CBSE Important Questions Class 10 Maths Ch5 Arithmetic Progressions 2025-26

The chapter Arithmetic Progressions (A.P.) is one of the most straightforward and scoring topics in Class 10 Mathematics Syllabus. It introduces you to sequences where the difference between consecutive terms remains constant.

This chapter strengthens logical reasoning and algebraic manipulation skills, helping you solve numerical and application-based problems easily. Since formulas are simple and repetitive, consistent practice ensures full marks in exams.

The Arithmetic Progressions Important Questions for Class 10 Mathematics help you:

• Master formulas related to nth term and sum of n terms.
• Solve both direct and real-life application-based problems.
• Build speed and accuracy for exam-oriented questions

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IMPORTANT QUESTIONS CLASS 10

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In the table given below, we have provided the links to downloadable Important Questions of Arithmetic Progression Class 10 PDFs. Now you can download them without requiring to login.

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Ch5 AP Important Questions Class 10 Maths 2025-26

1. Given below are the details of an experiment using a bucket and a mug to understand water consumption. 

The bucket's volume is 30 litres and the mug's volume is 1/20 of the bucket. The bucket has 1 litre of water before the tap is turned on. The tap is filling the bucket at a constant rate of 0.1 litres per second. Every 30 seconds, he takes a mug full of water from the bucket.

i) Write an arithmetic progression for the volume of water in the bucket every 30 seconds.

ii) Find the volume of water in the bucket after exactly 7.5 minutes. Show your work.

(Note: Assume no spillage of water.)

Soln. In every 30 seconds:

• Water added by the tap: 

0.1 litres/second×30 seconds

=3 litres

• Water removed using the mug: 

1.5 litres.

The net gain in water every 30 seconds:

Net water added

=3−1.5

=1.5 litres

Let the initial volume of water in the bucket be a₁=1 litre

After 30 seconds(1st term): a₂ =1+1.5

=2.5 litres

After 60 seconds(2nd term): a₃ =2.5 +1.5

=4.0 litres

a₁=1 

Common difference d= 1.5

AP= 1,2.5,4.0,5.5,…

⇒7.5minutes=7.5×60

=450seconds.

Each interval is 30 seconds.

∴ n= 450/30

n= 15

a_n​=a_1​+(n−1)⋅d

a₁₅=1+(15−1).1.5

a₁₅=1+(14).1.5

a₁₅=1+21

a₁₅=22

i) Arithmetic progression for the volume of water in the bucket every 30 seconds is 

AP= 1,2.5,4.0,5.5,…

ii) The volume of water in the bucket after exactly 7.5 minutes is 22 litres

2. The ratio of the sum of the first 11 terms of an arithmetic progression to the sum of its first 21 terms is given by 1:4.

i) Show that 23a + 10d = 0, where a is the first term and d is the common difference of the arithmetic progression.

ii) Write an expression for the nth term of the above arithmetic progression only in terms of a and n.Show your work.

Soln. 

S₁₁/S₂₁= ¼

i) Show that 23a + 10d = 0,

S_n​=n/2​(2a+(n−1)d),

Using the given ratio

S₁₁​=11/2​(2a+(11−1)d),

S₁₁​=11/2​(2a+(10)d),…………………1

S₂₁​=21/2​(2a+(21−1)d),

S₂₁​=21/2​(2a+(20)d), …………………2

Put equation 1 and 2 in the given ratio

S₁₁/S₂₁= ¼

(11/2​(2a+10d))/(21/2​(2a+20d))= ¼

4 (11/2​(2a+10d)) = (21/2​(2a+20d))

44(2a+10d)=21(2a+20d).

88a+440d=42a+420d.

88a−42a+440d−420d=0.(Combine like terms)

46a+20d=0. (divide by 2)

23a+10d=0. 

Hence proved

ii) Write an expression for the nth term

23a+10d=0.

