Previous Year Questions: Arithmetic Progressions

# Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions

## 2023

Q1: If a, b,  form an A.P. with common difference d. then the value of a- 2b-c is equal to
(a) 2a + 4d
(b) 0
(c) -2a- 4d
(d) -2a - 3d    [2023, 1 Mark]

Ans: (c)
We have, a, b, c are in A.P.
b = a + d, and c =  a + 2d
Now, a - 2b - c = a - 2(a + d) -  (a + 2d)
= a - 2a - 2d - a - 2d
= -2a- 4d

Question for Previous Year Questions: Arithmetic Progressions
Try yourself:Q2: The next term of the A.P.: √7, √28, √63 is

Q3: If k + 2, 4k - 6. and 3k - 2 are three consecutive terms of an A.P. then the value of k is
(a) 3
(b) -3
(c) 4
(d) -4    [2023, 1 Mark]

Ans: (a)
Since, k + 2, 4k - 6 and 3k - 2 are three consecutive terms of A.P.
a2 - a1 = a3 - a2
⇒ (4k - 6)- (k + 2) = (3k - 2) - (4k - 6)
⇒ 4k -6 - k - 2= 3k - 2 - 4k + 6
⇒ 3k - 8 = -k + 4
⇒ 4k = 12
⇒ k = 3

Q4: How many terms are there in A.P. whose first and fifth term are -14 and 2, respectively and the last term is 62.    [2023, 3 Marks]

Ans: We have
First term, a1 = - 14
Fifth term, a5 = 2
Last term, an = 62
Let d be the common difference and n be the number of terms.
∵ a5 = 2
⇒ -14 +(5 - 1)d = 2
⇒  4d = 16
⇒  d =4
Now, a= 62
-14 + (n - 1)4 = 62
⇒ 4n - 4 = 76
⇒ 4n = 80
⇒ n= 20
There are 20 terms in A.P.

Q5: Which term of the A.P. : 65, 61, 57, 53, _____ is the first negative term ?   [2023, 3 Marks]

Ans: Given, A.P. is 65, 61, 57, 53,.....
Here, first term a = 65 and  common difference, d = -4
Let the nth term is negative.
Last term, an = a + (n - 1) = 65 + (n  - 1)(-4)
= 65 - 4n + 4
= 69 - 4n, which will be negative when n = 18
So, 18th term is the first negative term.

Q6: Assertion:  a, b, c are in AP if and only if 2b = a + c
Reason: The sum of first n odd natural numbers is n2.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.   [2023, 1 Mark]

Ans: (b)
Since, a, b ,c are in A.P. then b - a = c -b
⇒ 2b = a + c
First n odd natural number be 1, 3, 5 ..... (2n - 1).
which form an A.P. with a = 1 and d = 2
Sum of first n odd natural number =

Hence, assertion and reason are true but reason is not the correct expiation of assertion.

Q7: The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.    [2023, 3 Marks]

Ans: Here, a = 15 and S15 = 750

⇒ 15(15 + 7d) = 750
⇒ 15 + 7d = 50
⇒ 7d = 35
⇒ d = 5
Now, 20th term = a + (n - 1)d
= 15 + (20 - 1] 5
= 15 + 95
= 110

Q8: Rohan repays his total loan of Rs. 1,18,000 by paying every month starting with the first instalment of Rs. 1,000. If he increase the instalment by Rs. 100 every month. what amount will be paid by him in the 30th instalment? What amount of loan has he paid after 30th instalment?   [2023, 3 Marks]

Ans: Total amount of loan Rohan takes = Rs. 1,18,000
First instalment paid by Rohan = Rs. 1000
Second instalment paid by Rohan = 1000 + 100 = Rs. 1100
Third instalment paid by Rohan = 1100 + 100 = Rs. 1200 and so on.
Let its 30th instalment be n.
Thus, we have 1000,1100,1200, which forms an A.P. with first term (a) = 1000
and common difference (d) = 1100 - 1000 = 100
nth term of an A.P. an= a + (n - 1)d
For 30th instalment, a30 = a + (30 - 1)d
= 1000 + (29) 100 = 1000 + 2900 = 3900
So Rs. 3900 will be paid by Rohan in the 30th instalment.
Now, we have a = 1000, last term (l)= 3900

⇒ S30 = 15(1000 + 3900) = Rs. 73500
Total amount he still have to pay after the 30th instalment = (Amount of loan) - (Sum of 30 instalments)
= Rs. 1,18,000 - Rs. 73,500 = Rs. 44,500
Hence, Rs. 44,500 still have to pay after the 30th instalment.

Q9: The ratio of the 11th term to 17th term of an A.P. is 3 : 4. Find the ratio of 5th term to 21st term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms.   [2023, 4,5,6 Marks]

Ans:  Given,

Required ratio =

Q10: 250 logs are stacked in the following manner:
22 logs in the bot tom row. 21 in the next row . 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row?   [2023, 4,5,6 Marks]

Ans: Number of logs in 1st row = 22
Number of logs in 2nd row = 21
Number of logs in 3rd row = 20
The number of Logs i.e., 22, 21, 20, __ forms an A.P.. where
a = 22 ,d = a2 - a1 ,= 21 -22 = - 1
Let the number of rows be n.
Now,

∴ Number of logs in the 20th (top) row is 3.

