Arithmetic Progressions
Virtual Teacher + Formula Revision + 95-Entry Solved Question Bank + CBSE Competency Practice + HOTS & Olympiad Foundation Module
CBSE Class 10
NCERT
NCERT Exemplar Type
CBSE Pattern
Competency Based
HOTS
Olympiad Foundation
An arithmetic progression is one of the simplest examples of a predictable mathematical pattern. Once you understand the first term and the fixed change between consecutive terms, you can determine distant terms, sums, missing values and even model real-life situations without writing the entire sequence.
Virtual Teacher: Arithmetic Progressions from Basics to Olympiad Reasoning
We begin with simple number patterns and gradually develop the nth-term and sum formulas.
1. Sequences and Number Patterns
A sequence is an ordered list of numbers or mathematical objects that follow some rule or pattern.
2, 4, 6, 8, 10, … 5, 10, 15, 20, … 20, 17, 14, 11, …Term
Each individual member of a sequence.
Finite Sequence
A sequence containing a fixed number of terms.
Infinite Sequence
A sequence that continues indefinitely.
Progression
A sequence whose terms follow a particular mathematical rule.
Not every sequence is an AP
1, 4, 9, 16, …
has consecutive differences 3, 5, 7, … . Since these are not constant, the sequence is not an arithmetic progression.
2. Meaning of an Arithmetic Progression
A sequence is an Arithmetic Progression (AP) if the difference between every pair of consecutive terms is constant.
a2−a1 = a3−a2 = a4−a3 = … = dFirst terma
Common differenced
nth terman
Number of termsn
Increasing AP
3, 7, 11, 15, … d = 4Decreasing AP
20, 16, 12, 8, … d = −4Constant AP
5, 5, 5, 5, … d = 0
Important
The common difference may be positive, negative, zero, fractional or decimal.
3. How to Check Whether a Sequence is an AP
Consider:
4, 9, 14, 19, … 9 − 4 = 5 14 − 9 = 5 19 − 14 = 5Since all consecutive differences are equal, the sequence is an AP with:
d = 5
Exam Tip
Do not judge an AP merely by appearance. Explicitly calculate at least two consecutive differences.
4. General Form of an AP
If the first term is a and the common difference is d, the terms are:
a, a+d, a+2d, a+3d, … a1 = a a2 = a+d a3 = a+2d a4 = a+3d
Why does the nth term contain n−1?
The first term requires zero additions of d. The second term requires one addition, the third requires two, and therefore the nth term requires exactly n−1 additions.
an = a + (n−1)d
5. The nth Term of an AP
The fundamental nth-term formula is:
an = a + (n−1)dExample: Find the 20th term of 5, 8, 11, 14, …
a = 5 d = 8−5 = 3 n = 20 a20 = 5+(20−1)(3) = 5+57 = 62
Common Mistake
Incorrect:
an = a+nd
Correct:
an = a+(n−1)d
6. Finding Missing Terms and Arithmetic Means
If 5, x, 17 are consecutive terms of an AP:
x−5 = 17−x 2x = 22 x = 11If a, A, b are consecutive terms of an AP, then:
A−a = b−A 2A = a+b A = a+b2
Arithmetic Mean
The number exactly midway between two numbers in an AP is their arithmetic mean.
7. Finding the Number of Terms
How many terms are there in:
7, 11, 15, …, 95? a = 7, d = 4, an = 95 95 = 7+(n−1)4 88 = 4(n−1) 22 = n−1 n = 23
Important Check
If solving for n gives a non-positive or non-integral answer, the proposed number cannot occupy that position in the AP.
8. Terms from the End
If a finite AP has n terms, then the rth term from the end is:
an−r+1Alternatively, reverse the AP. The last term becomes the first term and the common difference becomes −d.
Which method is faster?
If n is already known, use an−r+1. If the last term is immediately available, treating the AP in reverse can be faster.
9. Useful Relations Between Terms
ap = a+(p−1)d aq = a+(q−1)dSubtracting:
ap−aq = (p−q)d d = ap−aqp−q
HOTS / Olympiad Insight
Terms equally distant from a central term satisfy:
an−r + an+r = 2an
because their d-components cancel.
10. Sum of the First n Terms
Let:
Sn = a+(a+d)+(a+2d)+…+[a+(n−1)d]Write the same sum in reverse:
Sn = [a+(n−1)d]+[a+(n−2)d]+…+aAdding vertically, every pair equals:
2a+(n−1)dThere are n such pairs:
2Sn = n[2a+(n−1)d] Sn = n2[2a+(n−1)d]If the last term is l:
l = a+(n−1)dtherefore:
Sn = n2(a+l)11. Choosing the Correct Sum Formula
If a, d and n are known
Sn = n2[2a+(n−1)d]
If a, l and n are known
Sn = n2(a+l)
Exam Strategy
Use the formula requiring the fewest preliminary calculations.
12. Finding n from the Sum
When Sn is given, substituting in the sum formula may produce a quadratic equation.
For example, for 3, 7, 11, … :
Sn = n/2[6+4(n−1)] = n/2(4n+2)
Root Check
Since n represents a number of terms, reject negative and non-integral roots.
