instructions
- This paper consists of four sections: A, B, C, and D.
- All questions are compulsory unless stated otherwise.
- Section A contains 10 MCQs of 1 mark each (10 marks).
- Section B contains 5 short answer questions of 2 marks each (10 marks).
- Section C contains 8 short answer questions of 4 marks each (32 marks).
- Section D contains 4 long answer questions of 7 marks each (28 marks).
- Use of calculator is not permitted. Show all working clearly.
- Marks for each question are indicated at the right-hand side.
| Section | Type | No. of Questions | Marks Each | Total |
|---|---|---|---|---|
| A | Multiple Choice | 10 | 1 | 10 |
| B | Short Answer I | 5 | 2 | 10 |
| C | Short Answer II | 8 | 4 | 32 |
| D | Long Answer | 4 | 7 | 28 |
| Grand Total | 80 | |||
Section A
Multiple Choice Questions — 1 mark each
1.
[1]
The common difference of the AP 3, 7, 11, 15, ... is:
(A) 1
(B) 4
(C) 3
(D) 7
2.
[1]
The nth term of an AP with first term a and common difference d is:
(A) a + nd
(B) a + (n+1)d
(C) a + (n-1)d
(D) a - (n-1)d
3.
[1]
The 10th term of the AP 2, 7, 12, 17, ... is:
(A) 42
(B) 47
(C) 52
(D) 45
4.
[1]
Which of the following is NOT an AP?
(A) 2, 4, 6, 8, ...
(B) 1, 1, 2, 3, 5, ...
(C) -3, -1, 1, 3, ...
(D) 0, 0, 0, 0, ...
5.
[1]
The sum of the first n terms of an AP is given by S_n = n(n + 3). The common difference is:
(A) 1
(B) 2
(C) 3
(D) 4
6.
[1]
The sum of the first 20 terms of an AP whose first term is 1 and common difference is 2 is:
(A) 360
(B) 380
(C) 400
(D) 420
7.
[1]
If the arithmetic mean of a and b is 12, then a + b equals:
(A) 6
(B) 12
(C) 24
(D) 144
8.
[1]
How many terms are there in the AP 7, 13, 19, ..., 205?
(A) 30
(B) 33
(C) 34
(D) 35
9.
[1]
The 4th term from the last of the AP -11, -8, -5, ..., 49 is:
(A) 37
(B) 40
(C) 43
(D) 46
10.
[1]
If the sum of the first n terms of an AP is 4n - n², then the first term and common difference respectively are:
(A) 3 and -2
(B) 4 and -2
(C) 3 and -1
(D) 2 and -1
✦
Section B
Short Answer I — 2 marks each
11.
[2]
Find the common difference and the next two terms of the AP: 3/2, 1/2, -1/2, -3/2, ...
12.
[2]
Check whether -150 is a term of the AP 11, 8, 5, 2, ... Give reason for your answer.
13.
[2]
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
14.
[2]
Write the first four terms of the AP when: a = -1.25 and d = -0.25
15.
[2]
Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?
✦
Section C
Short Answer II — 4 marks each
16.
[4]
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
17.
[4]
An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.
18.
[4]
Find the sum of first 51 terms of an AP whose 2nd term is 14 and 3rd term is 18.
19.
[4]
How many multiples of 4 lie between 10 and 250? Find their sum.
20.
[4]
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.
21.
[4]
If the sum of the first n terms of an AP is 4n - n², what is the first term? What is the sum of first two terms? Find the 3rd, 10th and nth terms.
22.
[4]
The 3rd and 9th terms of an AP are 4 and -8 respectively. Which term of this AP is zero?
23.
[4]
Find the sum of all odd numbers between 0 and 50. Also verify using the formula for sum of first n natural numbers.
✦
Section D
Long Answer — 7 marks each
24.
[7]
Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week her weekly savings become Rs. 20.75, find n. Also find her total savings at the end of that week.
25.
[7]
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.
(a) In how many rows are the 200 logs placed?
(b) How many logs are in the top row?
(c) What is the total number of logs if the stacking continued until only 1 log is in the top row?