## 2022

Q1: Find a and b so that the numbers a, 7, b,  23 are in A.P.      [2022, 2 Marks]

Ans: Since a, 7, b,  23 are in A.P.
∴ Common difference is same.
∴ 7 - a = b -7 = 23 - b
Taking second and third terms, we get
b - 7 = 23 - b
⇒ 2b = 30
⇒ b = 15
Taking first and second terms, we get
⇒ 7 - a = b - 7
⇒ 7 - a = 15 - 7
⇒ 7 - a = 8
⇒ a = -1
Hence, a = -1, b = 15.

Q2: For the A.P. ; a1, a2, a3, .....a4/a7 = 2/3, then find a6/a8.      [2022, 2 Marks]

Ans: Let a be the first term and d be the common difference of given A.P.
We have, a4/a7 = 2/3

Q3: Find the number of terms of the A.P. : 293, 235, 27,....., 53      [2022, 2 Marks]

Ans: Given, 293, 285, 277..... 53 be an A.P.
a = 293, d = 285 - 293 = -8
We know. an = a + (n - 1 )d
⇒ 53 = 293 =  (n - 1)(-8)
⇒ 53 - 293 = (n - 1) (-8)
⇒ -240 =  (n - 1) (-8)
⇒ 30 = n - 1
⇒ n = 31

Q4: For what value of n are the nth terms of the A.P.'s : 9, 7, 5, .... and 15, 12, 9.... the same?      [2022, 2 Marks]

Ans:

Q5: Which term of the A .P.       [2022, 2 Marks]

Ans: Let nth term of the given A.P. be 49/2.
Here a = -11/2
⇒ a1 = -11/2 and a2 = -3

Q6: Determine the A.P. whose third term is 5 and seventh term is 9.     [2022, 2 Marks]

Ans: Let the first term and common difference of an A.P. be a and d, respectively.
Given a3= 5 and a7 = 9
a + (3 - 1 ) d = 5 and a + (7 - 1)d  =  9
a + 2d = 5 --------------(i)
and a + 6d = 9--------------(ii)
On subtracting (i) from (ii), we get
⇒ 4d  = 4
⇒ d = 1
From (i), a + 2(1) = 5 ⇒ a + 2 = 5 ⇒ a = 3
So. required A.P. is a, a + d, a + 2d, a + 3d......
i.e. 3, 3 +1, 3 + 2(1), 3 + 3(1), .....,  i.e.. 3,  4,  5, 6, .....

Q7: In Mathematics, relations can be expressed in various ways. The matchstick patterns are based on linear relations. Different strategies can be used to calculate the number of matchsticks used in different figures. One such pattern is shown below. Observe the pattern and answer the following question using Arithmetic Progression:

(a) Write the AP for the number of triangles used in the figures. Also, write the nth term of this AP.
(b)Which figure has 61 matchsticks.     [2022, 2 Marks]

Ans: (a): 20
(b) Numbers of matchsticks used in figure 1 = 12
Number of matchsticks used in figure 2 = 19
Number of matchsticks used in 3 = 26
Thus Required A.P. be 12 , 19 , 26 .....

Hence, figure 8 has 61 matchsticks.

Q8: In an AP if Sn = n (4n + 1), then find the AP.      [2022, 2 Marks]

Ans: Given Sn = n (4n + 1)
S=1(4 + 1) = 5 = a1
S2 = 2(8 + 1) =  2(9) = 18
S= 3(4(3)+1)= 3(13) = 39
S4 = 4(4(4) + 1) =4(17) = 68
We know that an = Sn - Sn-1

Q9: Find the common difference 'd' of an A.P. whose first term is 10 and sum of the first 14 terms is 1505.      [2022, 2 Marks]

Ans:

Question for Previous Year Questions: Arithmetic Progressions
Try yourself:Q10: Find the sum of first 20 terms of an A.P. in which = 5 and a20 = 135.

Q11: Find the sum of first 20 terms of an A.P whose nth term is given as an = 5 - 2n.      [2022, 2 Marks]

Ans: Given, nth term of the A.P. series is

So, the series becomes 3, 1, -1, - 3 .....
Here, a = 3 and d = a2 - a1 , = 1 - 3 = - 2
We know that, sum of n terms of an A.P. is

∴ Sum of first 20 terms of an A.P. is

Hence, the required sum of 20 terms of given A.P. is -320

## 2020

Q1: If -5/7, a, 2 are consecutive terms in an Arithmetic  Progression, then the value of a' is
(a) 9/7
(b) 9/14
(c) 19/7
(d) 19/14     [2020, 1 Mark]

Ans: (b)
Given, -5/7, a, 2 are in A.P. therefore common difference is same.
∴ a2 - a1 = a3 - a2

Q2: Which of the following is not an A.P?     [2020, 1 Mark]
(a) -1.2, 0.8.2.8, ....
(b) 3, 3+√2, 3+2√2,3 + 3√2,...
(c)

(d)

Ans: (c)
In option (c), We have

As a2 - a1 ≠  a3- a2 the given list of numbers does not form an A.P.

