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Chapter 5 Class 10 Arithmetic Progressions

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Updated for new NCERT Book - 2023-24 Edition

Get solutions of all NCERT Questions with examples of Chapter 5 Class 10 Arithmetic Progressions (AP). Video of all questions are also available. 

In this chapter, we will learn

• What is an AP - and what is First term (a) and Common Difference (d) of an Arithmetic Progression
• Finding n th term of an AP (a n )
• Finding n using a n formula
• Finding AP when some terms are given
• Finding n th term from the last term
• Finding Sum of n terms of an AP ( S n ) - both formulas
• Finding number of terms when Sum is given
• If n th term is given, finding S
• Some statement questions using a n and S n formulas, including optional exercise

Check out the answers to the exercises (including examples and optional) by clicking a link below, or learn from the concepts.

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Arithmetic Progression

If we observe then we may notice that there is some pattern in some things like 2,4,6,8 are all even numbers and the difference is same that is two. It means on adding 2 in each even number we get a new even number. Same things happen with odd numbers also. Like this there a lot of patterns. So, let’s learn more about the topic and patterns.

As the name suggest Arithmetic means adding or subtracting and progression means series.

A sequence of numbers such that the difference between the consecutive term is constant then the sequence is said to be in Arithmetic Progression.

For example 3,7,11,15,19,23. Here each term differs the previous term by 4 and since the difference is constant therefore it is a arithmetic progression.

• The constant difference is actually known as common difference and is denoted by d.
• The first term of an AP is denoted by a.

Therefore the general form of an Arithmetic Progression is a, a+d, a+2d, a+3d, a+4d………….n times.

• Here we can notice that the common difference between each term is d.

Common Difference:

The common difference can be positive, negative or even zero.

• If the difference is positive then the AP is called increasing AP. For example 2, 12, 22, 32, 42, 52.
• If the difference is negative then the AP is called decreasing AP. For example 12, 10, 8, 6, 4.
• If the difference is zero then it is called as constant AP. For example 5, 5, 5, 5, 5, 5, 5, 5.

If we denote the first term by a 1, second term by a 2  and so on therefore the n th  term will be a n . Then the AP hence formed will be a 1 , a 2 , a 3 , ……………..a n .

So, a 2 -a 1 = a 3 -a 2 = a 4 -a 3 ………..= a n -a n-1 = d.

N th Term of an AP

It is not always possible to determine all the terms of an AP if it contains a lot of terms. For example if we have to determine the 1000 th term of an AP of even nos then it is not that easy and also time taking.

Hence we generalize the result to find the n th term of an AP

Let a 1 , a 2 , a 3 , ……………..a n , be an AP of n terms and here a 1 is the first term and d is common difference.

a 2 = a + d = a + (2-1)d

a 3 = a+2d = a + (3-1)d

a 4 = a+3d= a + (4-1)d

General Term a n =  a + (n-1)d  

And the last term of AP l = a + (n-1)d if there are n term

Example1 :- Find the 10 th term of the AP 2,12, 22, 32….

Solution :- Here first term a = 2 and common difference d is

12-2 = 10 = d.

Using the formula and putting n=10 we get

a 10 = 2+ (10-1)10 = 2 + 90 = 92.

Example 2 :- Which term of the AP 15,10, 5, 0 … will be -40?

Solution :- Here a =15 and d = -5.

And here as we know that a n = -40.

On putting the values in the formula we get

-40 = 15 + (n-1)(-5)

-55 = (n-1)(-5)

Therefore (n-1) = 11 and hence n = 12.

So the 12 th term will be -40.

Sum of N terms of an AP

Let us consider an AP with first term a and common difference d

a, a+d, a+2d, a+3d, a+4d………………………a+(n-1)d.

Let S denotes the sum of the first n terms and therefore

• Here an important point is to remembered that a n = S n -S n-1.

It means that the n th term of an AP is equal to the difference of its sum of n th and sum of (n-1) th term.

Example 3 :- Find the sum of 10 terms of the AP -3, 1, 5, 9,13..?

Solution :-  Here a = -3 and d = 4

Therefore on putting the values in the formula we get

Hence the sum of first 10 terms is 150.

Word Problems

Example 4 :- Raman started working at a company at an annual salary of Rs 5000 per year and he received an annual increment of Rs 200 per year. After how many years his salary will be Rs 8000?

Solution :- Here the sequence will be 5000, 5200, 5400, 5800, 6000……….8000

Here the first term a =5000 and d=200, a n = 8000

On using the formula and putting the values in it, we get

8000 = 5000 +(n-1)200

3000 = (n-1)200 and hence (n-1) = 15

Therefore n= 16 which means after 16 years Raman’s annual salary will be Rs 8000 per year.

Example 5 :- A contract company has to pay penalty if the works get delayed. Rs 100 for first day, Rs 150 for second, Rs 200 for third and so on. Find out how much money the contractor has to pay if the work gets delayed by 30 days.

Solution :- The sequence formed will be 100 , 150, 200, 250 …

 Here a=100 and d=50 and n=30

Therefore on putting the values in the formula we get,

S n = 15[200+1450]

S n = 15 x 1650 = 24750.

Hence the contractor has to pay Rs 24,750 as penalty for the delay of work for 30 days.

