Tough questions on Arithmetic Progression for class 10th

Tough and Hots questions of Arithmetic Progression class 10

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The arithmetic progression chapter is essential for your upcoming board, class 11 and class 12. The arithmetic progression is a basic chapter of sequence and series. You should focus on this important chapter regarding board purpose and the next class. 

In this article, I have provided only tough and previous year's questions based on the arithmetic progression with digestible solutions. I hope, this clears your doubts regarding the question's difficulties and helps in board preparation. 

Why is it necessary to solve all the previous year's questions on Arithmetic Progression?

Arithmetic Progression of  Chapter 5 of Class 10th is an interesting chapter for Class 10th students.  In this chapter, each question has a different difficulty level. Overall, Due to only a limited formula used in this chapter. this chapter is interesting for students. but, from the board's perspective. students do not build their confidence regarding this chapter due to the right questions concepts clarity and practices.

If you are trying to find a short way to handle the difficulties of Arithmetic progression, chapter 5. then, you are at the right place. where you can get premium suggestions or contact us for our A plus support.

It does not matter, how many questions you solve? the matter is whether the questions solved by you are based on the exam perspective or not. If you think it is worthless, remember you have limited time to concentrate on a specific topic (less than a whole year), due to other subjects, and difficult topics.

While preparing this chapter for the board, they used different sources for question practices such as sample papers, question banks, and reference books. they increase their stress and unwanted confusion regarding HOTs questions. The question of Arithmetic Progression identities requires a unique concept of different sequences. Due to this, the students are not able to say that he/she is completely prepared for Arithmetic Progression.

It reduces your stress, our mathematics expert separates the question from previous year's papers and trending questions. these questions are categorised by difficulty level, 1 mark, 2 marks, 3 marks and 5 marks as well as by the year they appeared in previous years' papers with detailed solutions. These levels are designed to address common difficulties every year. 

By practising these questions, you get an insight into the trends of Arithmetic Progression questions in previous year's papers that improve your understanding of the approach and limitations of the CBSE Board. It will boost your confidence before and during the Board exams.

These questions are profitable for your board preparation. This is designed to help you understand the new approach and limitations of the upcoming CBSE board examinations.

We have provided detailed solutions with all required concepts, theorems and formulas related to Tough questions on Arithmetic Progression. All the questions are selected from previous years, sample papers and competitive exam papers. 

You need to understand and learn some important formulas of Arithmetic Progression.

All the previous year's questions were based on arithmetic progression formulas.

Question 1 Find the 6th term from the end of the A.P 17, 14, 11, ...., -40

[CBSE 2005]

Solution

we have,

last term = l = -40

common difference = d  = -3 

we know that,

⇒nth term from the end = l - (n - 1) d

⇒6th term from the end = l - (6 -1)(-3)

⇒6th term from the end = -40 - (6 - 1)(-3)

⇒6th term from the end = -40 - 5(-3)

⇒6th term from the end = -40 + 15

⇒6th term from the end = - 25

Question 2 For what value of n is the n^th term of the following two A.P. are the same? 

(1) 69, 68, 67, .........      (2) 1, 7, 13, 19, ........

[CBSE 2006C]

Solution

We have,

First term = a = 69

Common difference = d = -1

⇒a_n = a + (n - 1)d

⇒a_n = 69 + (n - 1) x (-1)

⇒a_n = 69 - n + 1

⇒a_n = 70 - n    ..........(1)

Similarly,

For A.P

First term = a = 1

Common difference = d = 6

⇒a_n = a + (n - 1)d

⇒a_n = 1 + (n - 1) 6

⇒a_n = 1 + 6n - 6

⇒a_n = 6n - 5       ...............(2)

From (1) & (2) 

⇒a_n= a_n 

⇒70 - n = 6n - 5

⇒70 + 5 = 6n + n

⇒75 = 7n 

⇒n = 75/7

which is not a natural number.