13. Sum Between Two Positions
To find the sum from the 11th to the 25th terms:
a11+a12+…+a25 = S25−S10
Shortcut
Do not manually add a long block of consecutive terms. Use partial sums.
14. Real-Life Applications
Arithmetic progressions often appear whenever a quantity changes by the same fixed amount at regular stages.
Auditorium
Each row may contain a fixed number of seats more than the preceding row.
Savings
A deposit may increase by a fixed amount each month.
Salary
An employee may receive a fixed annual increment.
Staircases / Stacks
Each level may contain a fixed number more or fewer objects.
Translate Before Calculating
From a word problem identify:
a, d, n, an, Sn
15. Why an AP Produces a Straight-Line Term Pattern
Rewrite:
an = a+(n−1)d an = dn+(a−d)The term changes by the same amount whenever n increases by 1. Therefore the plotted points (n,an) lie in a straight-line pattern.
Explore an Arithmetic Progression
Change the first term, common difference and number of terms. This activity is supplementary; the complete lesson and question bank remain available without it.
Terms3, 7, 11, 15, 19, …
nth term39
Sum Sₙ210
TypeIncreasing AP
Arithmetic Progressions Formula Revision Sheet
Common Difference
d = a2−a1
nth Term
an = a+(n−1)d
Last Term
l = a+(n−1)d
Sum of First n Terms
Sn = n2[2a+(n−1)d]
Alternative Sum Formula
Sn = n2(a+l)
Arithmetic Mean
A = a+b2
rth Term from End
an−r+1
Difference from Two Terms
d = ap−aqp−q
Symmetric Terms — HOTS
an−r+an+r=2an
Sum Between Positions
ar+1+…+an = Sn−Sr
How to Solve Arithmetic Progression Questions
A specific term is required
Use an = a+(n−1)d.
Number of terms is required
Set the given last term equal to a+(n−1)d and solve for n.
A sum is required
Use one of the Sn formulas.
First and last terms are given
Prefer Sn = n(a+l)/2.
A middle term is missing
Use equality of consecutive differences or the arithmetic-mean relation.
Two different terms are given
Write two equations in a and d or subtract them directly.
Sum from one position to another
Subtract partial sums.
Word problem
Identify a, d and n before doing algebra.
10 Common Mistakes Students Make in Arithmetic Progressions
1. Writing a+ndThe first term needs zero additions of d, so use a+(n−1)d.
2. Wrong common differenceAlways calculate later term minus earlier term.
3. Ignoring negative dA decreasing AP has a negative common difference.
4. Confusing aₙ and Sₙan is one term; Sn is a sum.
5. Using the wrong nCount positions carefully, especially in partial sums.
6. Accepting fractional positionsA term position must be a positive integer.
7. Accepting negative nA negative number of terms has no meaning here.
8. Assuming every pattern is an APCheck the consecutive differences.
9. Poor translation of word problemsIdentify first term and fixed change before selecting a formula.
10. Adding terms manuallyUse Sn or partial-sum subtraction when many terms are involved.
Complete Solved Question Bank
95 main entries: 30 MCQs + 15 Two-Mark + 15 Three-Mark + 20 Four-Mark + 10 Five-Mark + 5 Case Studies.
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Section A — 30 MCQs × 1 Mark
MCQ 11 MarkLevel 1 — FoundationNCERT Type
Which of the following is an arithmetic progression?
Correct Answer: B9−4 = 514−9 = 519−14 = 5
The common difference is constant, d = 5.
MCQ 21 MarkLevel 1 — Foundation
Which sequence is NOT an AP?
Correct Answer: C4−1 = 39−4 = 516−9 = 7
The differences are not constant.
MCQ 31 MarkLevel 1 — Foundation
The common difference of 20, 16, 12, 8, … is:
Correct Answer: Bd = 16−20= −4
MCQ 41 MarkLevel 1 — Foundation
For the AP 5, 5, 5, 5, …, the common difference is:
Correct Answer: Cd = 5−5 = 0
A constant sequence is also an AP.
MCQ 51 MarkLevel 1 — Foundation
The 10th term of 3, 7, 11, 15, … is:
Correct Answer: Ba = 3, d = 4a10 = 3+9(4)= 39
MCQ 61 MarkLevel 1 — Foundation
47 is which term of 2, 7, 12, 17, …?
Correct Answer: B47 = 2+(n−1)545 = 5(n−1)n−1 = 9n = 10
MCQ 71 MarkLevel 1 — Foundation
If an = 5n−2, the common difference is:
Correct Answer: Can+1−an = [5(n+1)−2]−(5n−2)= 5
MCQ 81 MarkLevel 1 — Foundation
The arithmetic mean between 8 and 18 is:
Correct Answer: CA = (8+18)/2= 26/2= 13
MCQ 91 MarkLevel 2 — StandardExemplar Type
If x, 7, 15 are consecutive terms of an AP, x equals:
Correct Answer: A2(7) = x+1514 = x+15x = −1
MCQ 101 MarkLevel 2 — Standard
How many terms are there in 5, 8, 11, …, 32?