10d=−23a

d= (-23a/10)

T_n​=a+(n−1)d,

T_n​=a+(n−1)(-23a/10),

T_n​=a−(23a​/10)(n−1).

T_n​=a(1−(23/10)(n−1))

T_n​=a(10−(23n-23)/10

T_n​=a(10−23n+23)/10

T_n​=a(33−23n)/10

3. Answer the questions based on the given information.

Below is a house of cards, a structure created by stacking playing cards on top of each other in the shape of a pyramid. Each small triangle is made using 3 cards, and each layer has one less triangle than the layer below it.

Ankit and his friends were having a sleepover and wanted to do something fun. One of the friends suggested that they could make a house of cards.

i) Ankit and his friends want to use 3 cards in the top layer and 18 in the bottom layer.

Form an AP showing the number of cards in each layer starting from the top layer.

ii) Ankit is planning to make a pyramid with the top and bottom layer containing 15 and 138 cards respectively.

How many layers will such a pyramid have? Show your work.

iii) They have a total of 360 cards with them.

Find the maximum number of layers that Ankit and his friends can make using the cards they have, if they want to have 1 triangle (3 cards) at the top layer. Show your work.

Soln. i) Ankit and his friends want to use 3 cards in the top layer and 18 in the bottom layer.

Form an AP showing the number of cards in each layer starting from the top layer.

Top layer: 3 cards

Bottom layer: 18 cards

The number of cards in each layer forms an arithmetic progression (AP), where:

The first term a=3

The last term, l=18

The common difference d is to be determined.

The general formula for the n-th term of an AP is:

a_n​=a+(n−1)d

a_n=18

18=3+(n−1)d

(n−1)d=15

⟹d= 15/(n-1)

The AP showing the cards in each layer starting from the top layer is:

3,3+d,3+2d,…,18.

ii) Ankit is planning to make a pyramid with the top and bottom layer containing 15 and 138 cards respectively.

How many layers will such a pyramid have? Show your work.

a_n​=a+(n−1)d

Substitute the values a_n=138, a=15 , common difference is d=3(since each subsequent layer adds 3 cards

138=15+(n−1)(3)

138=15+3(n−1)

138=15+3n−3

138=3n+12

138−12=3n.

126=3n

n= 126/3

n=42

Number of layers in the pyramid is 42

iii) They have a total of 360 cards with them.

Find the maximum number of layers that Ankit and his friends can make using the cards they have, if they want to have 1 triangle (3 cards) at the top layer. Show your work.

The number of cards in the layers will be:

3,6,9,12,…

 Total number of cards:  S_n​=n/2 [2a+(n−1)d]

360=n/2 [2(3)+(n−1)(3)]

360=n/2 [6+3(n−1)].

360=n/2 [6+3n−3]

360=n/2 [3+3n].

720=n[3+3n]

720=3n+3n²

3n²+3n−720=0

n²+n−240=0

n= 15 or n = -16

The maximum number of layers that can be made is:15

4. The average of an Arithmetic Progression with 151 terms is zero. One of its terms is zero.

Which term of the Arithmetic Progression is zero? Show your steps.

Soln. Average= Sum of all terms / Number of terms

S₁₅₁​=0

S_n​= n/2 (a+l)

For n=151, S₁₅₁​=0

0 ​= 151/2 (a+l)

This implies:

a+l=0⇒l=−a

Thus, the first term, a and the last term, l are negatives.

The AP has 151 terms, which is an odd number. In an AP with an odd number of terms, the middle term is the average of the AP. Since the average is 0, the middle term must be 0.

Middle term=(151+1)/2

Middle term= 76

The 76th term of the AP is 0.

5. Parth was receiving spam calls from a telemarketing centre. He got the first call at 2:48 pm as shown below.

If the telemarketer continues to call using the same pattern, at what time will Parth receive the 13th call? Show your work.

