Important Questions Class 10 Maths Chapter 5: Arithmetic Progressions With Solutions

An arithmetic progression is a sequence of numbers in which each term increases or decreases by a fixed value. The fixed value is called the common difference, and it can be positive, negative, or zero.

Reena’s salary starts at ₹8000 and grows by ₹500 every year. Shakila puts ₹100 in a money box and adds ₹50 each birthday. Both are arithmetic progressions, and Chapter 5 gives students a direct way to calculate any term or total. CBSE 2026 tests Important Questions Class 10 Maths Chapter 5 through identifying sequences, finding the nth term, calculating the sum of n terms, and solving real-life word problems. These questions follow the NCERT Reprint 2026-27 scope.

Key Takeaways

• Quadratic Equations: Chapter 4 focuses on standard form, factorisation, discriminant, roots, and word problems.
• Main Rule: D = b² - 4ac decides whether roots are distinct, equal, or not real.
• Exam Trap: Move all terms to one side before identifying a, b, and c.
• NCERT Update: Standalone completing-the-square exercise questions are not in the current NCERT Reprint 2026-27.

Important Questions Class 10 Maths Chapter 5 Structure 2026

Important Questions Class 10 Maths Chapter 5: Key Formulas

These formulas cover the main calculation patterns in Arithmetic Progressions. Use the nth term formula for position-based questions and the sum formula for total-value questions.

Q1. What Is The General Form Of An Arithmetic Progression?

The general form of an arithmetic progression is a, a + d, a + 2d, a + 3d, .... Here, d is the common difference.

1. Given Data:
   First term = a
   Common difference = d
2. Formula Used:
   Terms = a, a + d, a + 2d, a + 3d, ...
3. Final Result:
   General form = a, a + d, a + 2d, a + 3d, ...

Q2. What Is The nth Term Of AP Class 10 Formula?

The nth term of AP class 10 formula is an = a + (n - 1)d. It gives the term at any required position.

1. Given Data:
   First term = a
   Common difference = d
   Position = n
2. Formula Used:
   an = a + (n - 1)d
3. Final Result:
   nth term = a + (n - 1)d

Q3. What Is The Sum Of n Terms Class 10 AP Formula?

The sum of n terms class 10 AP formula is Sn = n/2[2a + (n - 1)d]. Use it when the last term is not directly given.

1. Given Data:
   First term = a
   Common difference = d
   Number of terms = n
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Final Result:
   Sum of n terms = n/2[2a + (n - 1)d]

Q4. What Is The Sum Formula When The Last Term Is Known?

When the last term is known, use Sn = n/2(a + l). Here, l is the last term.

1. Given Data:
   First term = a
   Last term = l
   Number of terms = n
2. Formula Used:
   Sn = n/2(a + l)
3. Final Result:
   Sum = n/2(a + l)

Q5. What Is Arithmetic Mean Class 10 In AP?

The arithmetic mean class 10 formula is b = (a + c)/2. It gives the middle term when a, b, c are in AP.

1. Given Data:
   Three terms in AP = a, b, c
2. Formula Used:
   b = (a + c)/2
3. Final Result:
   Arithmetic mean = (a + c)/2

Arithmetic Progression Class 10 MCQ With Answers

These arithmetic progression class 10 MCQ questions test formula use, term identification, and sum calculation. Each question follows common CBSE 2026 patterns.

Q1. What Is The Common Difference Of The AP 1/3, 5/3, 9/3, 13/3?

The common difference is 4/3. Subtract any term from the next term.

1. Given Data:
   AP = 1/3, 5/3, 9/3, 13/3
2. Formula Used:
   d = second term - first term
3. Calculation:
   d = 5/3 - 1/3
   d = 4/3
4. Final Result:
   Answer: (B) 4/3

Q2. What Is The 30th Term Of The AP 10, 7, 4, ...?

The 30th term is -77. The common difference is -3.

1. Given Data:
   a = 10
   d = -3
   n = 30
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   a30 = 10 + (30 - 1)(-3)
   a30 = 10 + 29(-3)
   a30 = 10 - 87
4. Final Result:
   Answer: (C) -77

Q3. Three Numbers In AP Have Sum 30. What Is The Middle Term?

The middle term is 10. In three AP terms, the middle term equals one-third of the sum.