Correct Answer: C32 = 5+(n−1)327 = 3(n−1)n = 10
MCQ 111 MarkLevel 2 — Standard
The 5th term from the end of 2, 5, 8, …, 29 is:
Correct Answer: B29 = 2+(n−1)3 ⇒ n = 105th from end = a10−5+1 = a6a6 = 2+5(3)=17
MCQ 121 MarkLevel 2 — Standard
The sum 1+2+3+…+10 is:
Correct Answer: CS10 = 10/2(1+10)= 5×11 = 55
MCQ 131 MarkLevel 2 — Standard
The sum of the first 20 terms of 3, 3, 3, … is:
Correct Answer: Cd = 020 terms each equal 3S20 = 20×3 = 60
MCQ 141 MarkLevel 2 — StandardCBSE Pattern
The sum of the first 10 terms of 5, 8, 11, … is:
Correct Answer: CS10 = 10/2[2(5)+9(3)]= 5(10+27)= 185
MCQ 151 MarkLevel 2 — Standard
An AP has first term 7, last term 43 and 10 terms. Its sum is:
Correct Answer: CS10 = 10/2(7+43)= 5×50= 250
MCQ 161 MarkLevel 3 — Advanced
If an = 4n+1, the first term and common difference are respectively:
Correct Answer: Ca1 = 4(1)+1 = 5a2 = 9d = 9−5 = 4
MCQ 171 MarkLevel 3 — Advanced
If a7 = 19 and a13 = 37, the common difference is:
Correct Answer: Ba13−a7 = (13−7)d18 = 6dd = 3
MCQ 181 MarkLevel 3 — Advanced
For a = 2 and d = 3, the term equal to 50 is:
Correct Answer: C50 = 2+3(n−1)48 = 3(n−1)n−1 = 16n = 17
MCQ 191 MarkLevel 3 — Advanced
If Sn = 3n2+2n, then the nth term is:
Correct Answer: Can = Sn−Sn−1= 3n²+2n − [3(n−1)²+2(n−1)]= 6n−1
MCQ 201 MarkLevel 3 — AdvancedConceptual
Which statement is always true for an AP?
Correct Answer: C
Constant difference is the defining property of an arithmetic progression.
MCQ 211 MarkLevel 4 — HOTS
For the AP 2, 5, 8, …, the sum of the 11th to 20th terms is:
Correct Answer: AS20 = 20/2[4+19(3)] = 610S10 = 10/2[4+9(3)] = 155Required sum = 610−155 = 455
MCQ 221 MarkLevel 4 — HOTS
If x−1, x+2 and 2x+1 are consecutive terms of an AP, x equals:
Correct Answer: C(x+2)−(x−1) = 3(2x+1)−(x+2) = x−1x−1 = 3x = 4
MCQ 231 MarkLevel 4 — HOTS
Four arithmetic means are inserted between 3 and 18. The second inserted mean is:
Correct Answer: BThere are 6 terms in total.d = (18−3)/(6−1) = 3Sequence: 3, 6, 9, 12, 15, 18
The second inserted mean is 9.
MCQ 241 MarkLevel 4 — HOTS
The sum of the first 20 positive even integers is:
Correct Answer: Ca = 2, l = 40, n = 20S20 = 20/2(2+40)= 10×42 = 420
MCQ 251 MarkLevel 4 — HOTS
If the 7th term of an AP is 0 and the 13th term is 18, the first term is:
Correct Answer: A6d = 18 ⇒ d = 3a+6d = 0a+18 = 0a = −18
MCQ 261 MarkLevel 5 — Olympiad Challenge
If Sn = n(2n+3), the 10th term is:
Correct Answer: Ca10 = S10−S9= 10(23)−9(21)= 230−189= 41
MCQ 271 MarkLevel 5 — Olympiad Challenge
In the AP 10, 8, 6, …, 0 occurs as the:
Correct Answer: B0 = 10+(n−1)(−2)2(n−1)=10n−1=5n=6
MCQ 281 MarkLevel 5 — Olympiad Challenge
If a5+a15 = 40, then a10 equals:
Correct Answer: C5 and 15 are equally distant from 10.a5+a15 = 2a1040 = 2a10a10 = 20
MCQ 291 MarkLevel 5 — Olympiad Challenge
If Sn = n2, the 10th term is:
Correct Answer: Can = n²−(n−1)²= 2n−1a10 = 19
MCQ 301 MarkLevel 5 — Olympiad Challenge
If the pth term of an AP is q and the qth term is p, where p ≠ q, then the (p+q)th term equals:
Correct Answer: Ba+(p−1)d = qa+(q−1)d = p(p−q)d = q−pd = −1a = p+q−1ap+q = p+q−1−(p+q−1)=0
Section B — 15 Two-Mark Questions
2-Mark Q12 MarksLevel 1 — Foundation
Check whether 1/2, 1, 3/2, 2, … is an AP. If yes, state its common difference.
1−1/2 = 1/23/2−1 = 1/22−3/2 = 1/2
Yes. It is an AP with d = 1/2.
2-Mark Q22 MarksLevel 1 — Foundation
Find the 25th term of 7, 11, 15, 19, …
a = 7, d = 4, n = 25a25 = 7+24(4)= 7+96= 103
a25 = 103.
2-Mark Q32 MarksLevel 1 — Foundation
121 is which term of 1, 6, 11, 16, …?
121 = 1+(n−1)5120 = 5(n−1)24 = n−1n = 25
121 is the 25th term.