Soln. From the missed call times:

• 2:48 PM → 2:52 PM → 2:58 PM → 3:06 PM → 3:16 PM.

Let’s calculate the time differences between consecutive calls:

1. From 2:48 PM to 2:52 PM → 4 minutes
2. From 2:52 PM to 2:58 PM → 6 minutes
3. From 2:58 PM to 3:06 PM → 8 minutes
4. From 3:06 PM to 3:16 PM → 10 minutes

Thus, the intervals between calls are increasing by 2 minutes2 \, \text{minutes}2minutes each time. This forms an arithmetic progression (AP) for the time intervals..

The sequence of time intervals is:

4,6,8,10,……

Here:

• The first term (a) = 4 minutes
• The common difference (d) = 2 minutes

S_n​=n/2​[2a+(n−1)d],

Where n=12 (number of time intervals),

a=4 (first term),

d=2 (common difference).

S₁₂=12/2​[2(4)+(12−1)(2)],

S₁₂=6[8+11(2)]

S₁₂=6[8+22]

S₁₂=6×30

=180minutes

2:48PM+3hours=5:48PM

Parth will receive the 13th call at:5:48PM

7. Shivam bought a large quantity of marigold flowers to decorate his house for a family function. He used 630 flowers to recreate the pattern shown below. He used 7 flowers in the shortest garland and 35 flowers in the longest garland.

(Note: The figure is for visual representation only.)

If he kept the difference between two consecutive garlands the same, how many garlands did he make? Show your steps.

Soln. 

• First term a=7 (shortest garland),
• Last term l=35 (longest garland),
• Total sum S_n=630
• n is the total number of garlands (unknown),
• Common difference d (unknown).

The sum of the first n terms of an AP is given by:

S_n​=n/2(a+l)

630=n/2​(7+35)

630=n/2​(42)

1260=n⋅42.

n=1260​/42

n=30

The number of garlands Shivam made is:30

7. √2, √18, √50, √98...

Is the above pattern in AP? Justify your answer.

Soln.  d=a₂​−a₁​=a₃​−a_2​=a_4​−a₃​(constant)

First term: a₁=√2 ≈ 1.141

Second term:a₂=√18 

=√9.2

=3√2 ≈ 1.141

Third term:a₃=√50

=√25.2

=5√92 ≈ 7.071

Fourth term:a₄=√98

=√49.2

=7√2 ≈9.899

a₂​−a_1​=4.243−1.414

=2.829,

a₃−a₂=7.071−4.243

=2.828

a₄−a₃=9.899−7.071

=2.828

The differences are approximately:

2.829,2.828,2.828

While they are very close, they are not exactly the same due to rounding differences. However, notice that each term follows the pattern:

a_n​=(2n−1) √2

d=a_n+1​−a_n

a_n+1​=(2(n+1)−1)√2 and a_n=(2n−1)2a_n = (2n-1)√2:

d=[(2(n+1)−1)√2 ​]−[(2n−1)√2​].

=√2

Yes, the sequence is AP

8. Sana decided to start practicing for an upcoming marathon. She decided to gradually increase the duration. She ran for 10 mins on day 1 and increased the duration by 5 minutes every day.

From which day onwards will she be running for 2.5 hours or more? Show your work.

Soln.  On Day 1, she ran for 10 minutes10

• First term (a) = 10 minutes
• Common difference (d) = 5 minutes,
• We need to find the day (n) when the duration is 2.5 hours or more.

Convert 2.5 hours into minutes:

2.5 hours=2.5×60

=150 minutes

a_n​=a+(n−1)d,

150=10+(n−1)5.

140=5(n−1)

140/5=(n−1)

n−1=28

n= 29

From Day 29 onwards, Sana will be running for 2.5 hours (150 minutes) or more

9. Kevin is baking a tall layered wedding cake as shown below. The customer has ordered a 111 kg cake and 12 layers.

(Note: The image is for visual representation only.)