(a) In how many rows are the 200 logs placed?
(b) How many logs are in the top row?
(c) What is the total number of logs if the stacking continued until only 1 log is in the top row?
26.
[7]
A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find:
(a) the production in the 1st year,
(b) the production in the 10th year,
(c) the total production in the first 7 years.
(a) the production in the 1st year,
(b) the production in the 10th year,
(c) the total production in the first 7 years.
27.
[7]
The houses of a row are numbered consecutively from 1 to 49. Show that there exists a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find the value of x.
[Hint: Use S(x-1) = S_49 - S_x]
[Hint: Use S(x-1) = S_49 - S_x]
Detailed Answer Sheet
Model answers with full working — for teacher reference and student self-assessment
Section A — Answers
1 mark each — correct option only
Q11 mark
Common difference of AP: 3, 7, 11, 15, ...
Correct Answer: (B) 4 | d = 7 - 3 = 4
Q21 mark
nth term formula of an AP
Correct Answer: (C) a + (n-1)d
Q31 mark
10th term of AP: 2, 7, 12, 17, ...
WORKa = 2, d = 5. a10 = 2 + 9 × 5 = 2 + 45 = 47
Correct Answer: (B) 47
Q41 mark
Which is NOT an AP?
REASONIn 1, 1, 2, 3, 5 ... the differences are 0, 1, 1, 2 — not constant. This is the Fibonacci sequence.
Correct Answer: (B) 1, 1, 2, 3, 5, ...
Q51 mark
Sn = n(n+3), find common difference
WORKan = Sn - S(n-1) = n(n+3) - (n-1)(n+2) = n² + 3n - (n² + n - 2) = 2n + 2. So d = a2 - a1 = 6 - 4 = 2
Correct Answer: (B) 2
Q61 mark
Sum of first 20 terms, a=1, d=2
WORKS20 = (20/2)[2×1 + 19×2] = 10[2 + 38] = 10 × 40 = 400
Correct Answer: (C) 400
Q71 mark
Arithmetic mean of a and b is 12, find a+b
WORKAM = (a+b)/2 = 12, so a+b = 24
Correct Answer: (C) 24
Q81 mark
Number of terms in AP: 7, 13, 19, ..., 205
WORKa=7, d=6, an=205. 205 = 7 + (n-1)×6 → 198 = 6(n-1) → n-1 = 33 → n = 34
Correct Answer: (C) 34
Q91 mark
4th term from last of AP: -11, -8, -5, ..., 49
WORKReverse AP: a=-11, d=3, last=49. Total terms: 49=-11+(n-1)×3 → n=21. 4th from last = 18th term. a18 = -11+17×3 = -11+51 = 40
Correct Answer: (B) 40
Q101 mark
Sn = 4n - n², find a and d
WORKa1 = S1 = 4-1 = 3. S2 = 8-4 = 4, so a2 = S2-S1 = 1. d = a2-a1 = 1-3 = -2. So a=3, d=-2.
Correct Answer: (A) 3 and -2
Section B — Answers
2 marks each
Q112 marks
AP: 3/2, 1/2, -1/2, -3/2, ... — find d and next two terms
STEP 1d = a2 - a1 = 1/2 - 3/2 = -2/2 = -1
STEP 25th term = -3/2 + (-1) = -5/2. 6th term = -5/2 + (-1) = -7/2
d = -1; next two terms are -5/2 and -7/2
Q122 marks
Is -150 a term of AP: 11, 8, 5, 2, ...?
STEP 1a = 11, d = 8-11 = -3. Assume an = -150. Then: -150 = 11 + (n-1)(-3)
STEP 2-161 = (n-1)(-3) → n-1 = 161/3 = 53.67, which is NOT an integer.
Since n is not a positive integer, -150 is NOT a term of this AP.
Q132 marks
17th term exceeds 10th term by 7. Find d.
STEP 1a17 - a10 = [a+16d] - [a+9d] = 7d
STEP 2Given 7d = 7, so d = 1
Common difference d = 1
Q142 marks
First four terms: a = -1.25, d = -0.25
STEP 1a1 = -1.25, a2 = -1.25 + (-0.25) = -1.50
STEP 2a3 = -1.50 + (-0.25) = -1.75, a4 = -1.75 + (-0.25) = -2.00
The AP is: -1.25, -1.50, -1.75, -2.00, ...