2-Mark Q42 MarksLevel 1 — Foundation
Find x if 5, x, 17 are consecutive terms of an AP.
x−5 = 17−x2x = 22x = 11
x = 11.
2-Mark Q52 MarksLevel 2 — Standard
Insert three arithmetic means between 2 and 18.
Total terms = 518 = 2+4d16 = 4dd = 4AP: 2, 6, 10, 14, 18
The arithmetic means are 6, 10 and 14.
2-Mark Q62 MarksLevel 2 — Standard
Find the nth term of −4, −1, 2, 5, …
a = −4d = 3an = −4+(n−1)3= −4+3n−3= 3n−7
an = 3n−7.
2-Mark Q72 MarksLevel 2 — Standard
How many terms are there in 7, 11, 15, …, 95?
95 = 7+(n−1)488 = 4(n−1)n−1 = 22n = 23
23 terms.
2-Mark Q82 MarksLevel 2 — Standard
Find the 8th term from the end of 3, 7, 11, …, 79.
79 = 3+(n−1)4n = 208th from end = a20−8+1 = a13a13 = 3+12(4)=51
51.
2-Mark Q92 MarksLevel 2 — Standard
Find the sum of the first 15 terms of 2, 5, 8, …
a = 2, d = 3, n = 15S15 = 15/2[4+14(3)]= 15/2(46)= 345
S15 = 345.
2-Mark Q102 MarksLevel 2 — Standard
Find the sum of the AP 5, 10, 15, …, 60.
60 = 5+(n−1)5n = 12S12 = 12/2(5+60)= 6×65 = 390
390.
2-Mark Q112 MarksLevel 3 — Advanced
If a8 = 24 and a15 = 45, find d.
a15−a8 = 7d45−24 = 7d21 = 7dd = 3
d = 3.
2-Mark Q122 MarksLevel 3 — Advanced
The 12th term of an AP is 35 and d = 3. Find the first term.
35 = a+11(3)35 = a+33a = 2
First term = 2.
2-Mark Q132 MarksLevel 3 — Advanced
If 2x, 3x+1 and 5x−3 are consecutive terms of an AP, find x.
2(3x+1) = 2x+(5x−3)6x+2 = 7x−3x = 5
x = 5.
2-Mark Q142 MarksLevel 3 — Advanced
Is 202 a term of 3, 8, 13, 18, …?
202 = 3+(n−1)5199 = 5(n−1)n−1 = 199/5 = 39.8
Why reject this?A term position must be a positive integer.
No. 202 is not a term of the AP.
2-Mark Q152 MarksLevel 4 — HOTS
Find the sum of the 11th to 20th terms of 4, 7, 10, …
S20 = 20/2[8+19(3)] = 650S10 = 10/2[8+9(3)] = 175Required sum = 650−175= 475
475.
Section C — 15 Three-Mark Questions
3-Mark Q13 MarksLevel 2 — Standard
The 5th term of an AP is 17 and the 12th term is 45. Find the first term and common difference.
a+4d = 17a+11d = 457d = 28d = 4a+16 = 17a = 1
a = 1, d = 4.
3-Mark Q23 MarksLevel 2 — Standard
If a8 = 3 and a18 = 33, find a, d and a25.
a+7d = 3a+17d = 3310d = 30 ⇒ d = 3a+21 = 3 ⇒ a = −18a25 = −18+24(3)= 54
a = −18, d = 3, a25 = 54.
3-Mark Q33 MarksLevel 2 — Standard
Find the number of terms and the sum of 5, 9, 13, …, 101.
101 = 5+(n−1)496 = 4(n−1)n = 25S25 = 25/2(5+101)= 25×53= 1325
25 terms; sum = 1325.
3-Mark Q43 MarksLevel 2 — Standard
Find the sum of the first 30 terms of 7, 10, 13, …
a = 7, d = 3, n = 30S30 = 30/2[14+29(3)]= 15(14+87)= 15×101= 1515
1515.
3-Mark Q53 MarksLevel 3 — Advanced
Insert five arithmetic means between 4 and 22.
Total terms = 722 = 4+6d18 = 6dd = 3AP: 4, 7, 10, 13, 16, 19, 22
The five means are 7, 10, 13, 16 and 19.
3-Mark Q63 MarksLevel 3 — Advanced
Find the 6th term from the end of 2, 7, 12, …, 92.
92 = 2+(n−1)590 = 5(n−1)n = 196th from end = a19−6+1= a14= 2+13(5)=67
67.
3-Mark Q73 MarksLevel 3 — Advanced
Which term of 24, 21, 18, … is zero?
a = 24, d = −30 = 24+(n−1)(−3)3(n−1)=24n−1 = 8n = 9
0 is the 9th term.
3-Mark Q83 MarksLevel 3 — Advanced
Find the sum of all positive multiples of 7 from 7 to 140.
AP: 7, 14, 21, …, 140n = 140/7 = 20S20 = 20/2(7+140)= 10×147= 1470
1470.
3-Mark Q93 MarksLevel 4 — HOTS
If Sn = 2n²+3n, find the nth term, first term and common difference.
an = Sn−Sn−1= 2n²+3n − [2(n−1)²+3(n−1)]= 4n+1a1 = 5d = a2−a1 = 9−5 = 4
an = 4n+1, a = 5, d = 4.