For the cake to stand properly, he makes the bottom-most cake of 17.5 kg and reduced the weight of each layer such that the difference in the weights of the consecutive layers is the SAME.

i) By what weight does he reduce each subsequent layer?

ii) What is the weight of the lightest cake layer?

Show your work.

Soln.  i) By what weight does he reduce each subsequent layer?

• The total weight of the cake = 111 kg,
• The number of layers = 12,
• The bottom-most layer weighs 17.5 kg,
• The weight of each layer decreases by a constant amount.

Let the weight of each layer form an arithmetic progression (AP) where:

• The first term a=17.5 (weight of the bottom-most layer),
• The number of terms n=12
• The sum of all terms S_n=111

S_n​=n/2​[2a+(n−1)d],

111= 12/2 [2(17.5)+(12−1)d]

111=6[2(17.5)+11d].

111=6[35+11d]

(111/6)=35+11d.

18.5=35+11d.

11d=18.5−35.

11d=−16.5

d= −16.5/11

d= −1.5.

ii) What is the weight of the lightest cake layer?

a_n​=a+(n−1)d,

a_12​=17.5+(12−1)(−1.5).

a_12​=17.5+11(−1.5).

a_12​=17.5−16.5

a_12​=1.0.

Some More Important Question Answer of Class 10 Arithmetic progression

Q1. Find the 20th term of the AP 7, 11, 15, …

Ans. 

1. First term a = 7. Common difference d = 11 − 7 = 4.
2. nth term formula: t_n = a + (n − 1)·d.
3. For n = 20: t_20 = 7 + (20 − 1)·4 = 7 + 19·4 = 7 + 76 = 83.

Answer: 83.

Q2. Find the 25th term of the AP 12, 9, 6, …

Ans. 

1. a = 12. d = 9 − 12 = −3.
2. t_25 = a + (25 − 1)·d = 12 + 24·(−3) = 12 − 72 = −60.
   Answer: −60.

Q3. In the AP 10, 8, 6, 4, 2, 0, ... find (a) the term-number at which term = 0 and (b) the sum of terms up to that term.

Ans. (a) t_n = 10 + (n − 1)(−2). Put t_n = 0: 10 − 2(n − 1) = 0 → −2(n − 1) = −10 → n − 1 = 5 → n = 6.
(b) Terms up to n = 6 are: 10, 8, 6, 4, 2, 0. Sum S_6 = 10 + 8 + 6 + 4 + 2 + 0 = 30. (Or use S_n = n/2*(first + last) = 6/2*(10 + 0) = 3*10 = 30.)
Answer: (a) 6th term is 0. (b) Sum = 30.

Q4. Find the sum of the first 30 terms of the AP 3, 7, 11, …

Ans. 

1. a = 3, d = 4. Number of terms n = 30.
2. Sum formula: S_n = n/2 * [2a + (n − 1)d].
3. S_30 = 30/2 * [2·3 + 29·4] = 15 * [6 + 116] = 15 * 122 = 1830.

Answer: 1830.

Q5. A staircase has 10 rows. The first row has 6 bricks and each next row has 4 more bricks than the previous. How many bricks in all?

Ans. 

1. AP with a = 6, d = 4, n = 10.
2. S_10 = 10/2 * [2·6 + (10 − 1)·4] = 5 * [12 + 36] = 5 * 48 = 240.
   Answer: 240 bricks.

Q6. Rahul saves money every month: first month ₹150, then he increases his saving by ₹25 every month. (a) What is the total after 12 months? (b) How many months until his total savings exceed ₹5000?