Question 3 If the 8^th term of an A.P is 31 and the 15^th term is 16 more than the 11^th term, find the A.P

[CBSE 2006C]

Solution

Let a be the first term and d be the common term difference of the A.P

we have,

⇒a₈ = 31

⇒a + (8 - 1)d = 31                {n =8}

⇒a + 7d = 31       ........(1)

Similarly,

⇒a₁₅ = a₁₁ + 16

⇒a + (15 - 1)d = a + (11 - 1)d + 16                        {n = 15 and 11}

⇒a + 14d = a + 10d + 16

⇒a + 14d - 10d  - a = 16

⇒4d = 16

⇒d = 4

Put d = 4 in eq. (1)

⇒a + 7 (4) = 31

⇒a + 28 = 31

⇒a = 31 - 18

⇒a = 3 

Hence, the A.P is a, a + d, a + 2d, a + 3d, .......

⇒3, 7, 11, 15, 19, 23, ...... 

Question 4 Which term of the arithmetic progression 5, 15, 25, ...... will be 130 more than its 31^th term?

[CBSE 2006C]

Solution

we have,

a =-5 , n =31 and d = 10 

∴ a_n = a + (n - 1)d

⇒a₃₁ = 5 + (31 - 1)10

⇒a₃₁ = 5 + 30(10)

⇒a₃₁ = 5 + 300

⇒a₃₁ = 305

Let the n^th term of the given A.P. be 130 more than its 31^th term.

a_n = 130 + a₃₁

⇒a + (n - 1) d = 130 + 305

⇒5 + (n - 1) 10  = 435

⇒5 + (n -1) 10 = 435

⇒5 + 10n - 10 = 435

⇒10n - 5 = 435

⇒10n = 435 + 5

⇒10n = 440

⇒n = 44

Hence, the 44^th term of the given A.P. is 130 more than its 31^th term.

Question 5 If the 10^th term of an A.P. is 52 and the 17^th term is 20 more than the 13^th term. find the A.P.

[CBSE 2006C]

Solution

Let a be the first term and d be the common term difference of the A.P

we have,

⇒a₁₀ = 52

⇒a + (10 - 1)d = 52                {n = 10}

⇒a + 9d = 52       ........(1)

Similarly,

⇒a₁₇ = a₁₃ + 20

⇒a + (17 - 1)d = a + (13 - 1)d + 20                       {n = 17 and 13}

⇒a + 16d = a + 12d + 20

⇒a + 16d - 12d  - a = 20

⇒4d = 20

⇒d = 5

Put d = 5 in eq. (1)

⇒a + 9 (5) = 52

⇒a + 45 = 52

⇒a = 52 - 45

⇒a = 7 

Hence, the A.P is a, a + d, a + 2d, a + 3d, .......

⇒7, 12, 17, 22, 27, 32, ......  

Question 6 The sum of the 5^th and 9^th terms of an A.P. is 72 and the sum of the 7^th and 12^th terms is 97. Find the A.P.

[CBSE 2009]

Solution

Let a be the first term and d be the common difference of A.P.

It is given that,

⇒a₅ + a₉ = 72

⇒(a + 4d) + (a + 8d) = 72

⇒2a + 12d = 72   .................(1)

Similarly,

⇒a₇ + a₁₂ = 97

⇒(a + 6d) + (a + 11d) = 97

⇒2a + 17d = 97    ...................(2)

Both equation, 

⇒2a + 12d = 72   .................(1)

⇒2a + 17d = 97    ...................(2)

Subtracting (1) from (2), we get

⇒5d = 25

⇒d = 5

Put d = 5 in eq. (1), we get

⇒2a + 12(5) = 72

⇒2a + 60 = 72

⇒2a = 72 - 60

⇒2a = 12

⇒a = 6

Hence, the A.P is a, a + d, a + 2d, a + 3d, .......

⇒6, 11, 16, 21, 26, ........

Question 7 If the p^th term of an A.P. is q and the q^th term is p, prove that its n^th term is (p + q -n).