3-Mark Q103 MarksLevel 4 — HOTS
The 3rd term of an AP is 8 and the 9th term is 32. Find the sum of its first 12 terms.
a+2d = 8a+8d = 326d = 24 ⇒ d = 4a = 0S12 = 12/2[0+11(4)]= 6×44 = 264
264.
3-Mark Q113 MarksLevel 4 — HOTS
How many terms of 4, 7, 10, … have sum 175?
175 = n/2[8+3(n−1)]350 = n(3n+5)3n²+5n−350 = 0(3n+35)(n−10)=0n = 10 or −35/3
RejectNumber of terms cannot be negative.
n = 10.
3-Mark Q123 MarksLevel 4 — HOTS
Riya saves ₹200 in the first month and increases her saving by ₹50 every month. Find her saving in the 12th month and her total saving during the first 12 months.
a = 200, d = 50a12 = 200+11(50)=750S12 = 12/2[400+11(50)]= 6(950)= ₹5700
12th-month saving = ₹750; total = ₹5700.
3-Mark Q133 MarksLevel 4 — HOTS
An auditorium has 20 seats in the first row and 2 more seats in each successive row. Find the number of seats in the 30th row and the total seats in 30 rows.
a = 20, d = 2, n = 30a30 = 20+29(2)=78S30 = 30/2(20+78)= 15×98= 1470
30th row = 78 seats; total = 1470 seats.
3-Mark Q143 MarksLevel 5 — Olympiad Challenge
If the pth term of an AP is q and the qth term is p, p ≠ q, prove that d = −1 and find the first term.
a+(p−1)d = qa+(q−1)d = p(p−q)d = q−p(p−q)d = −(p−q)d = −1a−(p−1)=qa = p+q−1
d = −1 and a = p+q−1.
3-Mark Q153 MarksLevel 5 — Olympiad Challenge
If Sn = 3n²+5n, find the 15th term.
an = Sn−Sn−1= 3n²+5n − [3(n−1)²+5(n−1)]= 6n+2a15 = 6(15)+2= 92
92.
Section D — 20 Four-Mark Questions
4-Mark Q14 MarksLevel 2 — Standard
The 7th term of an AP is 20 and the 13th term is 38. Find the AP and the sum of its first 20 terms.
a+6d = 20a+12d = 386d = 18 ⇒ d = 3a = 20−18 = 2AP: 2, 5, 8, 11, …S20 = 20/2[4+19(3)]= 10×61 = 610
AP = 2,5,8,… and S20 = 610.
4-Mark Q24 MarksLevel 3 — Advanced
The 4th term of an AP is 10 and the 10th term is 34. Find its 20th term and the sum of the first 20 terms.
a+3d = 10a+9d = 346d = 24 ⇒ d = 4a = −2a20 = −2+19(4)=74S20 = 20/2(−2+74)= 10×72 = 720
a20 = 74, S20 = 720.
4-Mark Q34 MarksLevel 3 — Advanced
How many terms of 6, 10, 14, … have sum 966?
966 = n/2[12+4(n−1)]966 = n/2(4n+8)966 = 2n(n+2)n(n+2)=483n²+2n−483=0(n−21)(n+23)=0n = 21 or −23
There are 21 terms.
4-Mark Q44 MarksLevel 3 — Advanced
Find the sum of the 11th to the 25th terms of 8, 14, 20, …
S25 = 25/2[16+24(6)]= 25/2(160)=2000S10 = 10/2[16+9(6)]= 5(70)=350Required sum = 2000−350= 1650
1650.
4-Mark Q54 MarksLevel 3 — Advanced
Insert six arithmetic means between 5 and 33 and find the sum of the inserted means.
Total terms = 833 = 5+7d28 = 7d ⇒ d = 4AP: 5, 9, 13, 17, 21, 25, 29, 33Inserted means = 9+13+17+21+25+29= 114
Means: 9,13,17,21,25,29; their sum = 114.
4-Mark Q64 MarksLevel 3 — Advanced
If x−2, 2x+1 and 5x−8 are consecutive terms of an AP, find x and the three terms.
(2x+1)−(x−2) = x+3(5x−8)−(2x+1) = 3x−9x+3 = 3x−912 = 2xx = 6Terms: 4, 13, 22
x = 6; terms are 4,13,22.
4-Mark Q74 MarksLevel 3 — AdvancedCompetency Based
An employee earns ₹25,000 in the first year and receives a fixed annual increment of ₹1,500. Find the salary in the 10th year and the total salary received over the first 10 years, ignoring other benefits.
a = 25000, d = 1500a10 = 25000+9(1500)= ₹38,500S10 = 10/2[50000+9(1500)]= 5(63500)= ₹3,17,500
10th-year salary = ₹38,500; total = ₹3,17,500.
4-Mark Q84 MarksLevel 3 — Advanced
A staircase pattern contains 18 blocks in the bottom row, 16 in the next, then 14, and so on until the top row has 2 blocks. Find the number of rows and total blocks.
a = 18, d = −2, l = 22 = 18+(n−1)(−2)16 = 2(n−1)n = 9S9 = 9/2(18+2)= 90
9 rows and 90 blocks.