Ans. (a) AP with a = 150, d = 25, n = 12.
S_12 = 12/2 * [2·150 + (12 − 1)·25] = 6 * [300 + 275] = 6 * 575 = 3450.
Total after 12 months = ₹3450.
(b) We want the smallest integer n with S_n > 5000. Use S_n = n/2 * [2·150 + (n − 1)·25] > 5000.
Simplify: S_n = n/2 * (25n + 275) = (n(25n + 275))/2. Solve inequality 25n^2 + 275n − 10000 > 0 → n^2 + 11n − 400 > 0. Positive root ≈ 15.245. So the smallest integer n > 15.245 is n = 16.
Check S_15 = 4875 (<5000), S_16 = 5400 (>5000).
Answer: (a) ₹3450. (b) After 16 months he will exceed ₹5000.

Q7. The 5th term of an AP is 20 and the 12th term is 48. Find a, d and the sum of the first 20 terms.
Ans. 

1. t_5 = a + 4d = 20. (Eq 1)
2. t_12 = a + 11d = 48. (Eq 2)
3. Subtract Eq1 from Eq2: (a + 11d) − (a + 4d) = 48 − 20 → 7d = 28 → d = 4.
4. Substitute d in Eq1: a + 4·4 = 20 → a + 16 = 20 → a = 4.
5. Sum S_20 = 20/2 * [2a + (20 − 1)d] = 10 * [8 + 19·4] = 10 * [8 + 76] = 10 * 84 = 840.

Answer: a = 4, d = 4, S_20 = 840.

Q8. If 4x − 1, 2x + 7 and x + 13 are three consecutive terms of an AP, find x and write the three terms.
Ans. 

1. For three consecutive AP terms, middle term = average of neighbors: 2·(2x + 7) = (4x − 1) + (x + 13).
2. Left side: 4x + 14. Right side: 5x + 12. So 4x + 14 = 5x + 12.
3. Rearrange: 14 − 12 = 5x − 4x → 2 = x.
4. Substitute x = 2: terms = (4·2 − 1, 2·2 + 7, 2 + 13) = (8 − 1, 4 + 7, 15) = (7, 11, 15).

Answer: x = 2; the terms are 7, 11, 15.

Q9. The 10th term of an AP is 40 and the sum of the first 10 terms is 250. Find a and d.

Ans. 

1. t_10 = a + 9d = 40. (Eq A)
2. S_10 = 10/2 * [2a + 9d] = 250 ⇒ 5 * [2a + 9d] = 250 ⇒ 2a + 9d = 50. (Eq B)
3. From Eq A: a = 40 − 9d. Put into Eq B: 2(40 − 9d) + 9d = 50 → 80 − 18d + 9d = 50 → −9d = −30 → d = 10/3.
4. Then a = 40 − 9·(10/3) = 40 − 30 = 10.

Answer: a = 10, d = 10/3.

Q10. If S_n = n(3n + 1)/2 is the sum of the first n terms of a sequence, find the nth term t_n.

Ans. 

1. t_n = S_n − S_{n−1}. Given S_n = n(3n + 1)/2. Compute S_{n−1} = (n − 1)[3(n − 1) + 1]/2 = (n − 1)(3n − 2)/2.
2. t_n = [n(3n + 1) − (n − 1)(3n − 2)]/2. Expand numerator: (3n^2 + n) − (3n^2 − 5n + 2) = 3n^2 + n − 3n^2 + 5n − 2 = 6n − 2.
3. Divide by 2: t_n = (6n − 2)/2 = 3n − 1.

Answer: t_n = 3n − 1.

Q11. A classroom has 15 rows of chairs. The first row has 6 chairs and each next row has 2 more chairs than the previous. How many chairs in total, and how many chairs in the 15th row?

Ans. 

1. AP with a = 6, d = 2, n = 15.
2. 15th row (last term) = a + (15 − 1)·d = 6 + 14·2 = 6 + 28 = 34 chairs.
3. Total chairs: S_15 = 15/2 * (first + last) = 15/2 * (6 + 34) = 7.5 * 40 = 300.

Answer: 15th row = 34 chairs; total = 300 chairs.