1. Given Data:
   Three AP terms have sum 30.
2. Formula Used:
   Terms = a - d, a, a + d
3. Calculation:
   (a - d) + a + (a + d) = 30
   3a = 30
   a = 10
4. Final Result:
   Answer: (B) 10

Q4. Which Term Of The AP 3, 8, 13, 18, ... Is 78?

78 is the 16th term. Use the nth term formula.

1. Given Data:
   a = 3
   d = 5
   an = 78
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   78 = 3 + (n - 1)(5)
   75 = 5(n - 1)
   n - 1 = 15
   n = 16
4. Final Result:
   Answer: (C) 16th

Q5. If The nth Term Of An AP Is 4n - 3, What Is The Common Difference?

The common difference is 4. Find the first two terms and subtract.

1. Given Data:
   an = 4n - 3
2. Formula Used:
   d = a2 - a1
3. Calculation:
   a1 = 4(1) - 3 = 1
   a2 = 4(2) - 3 = 5
   d = 5 - 1
4. Final Result:
   Answer: (A) 4

Q6. What Is The Sum Of The First 10 Terms Of The AP 2, 7, 12, ...?

The sum of the first 10 terms is 245. Use the AP sum formula.

1. Given Data:
   a = 2
   d = 5
   n = 10
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Calculation:
   S10 = 10/2[2(2) + 9(5)]
   S10 = 5[4 + 45]
   S10 = 5 × 49
4. Final Result:
   Answer: (A) 245

Q7. If a = 5, d = 3, And an = 50, Find n.

The value of n is 16. Substitute values in the nth term formula.

1. Given Data:
   a = 5
   d = 3
   an = 50
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   50 = 5 + (n - 1)(3)
   45 = 3(n - 1)
   n - 1 = 15
   n = 16
4. Final Result:
   Answer: (C) 16

Q8. What Is The 11th Term Of The AP -3, -1/2, 2, ...?

The 11th term is 22. The common difference is 5/2.

1. Given Data:
   a = -3
   d = -1/2 - (-3) = 5/2
   n = 11
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   a11 = -3 + 10(5/2)
   a11 = -3 + 25
   a11 = 22
4. Final Result:
   Answer: (B) 22

Q9. Two APs Have The Same Common Difference. Their 100th Terms Differ By 100. What Is The Difference Between Their 1000th Terms?

The difference is 100. APs with the same common difference keep the same corresponding-term difference.

1. Given Data:
   Both APs have the same d.
   Difference between 100th terms = 100
2. Rule Used:
   Equal common differences keep term-wise difference constant.
3. Final Result:
   Answer: (B) 100

Q10. How Many Two-Digit Numbers Are Divisible By 3?

There are 30 two-digit numbers divisible by 3. They form the AP 12, 15, ..., 99.

1. Given Data:
   First two-digit multiple of 3 = 12
   Last two-digit multiple of 3 = 99
   d = 3
2. Formula Used:
   l = a + (n - 1)d
3. Calculation:
   99 = 12 + (n - 1)(3)
   87 = 3(n - 1)
   n - 1 = 29
   n = 30
4. Final Result:
   Answer: (B) 30

AP Class 10 Important Questions On Identifying An Arithmetic Progression

These AP class 10 important questions begin with checking consecutive differences. If the difference stays constant, the list forms an arithmetic progression.

Q1. Does 2, 5/2, 3, 7/2, ... Form An AP?

Yes, it forms an AP with common difference 1/2. The next two terms are 4 and 9/2.

1. Given Data:
   Sequence = 2, 5/2, 3, 7/2, ...
2. Formula Used:
   d = next term - previous term
3. Calculation:
   5/2 - 2 = 1/2
   3 - 5/2 = 1/2
   7/2 - 3 = 1/2
4. Final Result:
   It forms an AP. d = 1/2. Next terms = 4, 9/2

Q2. Does -1.2, -3.2, -5.2, -7.2, ... Form An AP?

Yes, it forms an AP with common difference -2. The next two terms are -9.2 and -11.2.