Practice Questions

Q1) Write the first four terms of the AP if

• a= 3 and d=5 ii) a=12 and d=-2         
• a=4 and d=-2 iv) a=10 and d= 10

Q2) Find a 1 , a 2 , a 4  and d of the following AP

• 10, 5, 0, -5,-10 ….. b) -20,-12,-4, 4…..

Q3) Find the n th term of the following AP

• 5, 11, 17, 23, 29…….
• 10, 15 ,20, 25 ……….

Q4) Determine the AP whose first term is 8 and 3 rd term is 24.

Q5) How many two digit numbers are divisible by 5?

Q6) Find the sum first 200 even integers.

• The sequence having a common difference is known as AP.
• The general term is a n = a+(n-1)d

• The common difference d can be positive or negative or zero also.
• The addition, subtraction, multiplication or division of two AP’s also results in an AP.

Quiz for Arithmetic Progression

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Class 10 Maths Chapter 5 Arithmetic Progressions Important Questions

Here are some important questions for Class 10 Mathematics Chapter 5, Arithmetic Progressions, carefully selected to help students prepare for the CBSE Class 10 Mathematics Examination in 2023-24. These questions cover various types of problems and are designed to assist students in understanding Arithmetic Progressions better. By practicing these diverse question types, students can clarify any doubts they may have and improve their problem-solving skills, leading to better performance in the chapter on Arithmetic Progressions. 

Table of Content

Introduction, what is an arithmetic progression (a.p.), class 10 arithmetic progressions important questions and answers, cbse class 10 maths chapter wise important questions, frequently asked questions.

In Chapter 5 of Class 10 Mathematics, Arithmetic Progressions, you will study the motivation for studying Arithmetic Progression, the derivation of the nth term and sum of the first n terms of an A.P., and there are applications in solving real-life problems.

Answer: An Arithmetic Progression is a sequence of numbers in which the difference between consecutive terms remains constant. This constant difference is known as the common difference (d).

Q 1. The sum of first 20 odd natural numbers is :

(a) 100 (b) 210 (c) 400 (d) 420.

Explanation: ‍ The given A.P. is1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39 Thus,    a = 1, d = 2 and n = 20 We know that, \(S_n = {\Large\frac{n}{2}}\space\{ 2a + (n – 1)d \}\\[4.5bp] =\space{\Large\frac{20}{2}}\{{ 2(1) + (20 – 1)2}\}\) =10{2 + 38} = 400

Q 2. The number of multiples of 4 that lie between 10 and 250 is

(a) 62 (b) 60 (c) 59 (d) 55.

Explanation: Multiples of 4 lying between 10 and 250 are 12 , 16, 20, 24……, 248 Here, a = 12, d = 16 - 12 = 4, l = 248  l = a n = a + (n - 1)d248 = 12 + (n - 1)× 4⇒ 248 - 12 = 4(n - 1) \(\Rightarrow \frac{236}{4}=n-1 \\[4.5bp] \Rightarrow n - 1 = 59\\[4.5bp] \therefore n = 59 + 1 = 60\)

Q 3. If the sum of n terms of an A.P. is S n = 3n 2 + 5n, then write its common difference.

Ans. common difference = 6

Explanation: We have,  S n = 3n 2 +5n ⇒ S n - 1 = 3(n - 1) 2 + 5(n - 1)  = 3(n 2 + 1 - 2n) + 5n - 5 = 3n 2 + 3 - 6n + 5n - 5 = 3n 2 - n - 2 General term of A.P. is given as, T n = S n - S n-1       = 3n 2 + 5n - 3n 2 + n + 2       = 6n + 2 Putting n = 1,2,3, we get T 1 = 6 × 1 + 2= 8 T 2 = 6 × 2 + 2 = 14 T 3 = 6 × 3 + 2 = 20 Now, T 2 - T 1 = 14 - 8 = 6 and T 3 - T 2 = 20 - 14 = 6 ∴ common difference = 6

Q4. \(\text{Which term of the progression 20 ,}\) \(19\frac{1}{8},18\frac{1}{2},17\frac{3}{4}, \) .......is the first negative term?

Ans. 28th term will be the first negative term of given A.P.

Explanation: \(Given, A.P. is 20, 19\frac{1}{4},18\frac{1}{2},17\frac{3}{4}......\) \(=20,\frac{77}{4},\frac{37}{2},\frac{71}{4},\) \(Here , a = 20, d=\frac{77}{4}-20=\frac{77-80}{4}=\frac{-3}{4}\) Let a n be its first negative term  Then a n + (n – 1)d < 0. \(\Rightarrow 20 + (n – 1)\begin{pmatrix}   -\frac{3}{4} \end{pmatrix}\text{\textless} 0\) \(\Rightarrow 20 -\frac{3}{4}n+\frac{3}{4}\text{\textless} 0\) \(\Rightarrow 20+\frac{3}{4}\text{\textless}\frac{3}{4}n\) \(\Rightarrow \frac{83}{4} \text{\textless}\frac{3}{4}n\) \(\Rightarrow n\text{\textgreater}\frac{83}{4}×\frac{4}{3}\) \(\Rightarrow n\text{\textgreater}\frac{83}{3}=27.66...\) 28th term will be the first negative term of given A.P.