Q12. A thief starts running at 50 m/min. After 1 minute a policeman starts chasing. The policeman runs 40 m in his first minute and then increases his distance by 10 m each minute (so minutes of policeman cover 40, 50, 60, ...). After how many minutes of the policeman's running will he catch the thief? (Find minutes after the policeman starts.)

Ans. 

1. Let m = minutes of policeman’s running when he catches the thief. Policeman’s total distance in m minutes = sum of AP 40 + 50 + ... + [40 + 10(m − 1)]. This sum = m/2 * [first + last] = m/2 * [40 + (40 + 10(m − 1))] = m/2 * (10m + 70) = 5m^2 + 35m.
2. Thief started 1 minute earlier, so time for thief at catch = m + 1 minute. Thief distance = 50*(m + 1).
3. At catch: policeman distance = thief distance: 5m^2 + 35m = 50(m + 1).
4. Simplify: 5m^2 + 35m = 50m + 50 → 5m^2 − 15m − 50 = 0 → divide by 5 → m^2 − 3m − 10 = 0.
5. Solve quadratic: (m − 5)(m + 2) = 0 → positive root m = 5.

Answer: The policeman catches the thief after 5 minutes of his own run (i.e., 6 minutes after the thief started).

Q13. If S_5 = 60 and S_10 = 150 for an AP, find the first term a and common difference d.

Ans. 

1. S_5 = 5/2 * [2a + 4d] = 60 ⇒ 5*(2a + 4d) = 120 ⇒ 2a + 4d = 24 ⇒ a + 2d = 12. (Eq 1)
2. S_10 = 10/2 * [2a + 9d] = 150 ⇒ 5*(2a + 9d) = 150 ⇒ 2a + 9d = 30. (Eq 2)
3. Subtract Eq1×2 from Eq2 (or subtract Eq1 from Eq2 after writing Eq1 as 2a + 4d = 24): (2a + 9d) − (2a + 4d) = 30 − 24 ⇒ 5d = 6 ⇒ d = 6/5.
4. From a + 2d = 12 ⇒ a = 12 − 2·(6/5) = 12 − 12/5 = (60/5 − 12/5) = 48/5.

Answer: a = 48/5, d = 6/5.

Q14. If the nth partial sum S_n = 5n^2 − n, find t_n (the nth term).

Ans. 

1. t_n = S_n − S_{n−1} = [5n^2 − n] − [5(n − 1)^2 − (n − 1)].
2. Expand: 5n^2 − n − (5n^2 − 10n + 5 − n + 1) = 5n^2 − n − 5n^2 + 10n − 5 + n − 1 = 10n − 6.

Answer: t_n = 10n − 6.

Q15. The 8th term of an AP is twice its 3rd term. If the 6th term is 22, find the AP (a and d).

Ans. 

1. t_8 = a + 7d; t_3 = a + 2d. Given a + 7d = 2(a + 2d) ⇒ a + 7d = 2a + 4d ⇒ move terms: a = 3d.
2. Given t_6 = a + 5d = 22. Substitute a = 3d: 3d + 5d = 22 ⇒ 8d = 22 ⇒ d = 22/8 = 11/4.
3. Then a = 3d = 3*(11/4) = 33/4.
    
4. Answer: a = 33/4, d = 11/4. (So AP: 33/4, 44/4, 55/4, ...)

Q16. What is the smallest natural number n such that 1 + 2 + ... + n ≥ 1000?

Ans. 

1. Sum of first n natural numbers S_n = n(n + 1)/2. We need n(n + 1)/2 ≥ 1000. Multiply by 2: n^2 + n − 2000 ≥ 0.
2. Solve n = [−1 + sqrt(1 + 8000)]/2 = [−1 + sqrt(8001)]/2 ≈ (−1 + 89.449)/2 ≈ 44.224. So the smallest integer n ≥ this is n = 45.
3. Check: S_44 = 44·45/2 = 990 < 1000; S_45 = 45·46/2 = 1035 ≥ 1000.