1. Given Data:
   Sequence = -1.2, -3.2, -5.2, -7.2, ...
2. Formula Used:
   d = next term - previous term
3. Calculation:
   -3.2 - (-1.2) = -2
   -5.2 - (-3.2) = -2
4. Final Result:
   It forms an AP. d = -2. Next terms = -9.2, -11.2

Q3. Does -10, -6, -2, 2, ... Form An AP?

Yes, it forms an AP with common difference 4. The next two terms are 6 and 10.

1. Given Data:
   Sequence = -10, -6, -2, 2, ...
2. Formula Used:
   d = next term - previous term
3. Calculation:
   -6 - (-10) = 4
   -2 - (-6) = 4
   2 - (-2) = 4
4. Final Result:
   It forms an AP. d = 4. Next terms = 6, 10

Q4. Does 0, -4, -8, -12, ... Form An AP?

Yes, it forms an AP with common difference -4. The next two terms are -16 and -20.

1. Given Data:
   Sequence = 0, -4, -8, -12, ...
2. Formula Used:
   d = next term - previous term
3. Calculation:
   -4 - 0 = -4
   -8 - (-4) = -4
   -12 - (-8) = -4
4. Final Result:
   It forms an AP. d = -4. Next terms = -16, -20

Q5. Does Taxi Fare With ₹15 For First km And ₹8 For Each Additional km Form An AP?

Yes, the taxi fare forms an AP with common difference ₹8. The fares increase by a fixed amount.

1. Given Data:
   First km fare = ₹15
   Additional km charge = ₹8
2. Sequence:
   15, 23, 31, 39, ...
3. Rule Used:
   Consecutive difference must remain constant.
4. Final Result:
   It forms an AP with d = 8

Q6. Does Compound Interest Amount Form An AP?

No, compound interest amount does not form an AP. It grows by a fixed percentage, not a fixed difference.

1. Given Data:
   Amount grows by compound interest.
2. Rule Used:
   AP needs a fixed difference.
3. Final Result:
   Compound interest amount does not form an AP

Q7. Write The First Four Terms Of The AP When a = 4 And d = -3.

The first four terms are 4, 1, -2, -5. Add d each time.

1. Given Data:
   a = 4
   d = -3
2. Formula Used:
   Terms = a, a + d, a + 2d, a + 3d
3. Calculation:
   a1 = 4
   a2 = 4 - 3 = 1
   a3 = 1 - 3 = -2
   a4 = -2 - 3 = -5
4. Final Result:
   4, 1, -2, -5

nth Term Of AP Class 10 Important Questions With Solutions

These nth term of AP class 10 important questions use an = a + (n - 1)d. They ask for a term, position, or unknown first term.

Q1. Find The 10th Term Of The AP 2, 7, 12, ...

The 10th term is 47. The common difference is 5.

1. Given Data:
   a = 2
   d = 5
   n = 10
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   a10 = 2 + (10 - 1)(5)
   a10 = 2 + 45
   a10 = 47
4. Final Result:
   10th term = 47

Q2. Which Term Of The AP 21, 18, 15, ... Is -81? Is 0 A Term?

-81 is the 35th term, and 0 is the 8th term. Both positions are positive integers.

1. Given Data:
   a = 21
   d = -3
2. Formula Used:
   an = a + (n - 1)d
3. For -81:
   -81 = 21 + (n - 1)(-3)
   -102 = -3(n - 1)
   n - 1 = 34
   n = 35
4. For 0:
   0 = 21 + (n - 1)(-3)
   -21 = -3(n - 1)
   n - 1 = 7
   n = 8
5. Final Result:
   -81 is the 35th term, and 0 is the 8th term

Q3. Determine The AP Whose 3rd Term Is 5 And 7th Term Is 9.

The AP is 3, 4, 5, 6, 7, .... Use the two term equations to find a and d.

1. Given Data:
   a3 = 5
   a7 = 9
2. Formula Used:
   an = a + (n - 1)d
3. Equation Formation:
   a + 2d = 5
   a + 6d = 9
4. Calculation:
   Subtract first equation from second.
   4d = 4
   d = 1
   a + 2 = 5
   a = 3
5. Final Result:
   AP = 3, 4, 5, 6, 7, ...

Q4. Check Whether 301 Is A Term Of 5, 11, 17, 23, ...

301 is not a term of this AP. The value of n does not become a positive integer.