Q 5. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that the sum of numbers of houses proceeding the house numbered X is equal to sum of the numbers of houses following X.

Ans. X = 35

Explanation: Given, the houses in a row numbered consecutively from 1 to 49. Now, sum of numbers preceding the number X \(=\frac{x(x-1)}{2}\) and sum of numbers following the number X \(=\frac{49(50)}{2}-\frac{x(x-1)}{2}-X\) \(=\frac{2450-x^2+x-2x}{2}\) \(=\frac{2450-x^2-x}{2}\) According to the question,  Sum of no’s preceding X = Sum of no’s following X \(\frac{x(x-1)}{2}=\frac{2450-x^2-x}{2}\) \(\Rightarrow X^2 – X = 2450 – X^2 – X\) \(\Rightarrow 2X  = 2450\)  \(\Rightarrow X = 1225\) \(\Rightarrow X = 35\) Hence, at X = 35, sum of number of houses preceding the house no. X is equal to the sum of the number of houses following X.

If you want to improve your understanding of the concepts in this chapter, you can visit oswal.io. They have a wide range of questions that will help you practice and reinforce what you've learned. By solving these questions, you can strengthen your knowledge and become better at solving problems.

Q1:  How can  identify if a given sequence of numbers forms an Arithmetic Progression?

Ans: To identify an Arithmetic Progression, check if the difference between consecutive terms is constant. If the difference between any two consecutive terms is the same throughout the sequence, then it is an A.P.

Q2: State the formula for the nth term (a n ) of an Arithmetic Progression.

Ans: The formula for the nth term of an A.P. is given as: a n = a + (n - 1)d, where 'a' is the first term and 'd' is the common difference.

Q3: Define the sum of the first n terms (S n ) of an Arithmetic Progression.

Ans: The sum of the first n terms of an A.P. is denoted as S n and can be calculated using the formula: S n = (n/2) [2a + (n - 1) d], where 'a' is the first term and 'd' is the common difference.

Q4: Can the common difference (d) in an A.P. be negative?

Ans: Yes, the common difference in an A.P. can be negative. It means that the sequence is decreasing, and each term is obtained by subtracting the absolute value of 'd' from the previous term.

Q5: In an Arithmetic Progression, what is the condition for finding the nth term (a n ) or the sum of the first n terms (S n ) directly?

Ans: To find the nth term (a n ) or the sum of the first n terms (S n ) directly, the common difference (d) must be known, and the first term (a) should be given or easily determined.

Chapter Wise  Important Questions for CBSE Board Class 10 Maths

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Learn Class 10 Math - Arithmetic Progressions

The word arithmetic progression is a series of terms in which the numbers are present in the order sequence. And the difference between the consecutive numbers always remains constant. This sequence is also known as Arithmetic progression.

• Natural number series include the numbers from 1, 2, 3, 4, 5, ….so on.
• The difference between the two consecutive numbers is the same, i.e., 1
• The difference between any even number and an odd number always remains the same.

Types of Arithmetic Progression

When we see the types, then they are three different types. The types are discussed below:

• Geometric progression
• Harmonic progression
• Arithmetic progression

Sum of Arithmetic Progression

The formulas of an arithmetic progression are given below:

• AP nth term.
• And their sum,i.e., the sum of n terms.

Question 1:

Evaluate n, when a = 10, d = 5, an = 95..

It is given that a = 10, d = 5, & an = 95

With the help of the formula, we see that.

an = a + (n − 1) (d)

95 = 10 + (n − 1) (5)

(n − 1) (5) = 95 – 10 = 85

(n − 1) = 85/ 5

(n − 1) = 17

With the help of this formula, we will get the value n. And the value of n is equal to 18.

• Real Numbers
• Polynomials
• Pair of Linear Equations in Two Variables
• Quadratic Equations
• Arithmetic Progressions
• Coordinate Geometry
• Introduction to Trigonometry
• Some Applications of Trigonometry
• Constructions
• Areas Related to Circles
• Surface Areas and Volumes
• Probability
• General Knowledge

Practice Test: Arithmetic Progressions - Class 10

25 questions mcq test - practice test: arithmetic progressions, if the sum of first n terms of an ap be 3n 2  – n and it's common difference is 6, then its first term is :.

If 7 th  and 13 th  terms of an A.P. be 34 and 64, respectively, then it's 18 th  term is :

The sum of all 2-digit odd positive numbers is :.

Here a = 11 and d = 2, t n = 99, n = ? Sum of the n terms = (n/2)[2a+(n -1)d] But t n  = a + (n -1)d ⇒ 99 = 11+ (n-1)2 ⇒ 99 -11 = (n-1)2 ⇒ 88/2 = (n-1) ∴ n = 45. subsitute n = 45  in sum of the n terms we obtain ⇒ s 45 =  (45/2)(2×11 + (45 -1)2) ⇒ s 45 =  (45/2)(110) ⇒ s 45  = 45×55. ⇒  s 45  = 2475. ∴ sum of all two digit odd positive numbers = 2475.