[CBSE 2008]

Solution

Let a be the first term and d be the common difference of the given A.P. Then,

the p^th term = q

⇒a + (p - 1)d = q  ............(1)

Similarly,

the q^th term = p

⇒a + (q - 1)d = p  ....................(2)

Both equations

⇒a + (p - 1)d = q  ............(1)

a⇒ + (q - 1)d = p  ....................(2)

Subtracting equation (1) from equation (2), we get

⇒(p - q)d = q - p

⇒d = -1

Putting  d = -1 in equation (1), we get

⇒a + (p - 1)(-1) = q

⇒a - p + 1 = q

⇒a = (q + p - 1)

If a = (q + p - 1) and d = -1

⇒n^th term = (n -1)d

⇒n^th term = (q + p - 1) + (n - 1)(-1)

⇒n^th term = (p + q - n) 

Question 8 If m times the m^th term of A.P. is equal to n times its n^th term, show that the (m + n)^th term of an A.P. is zero.

[CBSE 2008]

Solution

Let a be the first term and d be the common difference of the given A.P. Then,

⇒(m times m^th term) = (n times n^th term)

⇒m a_m = n a_n

⇒m{a + (m - 1)d} = n{a + (n - 1)d}

⇒m{a + (m - 1)d} -  n{a + (n - 1)d} = 0

⇒a m + m(m - 1)d - a n - n(n -1)d = 0

⇒(m - n)a + {m(m - 1)d - n(n - 1)d} = 0

⇒(m - n)a + {(m² - n²) - (m - n)}d = 0

⇒(m - n)a + {(m + n)(m - n) - (m - n)d} = 0

⇒(m - n)a + {(m - n)(m + n -1)d}= 0

⇒(m - n){a + (m + n -1)d} = 0

⇒a + (m + n -1)d = 0                                          [∵ m ≠ n]

⇒a_m+n = 0

Hence, the (m + n)^th term of the given A.P. is zero.

Question 9 The 7^th term of an A.P. is 32 and its 31^th term is 62. find the A.P.

[CBSE 2004]

Solution

Given,

⇒7^th term of an A.P. = 32

⇒a₇ = 32

⇒a + 6d = 32      ...........................(1)

Similarly,

13^th term of A.P. = 62

⇒a₁₃ = 62

⇒a + 12d = 62    ................................(2)

Subtracting equation (1) from equation (2)

⇒6d = 30

⇒d = 5

Put d = 5 in eq. (1)

⇒a + 6(5) = 32

⇒a + 30 = 32

⇒a = 32 - 30

⇒a = 2

Hence, the A.P is a, a + d, a + 2d, a + 3d, .......

⇒2, 7, 12, 17, .................

Question 10 Which term of the A.P. 3, 10, 17, ..... will be 84 more than its 13^th term?

[CBSE 2004]

Solution

Given.

A.P.  3, 10, 17, .............

From A.P

First term is = a = 3

Common difference = d = 7

Let the n^th term of A.P. be 84 more than the 13^th term

⇒a_n = a₁₃ + 84

⇒a + (n - 1)d = a + (13 - 1)d + 84

⇒3 + (n - 1)7 = 3+ (13 - 1)(7) + 84

⇒3 + (n - 1)7 = 3 + 12(7) + 84

⇒3 + (n - 1)7 = 3 + 84 + 84

⇒3 + (n - 1)7 = 171

⇒3 + 7n - 7 = 171

⇒7n = 171 + 7 - 3

⇒7n = 175

⇒n = 25

Hence, the 25^th term of A.P. will be 84 more than the 13^th term.

Question 11 For what value of n, the n^th term of the A.P. 63, 65, 67,..... and 3, 10, 17,... are equal?

[CBSE 2008]

Solution

Given,

First A.P. = 63, 65, 67, ........

First term = a = 63

Common difference = d = 2

⇒a_n = a + (n - 1)d

⇒a = 63 + (n - 1)2

⇒a = 63 + 2n - 2

⇒a = 61 + 2n

Second A.P. = 3, 10, 17, .............