1. Given Data:
   a = 5
   d = 6
   an = 301
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   301 = 5 + (n - 1)(6)
   296 = 6(n - 1)
   n - 1 = 296/6
   n - 1 = 148/3
4. Final Result:
   301 is not a term of this AP

Q5. Find The 31st Term Of An AP Whose 11th Term Is 38 And 16th Term Is 73.

The 31st term is 178. First find the common difference.

1. Given Data:
   a11 = 38
   a16 = 73
2. Formula Used:
   an = a + (n - 1)d
3. Equation Formation:
   a + 10d = 38
   a + 15d = 73
4. Calculation:
   Subtract equations.
   5d = 35
   d = 7
   a = 38 - 70
   a = -32
5. Find a31:
   a31 = -32 + 30(7)
   a31 = 178
6. Final Result:
   31st term = 178

Q6. The 17th Term Of An AP Exceeds Its 10th Term By 7. Find Common Difference.

The common difference is 1. Subtract the 10th term from the 17th term.

1. Given Data:
   a17 - a10 = 7
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   [a + 16d] - [a + 9d] = 7
   7d = 7
   d = 1
4. Final Result:
   d = 1

Q7. Fill The Missing Term In 2, __, 26.

The missing term is 14. The middle term is the arithmetic mean.

1. Given Data:
   Terms = 2, __, 26
2. Formula Used:
   Middle term = (first term + third term)/2
3. Calculation:
   Missing term = (2 + 26)/2
   Missing term = 28/2
   Missing term = 14
4. Final Result:
   2, 14, 26

Q8. Fill The Missing Terms In __, 13, __, 3.

The missing terms are 18 and 8. The common difference is -5.

1. Given Data:
   Terms = __, 13, __, 3
2. Formula Used:
   d = (fourth term - second term)/(4 - 2)
3. Calculation:
   d = (3 - 13)/2
   d = -10/2
   d = -5
4. Find Missing Terms:
   First term = 13 - (-5) = 18
   Third term = 13 + (-5) = 8
5. Final Result:
   18, 13, 8, 3

Q9. How Many Three-Digit Numbers Are Divisible By 7?

There are 128 three-digit numbers divisible by 7. They form the AP 105, 112, ..., 994.

1. Given Data:
   a = 105
   d = 7
   l = 994
2. Formula Used:
   l = a + (n - 1)d
3. Calculation:
   994 = 105 + (n - 1)(7)
   889 = 7(n - 1)
   n - 1 = 127
   n = 128
4. Final Result:
   128 three-digit numbers

Sum Of n Terms Class 10 AP Questions With Solutions

These sum of n terms class 10 AP questions use Sn = n/2[2a + (n - 1)d]. If the last term is known, use Sn = n/2(a + l).

Q1. Find The Sum Of The First 22 Terms Of The AP 8, 3, -2, ...

The sum of the first 22 terms is -979. The AP has common difference -5.

1. Given Data:
   a = 8
   d = -5
   n = 22
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Calculation:
   S22 = 22/2[2(8) + 21(-5)]
   S22 = 11[16 - 105]
   S22 = 11(-89)
4. Final Result:
   S22 = -979

Q2. If S14 = 1050 And a = 10, Find The 20th Term.

The 20th term is 200. First find the common difference using S14.

1. Given Data:
   S14 = 1050
   a = 10
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Calculation:
   1050 = 14/2[2(10) + 13d]
   1050 = 7[20 + 13d]
   150 = 20 + 13d
   d = 10
4. Find a20:
   a20 = 10 + 19(10)
   a20 = 200
5. Final Result:
   20th term = 200

Q3. How Many Terms Of The AP 24, 21, 18, ... Must Be Taken To Get Sum 78?

The required number of terms is 4 or 13. Both values satisfy the sum equation.

1. Given Data:
   a = 24
   d = -3
   Sn = 78
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Calculation:
   78 = n/2[48 + (n - 1)(-3)]
   156 = n(51 - 3n)
   156 = 51n - 3n²
   3n² - 51n + 156 = 0
4. Factorisation:
   n² - 17n + 52 = 0
   (n - 4)(n - 13) = 0
5. Final Result:
   n = 4 or n = 13

Q4. Find The Sum Of The First 40 Positive Integers Divisible By 6.

The sum is 4920. The AP is 6, 12, 18, ...