Answer: n = 45.

Q17. In an AP, the 3rd term is 15 and the 11th term is 47. Find a, d and the sum of the first 20 terms.

Ans. 

1. t_3 = a + 2d = 15. (Eq 1)
2. t_11 = a + 10d = 47. (Eq 2)
3. Subtract Eq1 from Eq2: 8d = 32 ⇒ d = 4. Then a = 15 − 2·4 = 7.
4. S_20 = 20/2 * [2a + 19d] = 10 * [14 + 76] = 10 * 90 = 900.

Answer: a = 7, d = 4, S_20 = 900.

Q18. The sum of the first 4 terms of an AP is 40 and the sum of the first 14 terms is 280. Find a, d and the sum of the first 20 terms.
Ans. 

1. S_4 = 4/2 * [2a + 3d] = 2*(2a + 3d) = 40 ⇒ 2a + 3d = 20. (Eq 1)
2. S_14 = 14/2 * [2a + 13d] = 7*(2a + 13d) = 280 ⇒ 2a + 13d = 40. (Eq 2)
3. Subtract Eq1 from Eq2: 10d = 20 ⇒ d = 2. Substitute in Eq1: 2a + 3·2 = 20 ⇒ 2a + 6 = 20 ⇒ a = 7.
4. S_20 = 20/2 * [2·7 + 19·2] = 10 * [14 + 38] = 10 * 52 = 520.

Answer: a = 7, d = 2, S_20 = 520.

Q19. The sum of the 3rd and 7th terms of an AP is 40, and the sum of the 8th and 14th terms is 100. Find a, d and the sum of the first 20 terms.

Ans. 

1. t_3 + t_7 = (a + 2d) + (a + 6d) = 2a + 8d = 40. (Eq 1)
2. t_8 + t_14 = (a + 7d) + (a + 13d) = 2a + 20d = 100. (Eq 2)
3. Subtract Eq1 from Eq2: 12d = 60 ⇒ d = 5. Then 2a + 8·5 = 40 ⇒ 2a + 40 = 40 ⇒ a = 0.
4. S_20 = 20/2 * [2·0 + 19·5] = 10 * 95 = 950.

Answer: a = 0, d = 5, S_20 = 950.

Q20. The sum of the first 15 terms of an AP is 750 and the first term is 15. Find the common difference and the 20th term.

Ans. 

1. S_15 = 15/2 * [2a + 14d] = 750. With a = 15 we have: 15/2 * [30 + 14d] = 750.
2. Multiply both sides by 2/15: 30 + 14d = 750 * 2/15 = 1500/15 = 100. So 14d = 70 ⇒ d = 5.
3. 20th term t_20 = a + 19d = 15 + 19·5 = 15 + 95 = 110.
   Answer: d = 5, t_20 = 110.

Extra Questions for Practice – Challenge Yourself! 

Q1. The 15th term of an AP is 60 and the 25th term is 100. Find the first term and common difference.

Q2. How many terms of the AP 7, 10, 13, ... are needed to make a sum of 301?

Q3. The 7th term of an AP is 32 and the 13th term is 62. Find the AP.

Q4. If the sum of the first n terms of an AP is 7n − n², find its nth term.

Q5. The sum of 6 terms of an AP is 42 and the sum of its next 6 terms is 102. Find the AP.

Q6. Find the sum of all multiples of 9 between 100 and 300.

Q7. How many terms of the AP 3, 7, 11, ... must be taken so that the sum is 406?

Q8. A sum of ₹1000 is to be paid in 10 installments. Each installment is ₹10 more than the preceding one. Find the first installment.

Q9. The 4th term of an AP is 21 and the 9th term is 46. Find the sum of its first 20 terms.

Q10. The sum of the first 10 terms of an AP is 210. The sum of the next 10 terms is 610. Find the AP.