Q152 marks
Two APs, same d. Difference of 100th terms = 100. Difference of 1000th terms?
STEP 1Let APs be a1, a1+d, ... and a2, a2+d, ... Their 100th terms differ by: (a1 + 99d) - (a2 + 99d) = a1 - a2 = 100
STEP 2Difference of 1000th terms = (a1 + 999d) - (a2 + 999d) = a1 - a2 = 100
The difference between their 1000th terms is also 100.
Section C — Answers
4 marks each
Q164 marks
11th term = 38, 16th term = 73. Find 31st term.
STEP 1a11 = a + 10d = 38 ...(i)
STEP 2a16 = a + 15d = 73 ...(ii)
STEP 3Subtract (i) from (ii): 5d = 35 → d = 7
STEP 4From (i): a + 70 = 38 → a = -32
STEP 5a31 = a + 30d = -32 + 30×7 = -32 + 210 = 178
The 31st term = 178
Q174 marks
AP of 50 terms: 3rd term = 12, last term = 106. Find 29th term.
STEP 1a3 = a + 2d = 12 ...(i)
STEP 2a50 = a + 49d = 106 ...(ii)
STEP 3Subtract (i) from (ii): 47d = 94 → d = 2
STEP 4From (i): a + 4 = 12 → a = 8
STEP 5a29 = 8 + 28×2 = 8 + 56 = 64
The 29th term = 64
Q184 marks
2nd term = 14, 3rd term = 18. Find sum of first 51 terms.
STEP 1d = a3 - a2 = 18 - 14 = 4
STEP 2a = a2 - d = 14 - 4 = 10
STEP 3S51 = (51/2)[2×10 + 50×4] = (51/2)[20 + 200] = (51/2)(220)
STEP 4S51 = 51 × 110 = 5610
Sum of first 51 terms = 5610
Q194 marks
Multiples of 4 between 10 and 250 — count and sum.
STEP 1AP: 12, 16, 20, ..., 248. a = 12, d = 4, an = 248
STEP 2248 = 12 + (n-1)×4 → 236 = 4(n-1) → n-1 = 59 → n = 60
STEP 3S60 = (60/2)(12 + 248) = 30 × 260 = 7800
There are 60 multiples of 4 between 10 and 250. Their sum = 7800.
Q204 marks
S7 = 49, S17 = 289. Find Sn.
STEP 1S7 = (7/2)[2a + 6d] = 7(a + 3d) = 49 → a + 3d = 7 ...(i)
STEP 2S17 = (17/2)[2a + 16d] = 17(a + 8d) = 289 → a + 8d = 17 ...(ii)
STEP 3Subtract (i) from (ii): 5d = 10 → d = 2. Then a = 7 - 6 = 1
STEP 4Sn = (n/2)[2×1 + (n-1)×2] = (n/2)(2n) = n²
Sum of first n terms = n²
Q214 marks
Sn = 4n - n². Find a1, S2, a2, a3, a10, an.
STEP 1a1 = S1 = 4(1) - 1² = 3
STEP 2S2 = 4(2) - 4 = 4. a2 = S2 - S1 = 4 - 3 = 1
STEP 3S3 = 12 - 9 = 3. a3 = S3 - S2 = 3 - 4 = -1
STEP 4S10 = 40 - 100 = -60. S9 = 36 - 81 = -45. a10 = -60 - (-45) = -15
STEP 5an = Sn - S(n-1) = (4n-n²) - (4(n-1)-(n-1)²) = 4n-n² - 4n+4+n²-2n+1 = 5 - 2n
a = 3, S2 = 4, a2 = 1, a3 = -1, a10 = -15, an = 5 - 2n
Q224 marks
3rd term = 4, 9th term = -8. Which term is zero?