1. Given Data:
   a = 6
   d = 6
   n = 40
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Calculation:
   S40 = 40/2[2(6) + 39(6)]
   S40 = 20[12 + 234]
   S40 = 20 × 246
4. Final Result:
   S40 = 4920

Q5. Find The Sum Of First 51 Terms Of An AP Whose 2nd Term Is 14 And 3rd Term Is 18.

The sum of the first 51 terms is 5610. First find d and a.

1. Given Data:
   a2 = 14
   a3 = 18
2. Formula Used:
   d = a3 - a2
   Sn = n/2[2a + (n - 1)d]
3. Calculation:
   d = 18 - 14 = 4
   a = 14 - 4 = 10
4. Sum:
   S51 = 51/2[2(10) + 50(4)]
   S51 = 51/2 × 220
   S51 = 5610
5. Final Result:
   S51 = 5610

Q6. Find The Sum Of Odd Numbers Between 0 And 50.

The sum of odd numbers between 0 and 50 is 625. The AP is 1, 3, 5, ..., 49.

1. Given Data:
   a = 1
   d = 2
   l = 49
2. Formula Used:
   n = [(l - a)/d] + 1
   Sn = n/2(a + l)
3. Calculation:
   n = [(49 - 1)/2] + 1
   n = 24 + 1
   n = 25
4. Sum:
   S25 = 25/2(1 + 49)
   S25 = 25/2 × 50
   S25 = 625
5. Final Result:
   625

Q7. If Sn = 4n - n², Find The 1st Term, 2nd Term, And nth Term.

The first term is 3, the second term is 1, and the nth term is 5 - 2n.

1. Given Data:
   Sn = 4n - n²
2. Formula Used:
   an = Sn - S(n - 1)
3. Calculation:
   S1 = 4(1) - 1² = 3
   S2 = 4(2) - 2² = 4
   Second term = S2 - S1 = 1
4. nth Term:
   an = Sn - S(n - 1)
   an = (4n - n²) - [4(n - 1) - (n - 1)²]
   an = 5 - 2n
5. Final Result:
   First term = 3, second term = 1, nth term = 5 - 2n

Word Problems On AP Class 10 With Solutions

These word problems on AP class 10 questions require students to identify the first term, common difference, and required value. They cover salaries, savings, production, rows, penalties, and divisible numbers.

Q1. A Manufacturer Produced 600 TV Sets In The 3rd Year And 700 In The 7th Year. Find Production Values.

The first year production is 550 sets, the 10th year production is 775 sets, and total production in 7 years is 4375 sets.

1. Given Data:
   3rd year production = 600
   7th year production = 700
2. Formula Used:
   an = a + (n - 1)d
3. Equation Formation:
   a + 2d = 600
   a + 6d = 700
4. Calculation:
   Subtract equations.
   4d = 100
   d = 25
   a = 550
5. Required Values:
   a10 = 550 + 9(25) = 775
   S7 = 7/2[2(550) + 6(25)]
   S7 = 7/2 × 1250
   S7 = 4375
6. Final Result:
   1st year = 550 sets, 10th year = 775 sets, total = 4375 sets

Q2. ₹1000 Is Invested At 8% Simple Interest. Find Interest At The End Of 30 Years Using AP.

The interest at the end of 30 years is ₹2400. Simple interest increases by the same amount each year.

1. Given Data:
   Principal = ₹1000
   Rate = 8% per annum
2. AP Formed:
   Yearly accumulated interest = 80, 160, 240, ...
3. Formula Used:
   an = a + (n - 1)d
4. Calculation:
   a = 80
   d = 80
   n = 30
   a30 = 80 + 29(80)
   a30 = 80 × 30
5. Final Result:
   Interest after 30 years = ₹2400

Q3. A Flower Bed Has 23 Rose Plants In The First Row, 21 In The Second, And So On. Last Row Has 5 Plants. Find Rows.

There are 10 rows. The number of plants decreases by 2 in each row.

1. Given Data:
   a = 23
   d = -2
   an = 5
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   5 = 23 + (n - 1)(-2)
   -18 = -2(n - 1)
   n - 1 = 9
   n = 10
4. Final Result:
   Number of rows = 10

Q4. Subba Rao Started Work In 1995 At ₹5000 Salary With ₹200 Increment. When Did His Income Reach ₹7000?