How to Ace These Class 10 Arithmetic Progressions Important Questions

This chapter is easy to master once you understand the meaning of a common difference and how to apply formulas systematically. Follow this approach to prepare effectively:

Step 1: Understand What an Arithmetic Progression Is

A sequence is an Arithmetic Progression if the difference between any two consecutive terms is constant.
For example: 2, 4, 6, 8, 10... is an A.P. with a common difference (d) = 2.

Step 2: Learn the Standard Formulas

Memorise the key formulas used in this chapter:

• nth term (Tₙ) = a + (n – 1)d
• Sum of first n terms (Sₙ) = n/2 [2a + (n – 1)d]
• Sum of n terms (alternate form) = n/2 (a + l), where l is the last term

Step 3: Practise Finding nth Term and Common Difference

Start with identifying a (first term) and d (common difference). Practice finding any term or checking if a given sequence is an A.P.

Step 4: Learn to Find the Sum of n Terms

Practise problems where you must find the total of a series of numbers using the sum formulas. This type of question appears frequently in exams.

Step 5: Understand Application-Based Problems

Work on word problems involving real-life scenarios like savings, installment payments, and patterns. These questions test your understanding of formula usage in practical contexts.

Step 6: Check Each Step Carefully

Since A.P. questions are formula-driven, calculation accuracy is very important. Write each step clearly and verify your final answer.

Benefits of Solving Class 10 Arithmetic Progression Extra Questions

I understand that you are already aware of the benefits of using the class 10 AP extra questions, but let me reiterate the importance of using these questions:

• Solving extra AP questions in class 10 helps improve analytical and logical thinking.  These questions aim to enhance your reasoning abilities, equipping you to tackle not only AP issues but also intricate mathematical problems.
• Regular practice of arithmetic progression class 10 important questions ensures that you're well-prepared for your board exams.
• By solving the important question of AP class 10, you’ll learn to manage time more effectively during exams.
• As arithmetic progression is an important chapter in Class 10, regularly practicing extra questions ensures that you perform well in exams, thus boosting your overall score in mathematics
• Solving complex problems related to AP helps in building your confidence. When you know that you've worked through the toughest problems, you can approach the exam with a calm and composed mindset.

How to Use Class 10 Arithmetic Progression Questions

When studying arithmetic progression, working through Class 10 AP extra questions is one of the most effective ways to solidify your understanding of the topic. Here’s how you can use them:

• Before diving into the extra questions, ensure that you have a clear understanding of the basic concepts of AP. Go through your NCERT Class 10 Mathematics textbooks and revise the formulas, especially the general formula for the nth term of an AP and the sum of the first n terms.
• Start by solving basic AP questions, where you are required to find the nth term or the common difference. Once you’re comfortable with these, move on to more tough problems.
• When you go through the extra questions in class 10 maths chapter 5, pay special attention to the problems that resemble previous exam papers. Solving these types of questions helps you familiarize yourself with the format and difficulty level of the actual exam.
• As you practice, identify which types of questions are more difficult for you, whether it's calculating the sum of an AP, finding a specific term, or solving word problems. Concentrate on these areas by revisiting similar additional questions and reviewing their solutions.

FAQs

Q1. How many marks are usually allotted to this chapter in Class 10 board exams?

Ans. This chapter generally carries 6 to 8 marks, usually including a mix of direct formula-based and word problems.

Q2. Which topics are most important for exams?

Ans. Finding the nth term, finding the sum of n terms, and solving application-based questions involving money, distance, or patterns are most important.

Q3. How can I remember the formulas easily?

Ans. Revise the formulas daily and understand their meaning. For example, the nth term formula gives any term, while the sum formula gives the total of all terms up to that point.

Q4. How do I check if a series is an A.P.?

Ans. Subtract consecutive terms; if the difference is the same throughout, it is an A.P.

Q5. How can I improve accuracy in A.P. problems?

Ans. Write each step, substitute values carefully, and double-check units and signs before finalising answers.