The fourth term of an A.P. is 4. Then the sum of the first 7 terms is :

In an A.P., s 1  = 6, s 7  = 105, then s n  : s n-3  is same as :

(n + 3) : (n – 3)

(n + 3) : n

n : (n – 3)

None of these

In an A.P. s 3  = 6, s 6  = 3, then it's common difference is equal to :

The number of terms common to the two A.P. s 2 + 5 + 8 + 11 + ...+ 98 and 3 + 8 + 13 + 18 +...+198

For first A.P 2+5+8+11+......+98 a=2,an​=98,d=3 an​=a+(n−1)d 98=2+(n−1)3 98=2+3n−3 3n=99 n=33 Number of term =33 For first A.P 3+8+13+18+......+198 a=3,an​=198,d=5 an​=a+(n−1)d 198=3+(n−1)5 198=3+5n−5 5n=200 n=40 No of terms =40 Common terms=40−33=7

(p + q)th and (p – q)th terms of an A.P. are respectively m and n. The p th  term is :

l=a+(n-1)d (p+q)th term is m m=a+(p+q-1)d m=a+pd+qd-d  ….1 (p-q)th term is n n=a+(p-q-1)d n=a+pd-qd-d   ….2 Adding 1 and 2 m+n=2a+2pd-2d m+n=2(a+pd-d) ½(m+n=a+(p-1)d So pth term is ½(m+n)

The first, second and last terms of an A.P. are a,b and 2a. The number of terms in the A.P. is :

A.P : a , b , . . . . . . . . . . . . . .2a 1st term= a1 = a 2nd term = a2= b nth term = an = 2a d = a2 - a1 = b-a an = a1 + (n-1)d = a + (n-1)(b-a) = 2a (n-1)(b-a) = a (n-1) = a / (b-a) n = a / (b-a) + 1 = b / ( b -a ) Sn = n / 2 * ( a1 + an) = b / 2(b-a) * ( a + 2a) = 3ab / 2(b-a)

Let s 1 , s 2 , s 3  be the sums of n terms of three series in A.P., the first term of each being 1 and the common differences 1, 2, 3 respectively. If s 1  + s 3  = λs 2 , then the value of λ is :

Sum of first 5 terms of an A.P. is one fourth of the sum of next five terms. If the first term = 2, then the common difference of the A.P. is :

If x,y,z are in A.P., then the value of (x + y – z) (y + z – x) is equal to :

8yz – 3y 2  – 4z 2

8yz – 3z 2  – 4y 2

8yz + 3y 2  – 4z 2

8yz – 3y 2  + 4z 2

The number of numbers between 105 and 1000 which are divisible by 7 is :

If the numbers a,b,c,d,e form an A.P. then the value of a – 4b + 6c – 4d + e is equal to :

If s n  denotes the sum of first n terms of an A.P., whose common difference is d, then s n  – 2s n-1  + s n-2  (n >2) is equal to :

s n ​−2s n−1 ​+s n−2  ​= (s n​ −s n−1​ )−(s n−1 ​−s n−2 ​)

= a n ​−a n−1  ​[∵(s n ​−s n− 1​)= a n ​]

= [a+(n−1)d]−[a+(n−2)d]

= a+nd−d−a−nd+2d

The sum of all 2-digited numbers which leave remainder 1 when divided by 3 is :

The first term of an A.P. of consecutive integers is p 2  + 1. The sum of 2p + 1 terms of this series can be expressed as :

(2p + 1) (p + 1) 2

p 3  + (p + 1) 3

If the sum of n terms of an AP is 2n 2  + 5n, then its nth term is –

If the last term of an AP is 119 and the 8th term from the end is 91 then the common difference of the AP is –

If {a n } = {2.5, 2.51, 2.52,...} and {b n } = {3.72, 3.73, 3.74,...} be two AP's, then a 100005  – b 100005  =

– 1.22

Observing both the AP’s we see that the common difference of both the AP’s is same ,so difference between their corresponding terms will be same ie,a 1 -b 1 =2.5-3.72=-1.22 a 2 -b 2 =2.51-3.73=-1.22  So , a 100005 -b 100005 =-1.22

If A 1  and A 2  be the two A.M.s between two numbers a and b, then (2A 1  – A 2 ) (2A 2  – A 1 ) is equal to :

a – b

If a,b,c are positive reals, then least value of (a + b + c) (1/a+1/b+1/c) is :

The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

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Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions

Chapter 5 arithmetic progressions important questions for cbse class 10 maths board exams.

Important Questions for Chapter 5 Arithmetic Progressions Class 10 Maths

Arithmetic progressions class 10 maths important questions very short answer (1 mark).

Here, First term, a = 5 Common difference, d = 9 – 5 = 4 Last term, 1 = 185 n th term from the end = l – (n – 1)d 9 th term from the end = 185 – (9 – 1)4 = 185 – 8 × 4 = 185 – 32 = 153

6. Find the common difference of the AP 1/p, 1−p/p, 1−2p/p,…

The common difference,

Arithmetic Progressions Class 10 Maths Important Questions Short Answer-I(2 Marks)

Here, d = 9 - 4 = 14 -9 = 19 – 14 = 5 ∴ Difference between consecutive terms is constant. Hence, it is an A.P.