His income reached ₹7000 in 2005. The annual salary forms an AP.

1. Given Data:
   Starting year = 1995
   a = 5000
   d = 200
   an = 7000
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   7000 = 5000 + (n - 1)(200)
   2000 = 200(n - 1)
   n - 1 = 10
   n = 11
4. Year:
   1995 + 10 = 2005
5. Final Result:
   Income reached ₹7000 in 2005

Q5. 200 Logs Are Stacked With 20 In Bottom Row, 19 In Next Row, And So On. Find Rows And Top Row Logs.

There are 16 rows, and the top row has 5 logs. The 25-row result is rejected because it gives negative logs.

1. Given Data:
   a = 20
   d = -1
   Sn = 200
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Calculation:
   200 = n/2[2(20) + (n - 1)(-1)]
   200 = n/2(41 - n)
   400 = 41n - n²
   n² - 41n + 400 = 0
4. Factorisation:
   (n - 16)(n - 25) = 0
   n = 16 or n = 25
   n = 25 gives an impossible negative top row.
5. Top Row:
   a16 = 20 + 15(-1)
   a16 = 5
6. Final Result:
   16 rows and 5 logs in the top row

Q6. A Contractor Pays ₹200 On Day 1, ₹250 On Day 2, ₹300 On Day 3. Find Total Penalty For 30 Days.

The total penalty is ₹27,750. The daily penalty forms an AP.

1. Given Data:
   a = 200
   d = 50
   n = 30
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Calculation:
   S30 = 30/2[2(200) + 29(50)]
   S30 = 15[400 + 1450]
   S30 = 15 × 1850
4. Final Result:
   Total penalty = ₹27,750

Arithmetic Progression Class 10 Extra Questions With Solutions

These arithmetic progression class 10 extra questions with solutions cover higher-order patterns. They combine nth term, sum, and real-life AP applications.

Q1. If S7 = 49 And S17 = 289, Find Sn.

The sum of n terms is Sn = n². First find a and d.

1. Given Data:
   S7 = 49
   S17 = 289
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Equation Formation:
   S7 = 7/2[2a + 6d] = 49
   7(a + 3d) = 49
   a + 3d = 7
4. Second Equation:
   S17 = 17/2[2a + 16d] = 289
   17(a + 8d) = 289
   a + 8d = 17
5. Calculation:
   Subtract equations.
   5d = 10
   d = 2
   a = 1
6. Final Result:
   Sn = n/2[2(1) + (n - 1)2] = n²

Q2. Ramkali Saves ₹5 In The First Week And Increases Savings By ₹1.75 Every Week. When Will Savings Become ₹20.75?

Ramkali’s weekly saving becomes ₹20.75 in the 10th week. The savings form an AP.

1. Given Data:
   a = 5
   d = 1.75
   an = 20.75
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   20.75 = 5 + (n - 1)(1.75)
   15.75 = 1.75(n - 1)
   n - 1 = 9
   n = 10
4. Final Result:
   10th week

Q3. Find The Sum 7 + 10.5 + 14 + ... + 84.

The sum is 1046.5. The AP has common difference 3.5.

1. Given Data:
   a = 7
   d = 3.5
   l = 84
2. Formula Used:
   l = a + (n - 1)d
   Sn = n/2(a + l)
3. Calculation:
   84 = 7 + (n - 1)(3.5)
   77 = 3.5(n - 1)
   n - 1 = 22
   n = 23
4. Sum:
   S23 = 23/2(7 + 84)
   S23 = 23/2 × 91
5. Final Result:
   1046.5

Q4. Which Term Of The AP 3, 15, 27, 39, ... Is 132 More Than Its 54th Term?

The required term is the 65th term. First find the 54th term.

1. Given Data:
   a = 3
   d = 12
2. Formula Used:
   an = a + (n - 1)d
3. Calculation:
   a54 = 3 + 53(12)
   a54 = 639
   Required term = 639 + 132
   Required term = 771
4. Find n:
   771 = 3 + (n - 1)(12)
   768 = 12(n - 1)
   n - 1 = 64
   n = 65
5. Final Result:
   65th term

Q5. For What Value Of n Are nth Terms Of 63, 65, 67, ... And 3, 10, 17, ... Equal?