4-Mark Q94 MarksLevel 3 — Advanced
Twelve prizes are distributed so that the first prize is ₹5,000 and each successive prize is ₹250 less. Find the 12th prize and the total prize money.
a = 5000, d = −250a12 = 5000+11(−250)= ₹2250S12 = 12/2(5000+2250)= 6×7250= ₹43,500
12th prize = ₹2,250; total = ₹43,500.
4-Mark Q104 MarksLevel 4 — HOTS
If the mth term of an AP is n and the nth term is m, where m ≠ n, prove that the (m+n)th term is zero.
a+(m−1)d = na+(n−1)d = m(m−n)d = n−md = −1a = m+n−1am+n = a+(m+n−1)d= m+n−1−(m+n−1)= 0
Hence proved.
4-Mark Q114 MarksLevel 4 — HOTS
If ap = aq for p ≠ q in an AP, prove that the AP is constant.
a+(p−1)d = a+(q−1)d(p−q)d = 0p ≠ qTherefore d = 0Hence every term is equal to a.
The AP is constant.
4-Mark Q124 MarksLevel 4 — HOTS
The sum of the first 15 terms of an AP is 300 and the sum of the first 25 terms is 1000. Find a and d.
300 = 15/2[2a+14d]2a+14d = 40a+7d = 20 … (1)1000 = 25/2[2a+24d]2a+24d = 80a+12d = 40 … (2)(2)−(1): 5d = 20d = 4a+28 = 20 ⇒ a = −8
a = −8, d = 4.
4-Mark Q134 MarksLevel 4 — HOTS
If Sn = 5n²−3n, find the AP and the sum of its 11th to 20th terms.
an = Sn−Sn−1= 10n−8a = a1 = 2, d = 10S20 = 5(20²)−3(20)=1940S10 = 5(10²)−3(10)=470Required sum = 1940−470= 1470
AP: 2,12,22,… ; required sum = 1470.
4-Mark Q144 MarksLevel 4 — HOTS
The 7th term of an AP is three times its 3rd term. If the 11th term is 50, find the AP.
a+6d = 3(a+2d)a+6d = 3a+6d2a = 0 ⇒ a = 0a+10d = 5010d = 50 ⇒ d = 5
AP: 0, 5, 10, 15, …
4-Mark Q154 MarksLevel 4 — HOTS
Three positive numbers are in AP. Their sum is 24 and product is 440. Find the numbers.
Let the numbers be 8−d, 8, 8+d.Their product is:(8−d)8(8+d)=4408(64−d²)=44064−d²=55d²=9d=3 for increasing order.
The numbers are 5, 8 and 11.
4-Mark Q164 MarksLevel 5 — Olympiad Challenge
Four consecutive terms of an increasing AP have sum 40. The product of the first and fourth terms is 64. Find all four terms.
For four consecutive AP terms, first + fourth equals second + third.
Total = 402(first+fourth)=40first+fourth=20first×fourth=64t²−20t+64=0(t−4)(t−16)=0First = 4, fourth = 163d = 12 ⇒ d = 44, 8, 12, 16.
4-Mark Q174 MarksLevel 5 — Olympiad Challenge
The sum of the first n terms of an AP equals the sum of the next n terms. Prove that the AP must be constant.
Sn = S2n−Sn2Sn = S2nn[2a+(n−1)d] = n[2a+(2n−1)d]2a+(n−1)d = 2a+(2n−1)dnd = 0n > 0 ⇒ d = 0
Therefore the AP is constant.
4-Mark Q184 MarksLevel 5 — Olympiad Challenge
The 3rd, 8th and 13th terms of an AP are considered. Prove that they form an AP. If the 3rd term is 7 and the 13th term is 37, find the 8th term and their sum.
a8−a3 = 5da13−a8 = 5dHence a3,a8,a13 form an AP.a8 = (a3+a13)/2= (7+37)/2 = 22Sum = 7+22+37 = 66
a8 = 22 and the sum is 66.
4-Mark Q194 MarksLevel 5 — Olympiad Challenge
A decorative tile pattern contains 7 tiles in the first row, 11 in the second and 15 in the third. If the pattern continues, find the 18th-row count and the total tiles in the first 18 rows.
a = 7, d = 4, n = 18a18 = 7+17(4)= 75S18 = 18/2(7+75)= 9×82= 738
18th row = 75 tiles; total = 738 tiles.
4-Mark Q204 MarksLevel 5 — Olympiad Challenge
In an AP, a1+a4+a7 = 45 and a2+a5+a8 = 54. Find a, d and a5.
a1+a4+a7 = a+(a+3d)+(a+6d)3a+9d=45 … (1)The second sum advances every corresponding term by d.Therefore it exceeds the first by 3d.54−45 = 3dd = 33a+27=45 ⇒ 3a=18 ⇒ a=6a5 = 6+4(3)=18
a = 6, d = 3, a5 = 18.
Section E — 10 Five-Mark Questions
5-Mark Q15 MarksLevel 3 — Advanced
The 4th term of an AP is 10 and the 9th term is 30. Find a, d, the 25th term and the sum of the first 25 terms.
a+3d=10a+8d=305d=20 ⇒ d=4a+12=10 ⇒ a=−2a25=−2+24(4)=94S25=25/2(−2+94)=25/2(92)=1150
a=−2, d=4, a25=94, S25=1150.