Given, First term, a = 4, d = 5, a n = 109 (Let) ∴ a n = a + (n – 1) d …[General term of A.P.] ∴ 109 = 4 + (n – 1) 5 ⇒ 109 – 4 = (n – 1) 5 ⇒ 105 = 5(n − 1) ⇒ n – 1 = 105/5 = 21 ⇒ n = 21 + 1 = 22 ∴ 109 is the 22 nd term

12. Find the sum of first ten multiple of 5.

Let the first term be a, common difference be d, nth term be a n  and sum of n term be S n .

a = 5, n = 10, d = 5

= 5[10+ 9×5]

= 5×55 = 275

Hence, the sum of first ten multiple of 5 is 275.

13. If the sum of first m terms of an AP is the same as the sum of its first n terms, show that the sum of its first (m+n) terms is zero.

Let a be the first term and d be the common difference of the given AP.

S m  = S n

Let 1 st term = a, Common difference = d a 4 = 0 a + 3d = 0 ⇒ a = -3d …(i) To prove: a 25 = 3 × a 11 a + 24d = 3(a + 10d) …[From (i)] ⇒ -3d + 24d = 3(-3d + 10d) ⇒ 21d = 21d From above, a 25 = 3(a 11 )

19. Find 10th term from end of the A.P. 4,9, 14, …, 254. (2011OD)

Common difference d = 9 – 4 = 14 – 9 = 5 Given: Last term, l = 254, n = 10 n th term from the end = l – (n – 1) d ∴ 10 th term from the end = 254 – (10 – 1) × 5 = 254 – 45 = 209

20. Find, 100 is a term of the AP  25, 28, 31 ,....  or not.

Let the first term of an AP be a, common difference

be d and number of terms be n.

Let a n  = 100

Here, a = 25, d = 28-25 = 31-28 = 3

a n  = a + (n-1)d,

100 = 25 + (n-1) 3

⇒ 100 - 25 = 75 = (n-1) 3

⇒ 25 = n - 1

⇒ n  = 26

Since, 26 is an whole number, thus 100 is a term of given A.P.

21. Find the 7th term from the end of AP 7, 10, 13,.... 184

Let us write AP in reverse order i.e., 184, .....13, 10, 7

Let the first term of an AP be a and common

difference be d.

d = 7-10 = -3

a = 184, n = 7

7th term from the original end.

a 7   = a +6d

⇒ a 7  = 184 + 6(-3)

= 184 -18 = 166

Hence, 166 is the 7th term from the end.

Here, a n = 45, S n = 400, a = 5, n = ?, d = ?

Let the first term be a, common difference be d, nth term be a n and sum of n term be Sn .

We have, a = 65, d = -5, S n  = 0

⇒ 135n - 5n 2 = 0

⇒ n(135 - 5n) = 0

⇒ 5n = 135

⇒ n = 27

26. How many terms of the AP 18, 16, 14.... be taken so that their sum is zero?

Let the first term be a, common difference be d, nth term be an and sum of n term be S n .

a = 18, d = -2, S n  = 0

To find: 110 + 121 + 132 + … + 990 Here, a = 110, d = 121- 110 = 11, a n = 990 ∴ a + (n – 1)d = 990 110 + (n – 1).11 = 990 ⇒ (n – 1). 11 = 990 – 110 = 880 ⇒ (n – 1) = 880 = 80 ⇒ n = 80 + 1 = 81 As, S n = n/2 (a 1 + a n ) ∴ S 81 = 81/2 (110 + 990) = 81/2 (1100) = 81 × 550 = 44,550

29. Find how many integers between 200 and 500 are divisible by 8.

Number divisible by 8 are 208, 2016, 224, .... 496. It is an AP

Let the first term be a, common difference be d and nth term be a n .

We have a = 208, d = 8, a n  = 496

a + (n-1)d = a n

⇒ 208 + (n-1)d = 496

⇒ (n-1)8 = 496 - 208

⇒ n = 36+1 = 37

Hence, required numbers divisible by 8 is 37.

30. The fifth term of an AP is 26 and its 10th term is 51. Find the AP

a 5  = a + 4d = 26 ...(1)

a 10  = a + 9d = 51 ...(2)

Subtracting (1) from (2) we have

⇒ d = 5

Substituting this value of d in equation (1) we get

Hence, the AP is 6, 11, 16, ....

32. Find whether -150 is a term of the AP 11, 8, 5, 2, ...

Let the nth term of given AP 11, 8, 5, 2... be -150

a = 11, d = 8-11 = -3 and an = -150

a + (n-1)d = an

⇒ 11 + (n-1)(-3) = -150

⇒ (n-1)(-3) = -161

which is not a whole number. Hence -150 is not a term of given AP.

Arithmetic Progressions Class 10 Maths Important Questions Short Answer-II (3 Marks)

Let the required term be n th term, i.e., a n Here, d = 14 – 3 = 11, a = 3 According to the Question, a n = 99 + a 25 ∴ a + (n – 1) d = 99 + a + 24d ⇒ (n – 1) (11) = 99 + 24 (11) ⇒ (n – 1) (11) = 11 (9 + 24) ⇒ n – 1 = 33 ⇒ n = 33 + 1 = 34 ∴ 34 th term is 99 more than its 25th term.