First term = a = 3

Common difference = d = 7

⇒a_n = a + (n -1)d

⇒a = 3 + (n - 1)7

⇒a = 3 + 7n - 7

⇒a = 7n - 4

According to question

n^th term of second  A.P. = n^th term of second A.P.

⇒61 + 2n = 7n - 4

⇒61 + 4 = 7n - 2n

⇒65 = 5n

⇒n = 13

Question 12 How many three-digit numbers are divisible by 7?

[NCERT, CBSE 2013]

Solution

The first three-digit number that is divisible by 7 = 105

Next number = 105 + 7 = 122

Therefore, A.P. = 105, 112, 119, 126, ........

The maximum possible three-digit number is 999. when we divide it by 7, the remainder will be 5.

 So, the maximum possible number three-digit number 999 - 5 = 994 divisible by 7.

The series is as follows.

⇒105, 112, 119, ......, 994

Let 994 be the n^th term of this AP

a = 105

d = 7

a_n = 994

n = ?

⇒a_n = a + (n - 1)d

⇒994 = 105 + (n - 1)7

⇒994 = 105 + 7n - 7

⇒994 - 105 + 7 = 7n

⇒889 = 7n

⇒n = 128

Hence, 128 three-digit numbers are divisible by 7.

Question 13 Which term of an A.P. 8, 14, 20, 26, .... will be 72 more than its 41^st term?

[CBSE 2006C]

Solution

Given,

A.P. = 8, 14, 20, 26, ............

First term = a = 8

Common difference = d = 6

Let the n^th term of A.P. be 72 more than the 41^st term.

⇒a_n = a₄₁ + 72

⇒a + (n - 1)d = a + (41 - 1)d + 72

⇒8 + (n - 1)6 = 8 + 40(6) + 72

⇒8 +6n - 6 = 8 + 240 + 72

⇒6n -6 = 240 + 72

⇒6n = 240 + 72 + 6

⇒6n = 318

⇒n = 53

∴ The 53^rd term of A.P. will be 72 more than the  41^st term.

Question 14 Find the term of A.P. 9, 12, 15, 18, .... will be 39 more than its 36^th term.

[CBSE 2006C]

Solution

Given,

A.P.  = 9, 12, 15, 18,...

First term = a = 9

Common difference = d = 3

Let the n^th term of A.P. be 39 more than the 36^th term.

⇒a_n = a₃₆ + 39

⇒a + (n - 1)d = a + (36 - 1)d + 39

⇒9 + (n - 1)3 = 9 + 35(3) + 39

⇒9 +3n - 3 = 9 + 105 + 39

⇒3n -3 = 105 + 39

⇒3n = 105 + 39 + 3

⇒3n = 147

⇒n = 49

∴ The 49^th term of A.P. will be 39 more than the 36^th term.

Question 15 Find the 8^th term from the end of the A.P. 7, 10, 13, ....... 184.

[CBSE 2005]

Solution

Given,

The A.P. is 7, 10, 13,... 184

Last term = l = 184

Common difference = d = 3

∴ 8^th term from the end

⇒8^th term from the end = l - (n - 1)d

⇒8^th term from the end = 184 - (8 - 1)3

⇒8^th term from the end = 184 - 7(3)

⇒8^th term from the end = 184 - 21

⇒8^th term from the end = 163

Question 16 The sum of the  4^th and 8^th terms of an A.P. is 24 and the sum of the 6^th and 10^th terms Of an A.P. is 44, Find the A.P.

[CBSE 2009]

Solution

Given,

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

∴ The sum of 4^th and 8^th term = 24

⇒a₄ + a₈ = 24

⇒a + 3d + a + 7d = 24

⇒2a + 10d = 24

⇒a + 5d = 12    .......................(1)

Similarly,

The sum of 6^th and 10^th term = 44

⇒a₆ + a₁₀ = 44

⇒a + 5d + a + 9d = 44

⇒2a + 14d = 44

⇒a + 7d = 22        ............................(2)

Subtract eq.(1) from eq.(2)

⇒(a + 7d) - (a + 5d) = 22 - 12

⇒2d = 10

⇒d = 5

Put d = 5 in eq.(1)

⇒ a + 5(5) = 12

⇒a + 25 = 12

⇒a = 12 - 25

⇒a = -13

Hence, the A.P is a, a + d, a + 2d, a + 3d, .......