The value of n is 13. Equate the nth terms of both APs.

1. Given Data:
   AP 1 = 63, 65, 67, ...
   AP 2 = 3, 10, 17, ...
2. Formula Used:
   an = a + (n - 1)d
3. nth Term Of AP 1:
   an = 63 + (n - 1)2
   an = 61 + 2n
4. nth Term Of AP 2:
   an = 3 + (n - 1)7
   an = 7n - 4
5. Calculation:
   61 + 2n = 7n - 4
   65 = 5n
   n = 13
6. Final Result:
   n = 13

Class 10 AP Previous Year Questions With Solutions

These class 10 AP previous year questions follow repeated board-style patterns. They test arithmetic mean, nth term, and sum of terms.

Q1. Three Numbers In AP Have Sum 30. Find The Middle Term.

The middle term is 10. For three AP terms, the middle term is one-third of the sum.

1. Given Data:
   Three AP terms have sum 30.
2. Formula Used:
   Terms = a - d, a, a + d
3. Calculation:
   (a - d) + a + (a + d) = 30
   3a = 30
   a = 10
4. Final Result:
   Middle term = 10

Q2. A School Charity Run Has 10 Rounds. First Round Is 300 m. Each Later Round Increases By 50 m. Find Total Distance.

The total distance is 5250 m. The round distances form an AP.

1. Given Data:
   a = 300 m
   d = 50 m
   n = 10
2. Formula Used:
   Sn = n/2[2a + (n - 1)d]
3. Calculation:
   S10 = 10/2[2(300) + 9(50)]
   S10 = 5[600 + 450]
   S10 = 5 × 1050
4. Final Result:
   5250 m

Q3. An AP Has 50 Terms. The 3rd Term Is 12 And The Last Term Is 106. Find The 29th Term.

The 29th term is 64. Use the 3rd and 50th terms to find a and d.

1. Given Data:
   a3 = 12
   a50 = 106
2. Formula Used:
   an = a + (n - 1)d
3. Equation Formation:
   a + 2d = 12
   a + 49d = 106
4. Calculation:
   Subtract equations.
   47d = 94
   d = 2
   a = 8
5. Find a29:
   a29 = 8 + 28(2)
   a29 = 64
6. Final Result:
   29th term = 64

Q4. Sum Of 4th And 8th Terms Is 24. Sum Of 6th And 10th Terms Is 44. Find First Three Terms.

The first three terms are -13, -8, -3. Form equations using the nth term formula.

1. Given Data:
   a4 + a8 = 24
   a6 + a10 = 44
2. Formula Used:
   an = a + (n - 1)d
3. Equation Formation:
   (a + 3d) + (a + 7d) = 24
   2a + 10d = 24
   a + 5d = 12
4. Second Equation:
   (a + 5d) + (a + 9d) = 44
   2a + 14d = 44
   a + 7d = 22
5. Calculation:
   Subtract equations.
   2d = 10
   d = 5
   a = -13
6. Final Result:
   First three terms = -13, -8, -3

Arithmetic Progression Class 10 Questions And Answers: Most Repeated Variations

These arithmetic progression class 10 questions and answers cover recurring CBSE 2026 patterns. Students should practise identifying APs, nth term, sum, and word problems.

Q1. What Are The Most Repeated Variations In Arithmetic Progressions Class 10 Important Questions?

The most repeated variations are common difference, nth term, sum of n terms, arithmetic mean, missing terms, and word problems.

1. Identification Pattern:
   Check equal consecutive differences.
2. nth Term Pattern:
   Use an = a + (n - 1)d.
3. Sum Pattern:
   Use Sn = n/2[2a + (n - 1)d].
4. Final Result:
   These six patterns cover most Chapter 5 board-style questions

Q2. How Should Students Solve Important Questions Class 10 Maths Chapter 5 With Solutions?

Students should identify whether the question asks for a term, total, position, or common difference. Formula choice decides the solution path.

1. Read The Question Type:
   Term, sum, missing value, or word problem.
2. Identify Values:
   Find a, d, n, l, or Sn.
3. Apply Formula:
   Use nth term or sum formula.
4. Final Result:
   Correct value identification decides AP answers