36. Which term of the AP 20, 19 1/4, 18 1/2, 17 3/4, ...is the first negative term.

Hence, the first negative term is 28th term.

37. If in an AP, the sum of first m terms is n and the sum of its first n terms is m, then prove that the sum of its first (m+n) terms is -(m+n)

Let 1st term of series be a and common difference be d, then we have

S m  = n

S n  = m

Given, a 19 = 3(a 6 ) ⇒ a + 18d = 3(a + 5d) ⇒ a + 18d = 3a + 15d ⇒ 18d – 15d = 3a – a

42.  The ninth term of an AP is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and the common difference.

a 9  = 7a 2

⇒ a + 8d = 7(a+d)

⇒ a + 8d = 7a + 7d

⇒ -6a + d = 0 ...(1)

a 12  = 5a 3  + 2

⇒ a + 11d = 5(a+2d) + 2

⇒ a + 11d = 5a + 10d + 2

⇒ -4a + d = 2 ...(2)

Subtracting (2) from (1), we get

⇒ a = 1

Substituting this value of a in equation (1) we get

⇒ d = 6

Hence first term is 1 and common difference is 6.

46. If the ratio of the sums of first n terms of two AP’s is (7n + 1): (4n + 27), find the ratio of their nth terms.

Let a and A be the first term and d and D be the common difference of two AP’s, then we have

First term, a = 5, Last term, a n = 45 Let the number of terms = n S n = 400 ⇒ n/2 (a + a n ) = 400 ⇒ n/2(5 + 45) = 400 ⇒ n/2 (50) = 400 ⇒ n = 400/25 = 16 = Number of terms Now, a n = 45 a + (n – 1)d = a n ⇒ 5+ (16 – 1)d = 45 ⇒ 15d = 45 – 5 ∴ d = 40/15=8/3

48. The sum of first n terms of an AP is 3n 2 + 4n. Find the 25 th term of this AP.

We have, S n = 3n 2 + 4n Put n = 25, S 25 = 3(25) 2 + 4(25) = 3(625) + 100 = 1875 + 100 = 1975 Put n = 24, S 24 = 3(24) 2 + 4(24) = 3(576) + 96 = 1728 + 96 = 1824 ∴ 25 th term = S 25 – S 24 = 1975 – 1824 = 151

49. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.

Let hundred’s place digit = (a – d) Let ten’s place digit = a Let unit’s place digit = a + d According to the Question, a – d + a + a + d = 15 ⇒ 3a = 15 ⇒ a = 5 Original number = 100(a – d) + 10(a) + 1(a + d) = 100a – 100d + 10a + a + d = 111a – 99d Reversed number = 1(a – d) + 10a + 100(a + d) = a – d + 10a + 100a + 100d = 111a + 99d Now, Original no. – Reversed no. = 594 111a – 99d – (111a + 99d) = 594 -198d = 594 ⇒ d = 594/-198 = -3 ∴ The Original no. = 111a – 99d = 111(5) – 99(-3) = 555 + 297 = 852

50. Find the sum of all multiples of 7 lying between 500 and 900. (2012OD)

To find: 504 + 511 + 518 + … + 896 a = 504, d = 511- 504 = 7, a n = 896 a + (n – 1)d = a n ∴ 504 + (n – 1)7 = 896 (n – 1)7 = 896 – 504 = 392

Arithmetic Progressions Class 10 Maths Important Questions Long Answer (4 Marks)

Let A be the first term and D be the common difference of the given A.P. p th term = A + (p – 1)D = a …(i) q th term = A + (q – 1)D = b …(ii) r th term = A + (r – 1)D = c …(iii) L.H.S. = (a – b)r + (b – c)p + (c – a)q = [A + (p – 1)D – (A + (q – 1)D)]r + [A + (q – 1)D – (A + (r – 1)D)]p + [A + (r – 1)D – (A + (p – 1)D)]q = [(p – 1 – q + 1)D]r + [(q – 1 – r + 1)D]p + [(r – 1 – p + 1)D]q = D[(p – q)r + (q – r)p + (r – p)q] = D[pr – qr + qp – rp + rq – pq] = D[0] = 0 = R.H.S.

59. A sum of ₹1,600 is to be used to give ten cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.

Here, S 10 = 1600, d = -20, n = 10 S n = n/2 (2a + (n – 1)d] ∴ 10/2[2a + (10 – 1)(-20)] = 1600 2a – 180 = 320 2a = 320 + 180 = 500 a = 250 ∴ 1 st prize = a = ₹250 2 nd prize = a 2 = a + d = 250 + (-20) = ₹230 3 rd prize = a 3 = a 2 + d = 230 – 20 = ₹210 4 th prize = a 4 = a 3 + d = 210 – 20 = ₹190 5 th prize = a 5 = a 4 + d = 190 – 20 = ₹170 6 th prize = a 6 = a 5 + d = 170 – 20 = ₹150 7 th prize = a 7 = a 6 + d = 150 – 20 = ₹130 8 th prize = a 8 = a 7 + d = 130 – 20 = ₹110 9 th prize = a 9 = a 8 + d = 110 – 20 = ₹590 10 th prize = a 10 = a 9 + d = 90 – 20 = ₹70 = ₹1,600

60. If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

Let a be the first term and d be the common difference.