5-Mark Q25 MarksLevel 4 — HOTS
The sum of the first 10 terms of an AP is 210 and the sum of the next 10 terms is 610. Find the AP and its 25th term.
S10=2105[2a+9d]=2102a+9d=42 … (1)S20=210+610=82010[2a+19d]=8202a+19d=82 … (2)(2)−(1): 10d=40 ⇒ d=42a+36=42 ⇒ a=3a25=3+24(4)=99
AP: 3,7,11,… and a25=99.
5-Mark Q35 MarksLevel 4 — HOTS
How many terms of 8, 12, 16, … have sum 920? Also find the last term included.
920 = n/2[16+4(n−1)]920 = n/2(4n+12)920 = 2n(n+3)n(n+3)=460n²+3n−460=0(n−20)(n+23)=0n=20a20=8+19(4)=84
20 terms; last term = 84.
5-Mark Q45 MarksLevel 4 — HOTSCompetency Based
An auditorium contains 24 seats in the first row and 3 additional seats in each successive row. There are 20 rows. Find the seats in the last row, total seating capacity and total ticket revenue if every seat is sold at ₹150.
a=24, d=3, n=20a20=24+19(3)=81S20=20/2(24+81)=10×105=1050Revenue=1050×150=₹1,57,500
Last row = 81 seats; capacity = 1050; revenue = ₹1,57,500.
5-Mark Q55 MarksLevel 5 — Olympiad Challenge
Two APs are 3, 7, 11, … and 10, 13, 16, … . Find their first common term greater than 10, the next two common terms and the sum of these three common terms.
First AP terms are 3 modulo 4.Second AP terms are 1 modulo 3.Testing numbers of the first AP above 10:11 is not in second AP.15 is not in second AP.19 = 10+3(3)So 19 is the first common term.Common terms repeat after LCM(4,3)=12.Next common terms: 31 and 43.Sum=19+31+43=93
Common terms: 19, 31, 43; sum = 93.
5-Mark Q65 MarksLevel 5 — Olympiad Challenge
The 10th term of an AP is 2 and the 20th term is 22. Find a, d, S30, and the sum of the 11th to 30th terms.
a+9d=2a+19d=2210d=20 ⇒ d=2a+18=2 ⇒ a=−16S30=30/2[−32+29(2)]=15(26)=390S10=10/2[−32+9(2)]=5(−14)=−7011th to 30th sum = 390−(−70)=460
a=−16, d=2, S30=390; required partial sum=460.
5-Mark Q75 MarksLevel 5 — Olympiad Challenge
Three positive consecutive terms of an AP have sum 27 and product 585. Find the terms. If the AP continues with the same difference, find the sum of its first 10 terms beginning from the smallest of these terms.
Let terms be 9−d, 9, 9+d.9(81−d²)=58581−d²=65d²=16d=4Terms = 5,9,13Continuing AP: a=5,d=4,n=10S10=10/2[10+9(4)]=5×46=230
Terms = 5,9,13; S10=230.
5-Mark Q85 MarksLevel 5 — Olympiad Challenge
If Sn = 4n²+2n, determine the AP, its 25th term and the sum of its 11th to 25th terms.
an=Sn−Sn−1=4n²+2n−[4(n−1)²+2(n−1)]=8n−2a=6, d=8a25=8(25)−2=198S25=4(625)+50=2550S10=4(100)+20=420Required sum=2550−420=2130
AP: 6,14,22,… ; a25=198; required sum=2130.
5-Mark Q95 MarksLevel 5 — Olympiad Challenge
If the mth term of an AP is n and the nth term is m, m ≠ n, prove that the sum of the first m+n terms is (m+n)(m+n−1)/2.
a+(m−1)d=na+(n−1)d=md=−1a=m+n−1Let N=m+n.SN=N/2[2(N−1)+(N−1)(−1)]=N/2[N−1]=N(N−1)/2= (m+n)(m+n−1)/2
Hence proved.
5-Mark Q105 MarksLevel 5 — Olympiad ChallengeRoot Analysis
A prize scheme begins with ₹2,000 for first place and each successive prize is ₹100 less. If the total budget used is ₹15,500, find the possible values of n algebraically and determine the physically meaningful number of prizes.
a=2000, d=−100, Sn=1550015500=n/2[4000−100(n−1)]15500=n/2(4100−100n)310=n(41−n)n²−41n+310=0(n−10)(n−31)=0n=10 or n=31For n=31, last prize = 2000−30(100)=−₹1000
Reject n=31A negative prize amount is not meaningful in this context.
The meaningful number of prizes is 10.
Section F — 5 Case Studies
Case Study 1 — Auditorium Seating4 Sub-QuestionsFoundation → HOTS
An auditorium has 18 seats in the first row. Every successive row has 2 more seats than the previous row. There are 25 rows.
(a) Identify a and d.
(b) Find the number of seats in the 10th row.
(c) Find the total number of seats in all 25 rows.
(d) Find the total number of seats from the 11th to 25th rows.