⇒-13, -8, -3, 2, .................

Question 17  Which term of A.P. is 3, 15, 27, 39, ..... will be 120 more than its 21^st term?

[CBSE 2009]

Solution

Given,

The A.P. is 3,15, 27, 39, ........

First term = a = 3

Common difference = d = 12

Let the n^th term be 120 more than its 21^st term.

Then a_n = a₂₁ + 120

⇒a + (n - 1)d = a + (21 - 1)d + 120

⇒3 + (n - 1)12 = 3 + 20(12) + 120

⇒3 + 12n -12 = 3 + 240 + 120

⇒12n = 3 + 240 + 120 - 3 + 12

⇒12n = 372

⇒n = 31

Hence the 31^st term will be 120 more than its 21^st term 

Question 18 The 17^th term of an A.P. is 5 more than twice its 8^th term. If the 11^th term of the A.P. is 43, find the A.P.

[CBSE 2012]

Solution 

Given,

⇒a₁₇ = 5 + 2a₈

⇒a +16d = 5 + 2(a + 7d)

⇒a + 16d = 2a + 14d + 5

⇒a - 2d + 5   

⇒a - 2d = -5     ..............(1)

Similarly,

⇒a₁₁ = 43

⇒a + 10d = 43  .................(2)

Subtract eq.(1) from eq.(2)

⇒(a + 10d) - (a - 2d) = 43 + 5

⇒12d = 48

⇒d = 4

Put d = 4 in eq.(1)

⇒a - 2(4) = -5

⇒a - 8 = -5

⇒a = -5 + 8

⇒a = 3

Thus, the n^th term is given by

⇒a_n = a + (n - 1)d

⇒a_n = 3 + (n -1)4

⇒ a_n = 3+ 4n -4

⇒a_n = 4n - 1

Question 19 Find all three-digit natural numbers that are divisible by 9.

[CBSE 2013]

Solution 

Given,

The smallest three-digit number divisible by 9 = 108

The largest three-digit number divisible by 9 = 999

Let's write the series of obtained an A.P.

108, 117, 126, ........,999

First term = a = 108

Common difference = d = 9

a_n = 999

⇒a_n = a + (n - 1)d

⇒999 = 108 + (n - 1)9

⇒999 = 108 + 9n - 9

⇒9n = 999 - 108 + 9 

⇒9n = 900

⇒n = 100

Number of the terms divisible by 9

The number of all three-digit naturals divisible by 9 is 100.

Question 20 The 19^th term of an A.P. equals three times its 6^th term. if the 9^th term is 19, find the A.P.

[CBSE 2013]

Solution 

Given,

Let the first term of the A.P. 'a'

and the common difference  'd'

⇒a₁₉ = 3a₆

⇒a + 18d = 3(a + 5d)

⇒a + 18d = 3a + 15d

⇒a - 3a = 15d - 18d

⇒-2a = -3d

⇒2a = 3d

⇒a = 3d/2 .........(1)

Similarly,

⇒a₉ = 19

⇒a + 8d = 19  ..............(2)

Put a = 3d/2 in the equation (2)

⇒3d/2 + 8d = 19

⇒3d + 16d = 38

⇒19d = 38

⇒d = 2

Put d = 2  in eq.(1)

⇒2a = 3(2)

⇒a = 3

Hence, the A.P is a, a + d, a + 2d, a + 3d, .......

⇒3, 5, 7, 9, .................

Conclusion  

I have provided tough and previous year's questions on Arithmetic Progression with digestible solutions. I hope you learn easily and enjoy it. Most of the questions are in detail based on the Previous years listed above. If you have any doubts regarding the A.P. solutions. you should contact us immediately. Our expert reach you very soon.