STEP 1a + 2d = 4 ...(i) and a + 8d = -8 ...(ii)
STEP 2Subtract (i) from (ii): 6d = -12 → d = -2
STEP 3From (i): a + 2(-2) = 4 → a = 4 + 4 = 8
STEP 4Set an = 0: 8 + (n-1)(-2) = 0 → 2(n-1) = 8 → n-1 = 4 → n = 5
The 5th term of this AP is zero.
Q234 marks
Sum of all odd numbers between 0 and 50.
STEP 1AP: 1, 3, 5, ..., 49. a = 1, d = 2, l = 49.
STEP 2Find n: 49 = 1 + (n-1)×2 → 48 = 2(n-1) → n = 25
STEP 3S25 = (25/2)(1 + 49) = (25/2)(50) = 625
VERIFYUsing Sn = n²: Sum of first 25 odd numbers = 25² = 625 ✓
Sum of all odd numbers between 0 and 50 = 625
Section D — Answers
7 marks each
Q247 marks
Ramkali saves Rs. 5 in week 1, increases by Rs. 1.75/week. When does saving reach Rs. 20.75? Find total savings.
GIVENa = 5, d = 1.75, an = 20.75. Find n.
STEP 120.75 = 5 + (n-1) × 1.75
STEP 215.75 = (n-1) × 1.75 → n-1 = 15.75 / 1.75 = 9
STEP 3n = 10 weeks
STEP 4Total savings = Sn = (n/2)(a + an) = (10/2)(5 + 20.75)
STEP 5S10 = 5 × 25.75 = Rs. 128.75
n = 10 weeks. Total savings by end of 10th week = Rs. 128.75
Q257 marks
200 logs stacked: 20 in bottom row, 19 in next, etc.
PART aSn = 200. AP: a = 20, d = -1. Sn = (n/2)[2×20 + (n-1)(-1)] = (n/2)(41-n)
STEP 1n(41-n)/2 = 200 → n(41-n) = 400 → 41n - n² = 400
STEP 2n² - 41n + 400 = 0 → (n-16)(n-25) = 0 → n = 16 or n = 25
CHECKIf n = 25: a25 = 20 + 24(-1) = -4 (impossible — negative logs). So n = 16 rows
PART bTop row = a16 = 20 + 15(-1) = 20 - 15 = 5 logs
PART cUntil 1 log at top: AP from 20 down to 1. n = 20 rows. S20 = (20/2)(20+1) = 10×21 = 210 logs
(a) 16 rows | (b) Top row has 5 logs | (c) Full stack = 210 logs
Q267 marks
TV sets: 600 in year 3, 700 in year 7. Find year 1, year 10, and total in 7 years.
GIVENa3 = a + 2d = 600 ...(i) and a7 = a + 6d = 700 ...(ii)
STEP 1Subtract (i) from (ii): 4d = 100 → d = 25
STEP 2From (i): a + 50 = 600 → a = 550
PART aProduction in year 1 = a = 550 sets
PART ba10 = 550 + 9×25 = 550 + 225 = 775 sets
PART cS7 = (7/2)[2×550 + 6×25] = (7/2)[1100+150] = (7/2)(1250) = 4375 sets
(a) Year 1: 550 sets | (b) Year 10: 775 sets | (c) First 7 years total: 4375 sets
Q277 marks
Houses numbered 1 to 49. Find x where sum before x = sum after x.
SETUPSum before x = S(x-1) = x(x-1)/2. Total S49 = (49×50)/2 = 1225.
STEP 1Sum after x = S49 - Sx = 1225 - x(x+1)/2
STEP 2Set S(x-1) = S49 - Sx: x(x-1)/2 = 1225 - x(x+1)/2
STEP 3x(x-1)/2 + x(x+1)/2 = 1225 → x[(x-1)+(x+1)]/2 = 1225
STEP 4x[2x]/2 = 1225 → x² = 1225 → x = ±35
STEP 5Since x must be a positive integer between 1 and 49, x = 35
VERIFYSum before 35 = (34×35)/2 = 595. Sum after 35 = 1225 - 595 - 35 = 595 ✓
x = 35. The sum of houses before house 35 equals the sum after house 35 (both = 595).