Sum of n terms of an AP,

Hence, sum of n terms is 6n + n 2 .

61. If Sn denotes the sum of first n terms of an AP, prove that, S 30  = 3(S 20  - S 10 )

Let the first term be a, and common difference be d.

Hence, S 30  = 3(S 20  - S 10 )

62. Find the 60th term of the AP 8, 10, 12,.... if it has a total of 60 terms and hence find the sum of its last 10 terms.

Let the first term be a, common difference be d, nth term be a n and sum of n term be S n

a = 8, d = 10 - 8 = 2

a n  = a + (n-1)d

a 60  = 8 + (60 - 1)2

= 8+ 59×2 = 126

and a 51  = 8 + 50×2

= 8 + 100 = 108

Sum of last 10 terms,

Hence sum of last 10 terms is 1170.

63. An AP consists of 37 terms. The sum of the three  middle most terms is 225 and the sum of the past  three terms is 429. Find the AP.

Let the middle most terms of the AP be (x-d), x and (x+d).

x-d + x + x+d = 225

⇒ 3x = 225

⇒ x = 75

and the middle term = 37+1/2 = 19th term.

Thus AP is,

(x-18d),...(x-2d), (x-d), x, (x+d), (x+2d),...(x+18d)

Sum of last three terms,

(x+18d) + (x+17d) + (x+16d) = 429

⇒ 3x + 51d = 429

⇒ 225 + 51d = 429

⇒ d = 4

First term, a 1  = x-18d = 75- 18×4 = 3

a 2  = 3+4 = 7

Hence, AP = 3, 7, 11,...,147.

64. The first and the last terms of an A.P. are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum?

Here, a = 8, a n = 350, d = 9 As we know, a + (n − 1) d = a 2 ∴ 8 + (n – 1) 9 = 350 ⇒ (n − 1) 9 = 350 – 8 = 342 ⇒ n – 1 = 342/9 = 38 ⇒ n = 38 + 1 = 39 ∴ There are 39 terms. ∴ S n = n/2(a + an) ∴ S 39 = 39/2 (8 + 350) = 39/2 × 358 = 39 × 179 = 6981

65. In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP.

Here, n = 50, S 10 = 210 = 10/2 (2a + 9d) = 210 …[S n = 1/2 [2a+(n – 1)2] ⇒ 5(2a + 9d) = 210 ⇒ 2a + 9d = 210/5 = 42 ⇒ 2a = 42 – 9d ⇒ a = 42−9d/2 …(i) Now, 50 = (1 + 2 + 3 + …) + (36 + 37 + … + 50) Sum = 2565 Sum of its last 15 terms = 2565 …(Given) S 50 – S 35 = 2565 ⇒ 50/2(2a + 49d) – 35/2 (2a + 34d) = 2565 ⇒ 100a + 2450d – 70a – 1190d = 2565 × 2 ⇒ 30a + 1260d = 5130 ⇒ 3a + 1260 = 513 …(Dividing both sides by 10)

Classes: 1 + I + II + … + XII Sections: 2(I) + 2(II) + 2(III) + … + 2(XII) Total no. of trees = 2 + 4 + 6 … + 24 = (2 × 2) + (2 × 4) + (2 × 6) + … + (2 × 24) = 4 + 8 + 12 + … + 48 :: S 12 = 12/2(4 + 48) = 6(52) = 312 trees

67. Find the sum of first 24 terms of an AP whose nth  term is given by an = 3+2n

Let the first term be a, common difference be d, nth term be an and sum of n term be Sn

a n  = 3+2n

a 1  = 3+ 2×1 = 5

a 2  = 3 + 2×2 = 7

a 3  = 3 + 2×3 = 9

Thus, the series is 5, 7, 9,... in which

a = 5 and d = 2

Hence, S 24  = 672

68. Ramkali required ₹500 after 12 weeks to send her daughter to school. She saved ₹100 in the first week and increased her weekly saving by ₹20 every week. Find whether she will be able to send her daughter to school after 12 weeks.

Money required by Ramkali for admission of her daughter = ₹2500 A.P. formed by saving 100, 120, 140, … upto 12 terms …(i) Let, a, d and n be the first term, common difference and number of terms respectively. Here, a = 100, d = 20, n = 12 S n = n/2 (2a + (n − 1)d) ⇒ S 12 = 12/2 (2(100) + (12 – 1)20) ⇒ S 12 = 12/2 [2(100) + 11(20)] = 6[420] = ₹2520

69. How many multiples of 4 lie between 10 and 250? Also find their sum. (2011D)

Multiples of 4 between 10 and 250 are: 12, 16, 20, … 248 Here, a = 12, d = 4, a n = 248 As we know, a + (n – 1) d = a n ∴12 + (n – 1) 4 = 248 ⇒ (n – 1) 4 = 248 – 12 = 236 ⇒ n – 1 = 236/4 = 59 ⇒ n = 59 + 1 = 60 ∴ There are 60 terms. Now, S n = n2 (a + a n ) ∴ S 60 = 60/2(12 + 248) = 30 (260) = 7800

70. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.