(a)a=18, d=2
(b)a10=18+9(2)=36
(c)S25=25/2[36+24(2)]=25/2(84)=1050
(d)S10=10/2[36+9(2)]=5(54)=270Required=1050−270=780
a=18, d=2; 10th row=36; total=1050; rows 11–25 total=780.
Case Study 2 — Monthly Savings4 Sub-QuestionsFoundation → HOTS
A student saves ₹500 in the first month and increases the monthly saving by ₹100 each month for one year.
(a) Write a and d.
(b) Find the saving in the 6th month.
(c) Find the total saving in 12 months.
(d) Find the average monthly saving and explain its relation to the first and last monthly deposits.
(a)a=₹500, d=₹100
(b)a6=500+5(100)=₹1000
(c)a12=500+11(100)=₹1600S12=12/2(500+1600)=6×2100=₹12,600
(d)Average=12600/12=₹1050(First+Last)/2=(500+1600)/2=₹1050
Average monthly saving is the arithmetic mean of the first and last deposits.
Case Study 3 — Staircase Blocks4 Sub-QuestionsStandard → HOTS
A stepped display uses 30 blocks in its bottom row, 28 in the next row, 26 in the next, and continues until the top row has 2 blocks.
(a) Identify a and d.
(b) Find the number of blocks in the 8th row.
(c) Determine the number of rows and total blocks.
(d) If the first three rows are removed, how many blocks remain?
(a)a=30, d=−2
(b)a8=30+7(−2)=16
(c)2=30+(n−1)(−2)n=15S15=15/2(30+2)=240
(d)First three rows=30+28+26=84Remaining=240−84=156
8th row=16; 15 rows; total=240; remaining after removal=156.
Case Study 4 — Tree Plantation4 Sub-QuestionsStandard → HOTS
A plantation project places 12 saplings in the first row and 3 more saplings in each successive row. There are 18 rows.
(a) Write the AP represented by the first four rows.
(b) Find the number of saplings in the 18th row.
(c) Find the total number of saplings.
(d) Find the number of saplings in rows 6 through 12 inclusive.
(a)12, 15, 18, 21, …
(b)a18=12+17(3)=63
(c)S18=18/2(12+63)=9×75=675
(d)Rows 6–12 = S12−S5S12=12/2[24+11(3)]=342S5=5/2[24+4(3)]=90Required=342−90=252
18th row=63; total=675; rows 6–12 contain 252 saplings.
Case Study 5 — School Prize Distribution4 Sub-QuestionsAdvanced → Olympiad
A school plans 12 merit prizes. The first prize is ₹5,000 and each successive prize is ₹250 less than the preceding one.
(a) Identify a and d.
(b) Find the 12th prize.
(c) Find the total amount needed for all 12 prizes.
(d) If only ₹40,000 is available, what is the greatest number of top prizes that can be awarded according to this pattern?
(a)a=5000, d=−250
(b)a12=5000−11(250)=₹2250
(c)S12=12/2(5000+2250)=₹43,500
(d)S10=10/2[10000−9(250)]=5(7750)=₹38,750S11=11/2[10000−10(250)]=11/2(7500)=₹41,250₹38,750 ≤ ₹40,000 but ₹41,250 > ₹40,000
At most 10 top prizes can be awarded.
60-Second Arithmetic Progressions Revision
Definition
AP ⇒ constant difference between consecutive terms.
Common Difference
d = later term − previous term.
nth Term
an = a+(n−1)d.
Last Term
l = a+(n−1)d.
Sum
Sn = n[2a+(n−1)d]/2.
Alternative Sum
Sn = n(a+l)/2.
Arithmetic Mean
For a, A, b in AP: A=(a+b)/2.
Term from End
rth from end = an−r+1.
Partial Sum
11th to 25th = S25−S10.
Final Check
Number of terms must be a positive integer.
Can You Solve These Without Looking at the Formula Sheet?
Five additional ungraded HOTS/Olympiad questions. They are separate from the required 95 main entries.
Challenge 1Olympiad
In an AP, a4+a8 = 30. Find a6.
Terms 4 and 8 are equally distant from term 6.a4+a8=2a630=2a6
a6=15.
Challenge 2Olympiad
If Sn=2n², find a50 without first writing the whole AP.
an=Sn−Sn−1=2n²−2(n−1)²=4n−2a50=200−2=198
198.
Challenge 3Olympiad
Five consecutive terms of an AP have sum 100. Find the middle term without determining a or d.
Write the five terms as m−2d, m−d, m, m+d, m+2d.
Sum = 5m5m=100m=20Middle term = 20.
Challenge 4Olympiad
If a2+a5+a8=36, find a5.
a2 and a8 are equally distant from a5.a2+a8=2a5Therefore total=3a53a5=36
a5=12.
Challenge 5Olympiad
The sum of the first n terms of an AP is 3n²−n. Find the first three terms and verify directly that their sum equals S3.
an=Sn−Sn−1=3n²−n−[3(n−1)²−(n−1)]=6n−4a1=2, a2=8, a3=14Direct sum=2+8+14=24S3=3(9)−3=24
First three terms are 2,8,14 and the sum is verified.
SK Tuitions • Class 10 Mathematics
Identify a and d → decide whether the question asks for a term or a sum → use n−1 carefully → check the meaning of every root and answer.
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