Here, A.P. is 1, 2, 3, …., 49 a = 1, d = 1, a n = 49

Now, ⇒ (X – 1) 2 (2 + (X – 2)) = 49(2 + 48) – X[2 + (x – 1)] ⇒ (X – 1). X = 2,450 – X(X + 1) ⇒ x 2 – X = 2,450 – X 2 – X ⇒ X 2 – X + X 2 + X = 2,450 ⇒ 2X 2 = 2,450 ⇒ X 2 = 1,225 ∴ X = √1225 = 35 …[X can not be -ve]

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Arithmetic Progression Questions

The Arithmetic Progression questions and answers are given here for students to help them better grasp the concept. Arithmetic Progressions (AP) is one of the most important concepts in Maths and it is included in higher education. NCERT standards will be followed for preparing these questions. The problems given here will cover both the basics and more complex topics, for students of all levels. Students can practise the arithmetic progression questions, and can then cross verify their answers with the provided solutions. To learn more about Arithmetic progression, click here .

Here, we have provided the Arithmetic Progression questions and answers with complete explanations.

Arithmetic Progression Questions with Solutions

1. Find the common difference for the following AP: 10, 20, 30, 40, 50.

Given AP: 10, 20, 30, 40, 50

Common difference:

d = 20 – 10 = 10

d = 30 – 20 = 10

d = 40 – 30 = 10

d = 50 – 40 = 10.

Hence, the common difference for the sequence, 10, 20, 30, 40, 50 is 10.

2. Is a, 2a, 3a, 4a, … an arithmetic progression?

Given sequence: a, 2a, 3a, 4a, …

To check whether the given sequence is AP or not, we have to find the common difference.

Hence, d = 2a – a = a

d = 3a – 2a = a

d = 4a – 3a = a

Hence, the common difference is “a”.

Therefore, the sequence a, 2a, 3a, 4a,… is an arithmetic progression.

3. Prove that 7, 11, 15, 19, 23 is an AP.

Given sequence: 7, 11, 15, 19, 23.

To prove that the sequence is AP, find the common difference between two consecutive terms.

d = 11 – 7 = 4

d = 15 – 11 = 4

d = 19 – 15 = 4

d = 23 – 19 = 4

Hence, 7, 11, 15, 19, 23 is an AP with a common difference of 4.

4. The sequence 28, 22, x, y, 4 is an AP. Find the values of x and y.

Given AP: 28, 22, x, y, 4

Here, first term, a = 28

Common difference, d = 22 – 28 = -6

Hence, x = 22 – 6 = 16

y = 16 – 6 = 10.

Hence, the values of x and y are 16 and 10, respectively.

5. What is the nth term of an AP 9, 13, 17, 21, 25, …?

Given AP: 9, 13, 17, 21, 25, …

Here, a = 9

d = 13 – 9 = 4

The formula to find the nth term of an AP is a + (n-1)d.

The nth term of an AP = 9 + (n-1)4

= 9 + 4n -4

Hence, the nth term of AP 9, 13, 17, 21, 25, … is 4n+5.

6. Find the 5th term of the arithmetic progression 1, 4, 7, ….

Given AP: 1, 4, 7, …

d = 4 – 1 = 3

As we know,

The nth term of AP = a + (n-1)d

Hence, 5th term of AP = 1 + (5-1)3

Hence, the 5th term of AP is 13.

Also, check: Arithmetic Progression Class 10 Notes .

7. Find the 17th term of AP 4, 9, 14, …

Given AP: 4, 9, 14, …

Here, a = 4

d = 9 – 4 = 5

Now, substitute the values in the formula a+(n-1)d,

17th term of AP = 4+(17-1)5

= 4+80 = 84

Hence, the 17th term of AP is 84.

8. If the first, second and last terms of the AP are 5, 9, 101, respectively, find the total number of terms in the AP.

Given: First term, a = 5

Common difference, d = 9 – 5 = 4

Last term, a n = 101

Now, we have to find the value of “n”.

Hence, a n = a+(n-1)d

Substituting the values, we get

5+(n-1)4 = 101

5+4n-4 = 101

n=100/4 = 25

Hence, the number of terms in the AP is 25.

9. What is the general term of the series, 4, 7, 10, 13, …?

Given sequence is 4, 7, 10, 13, …

d = 7-4 = 3

The general term of an AP is:

a n = a+(n-1)d

a n = 4 +(n-1)3

a n = 4 + 3n-3

Therefore, the general term of the series 4, 7, 10, 13 is 3n+1.

10. Which term of AP 27, 24, 21, … is 0?

Given AP: 27, 24, 21, …

Here, a = 27

d = 24 – 27 = -3.

Also given that a n = 0

Now, we have to find the value of n.

0 = 27 +(n-1)(-3)

0 = 27 -3n +3

0 = 30 – 3n

Hence, n = 10

Therefore, the 10th term of AP is 0.

Practice Questions

• The sequence 12b, 8b, 4b is in AP. Find the sum of the first 18 terms.
• The first three terms of a sequence are 8, y, 18. Find the value of y so that the sequence becomes an Arithmetic progression.
• In an Arithmetic progression, the ratio of the 7th term to the 10th term is -1. If the 16th term is -15, find the 3rd term.

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