Q3: The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an A.P, is
(a) 6
(b) -6
(c) 18
(d) -18     [2020, 1 Mark]

Ans: (a)
Given, 2x, (x + 10) and (3x + 2) are in A.P.
(x + 10) - 2x = (3x + 2) -(x + 10)
⇒ -x + 10= 2x - 8
⇒ - 3x = -18
⇒ x = 6

Q4: Show that (a - b)2, (a2 + b2) and (a + b)2 are in A.P.    [2020, 2 Marks]

Ans: Let a1 = (a - b)2, a2 = (a2 + b2) and a3= (a + b)2
Now. a2 - a1 = (a2 + b2) - (a - b)2
= a2 + b2 - (a2 + b2 - 2ab)
= a2 + b2- a2- b2 + 2ab = 2ab
Again a- a2 = (a + b)2 - (a2 + b2)
= a2 + b2 + 2ab - a2 - b2 = 2ab
∴ a2 - a1 = a3 - a2
So, (a - b)2, (a2 + b2) and (a + b)2 are in A.P.

Question for Previous Year Questions: Arithmetic Progressions
Try yourself:Q5: The first term of an A.P. is p and the common difference is q, then its 10th term is

Q6: The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the numbers.       [2020, 4,5,6 Marks]

Ans: Let the four consecutive numbers be (a - 3d), (a - d), (a + d), (a + 3d).

Sum of four numbers = 32 [Given]
⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32 ⇒  a = 8

If d = 2. then the numbers are (8 - 6), (8 - 2), (8 + 2) and (8 + 6) i.e., 2,6, 10, 14 .
If d = -2. then the numbers are (8 + 6), (8 + 2), (8 - 2). (8 - 6) i.e.,14 ,10 ,6 ,2 .
Hence, the numbers are 2, 6, 10, 14 or 14, 10, 6, 2.

Q7: Find the sum of the first 100 natural numbers.       [2020, 1 Mark]

Ans: First 100 natural numbers are 1 , 2 , 3 ...... 100 which form an A.P. with a = 1, d = 1.
Sum of n terms =

Q8: Find the sum of first 16 terms of an Arithmetic Progression whose 4th and 9th terms are - 15 and - 30 respectively.       [2020, 3 Marks]

Ans: Given, a4 = -15 and a9 = -30
a + 3d = - 15  (i)
a + 8d = -30   (ii)
On subtracting (ii)from (i), we have
-5d = 15
⇒ d = - 3
Put d = - 3 in (i), we have
a + 3(-3)= - 15
⇒ a - 9 = - 15
⇒ a = - 6
Now,

Q9:  in an A.P. given that the first term (a) = 54. the common difference (d) = -3 and the nth term (an) = 0. find n and the sum of first n terms (Sn) of the A.P.       [2020, 3 Marks]

Ans:  Given, d = - 3, a = 54 and an= 0
Since an = a + (n -1)d
∴ 0 = 54 + (n - 1)(-3)
⇒ 0 = 54 - 3n + 3
⇒  3n = 57
⇒ n = 19
Now,

Q10: Find the sum: ( - 5 ) + ( - 8 ) + (- 11)+...+ (- 230)      [2020, 3 Marks]

Ans: Let Sn = (- 5) + (- 8) + (- 11) + ...+ (-230)

Q11: For an A.P, it is given that the first term (a) - 5, common difference (d) = 3, and the nth term (an) = 50. Find n and sum of first n terms (S„) of the A.P       [2020, 3 Marks]

Ans: Given, a = 5, d = 3 and an= 50
Since, an = a + (n -1)d

## 2019

Q1: Find the common difference of the Arithmetic Progression (A.P.)        [2019, 1 Mark]

Ans: Given A.P. is

So. common difference

Q2: Write the common difference of A.P.       [2019, 1 Mark]
√3, √12, √27, √48, ......

Ans: Give A.P. is
√3, √12, √27, √48, ......
or √3, 2√3, 3√3,4√3, .......
∴ d = common difference = 2√3 - √3 = √3

Q3: If the nth term of an A.P. is pn + q. find its common difference.      [2019, 1 Mark]

Ans: Given, an = pn + q
⇒ d + (n - 1)d  = pn + q
⇒ (n - 1)d = pn + q - a
⇒

Q4: Which term of the A.P. 10, 7, 4, ...is -41 ?      [2019, 1 Mark]

Ans:  Let nth term of A.P. 10, 7, 4, ...is -41
∴ an = a + (n - 1)d

∴ 18th term of given A.P. is - 41

Q5: If in an A.P. a = 15, d = - 3 and an = 0, then find the value of n.       [2019, 1 Mark]

Ans: Given, a = 15,  d = - 3 and an = 0
∴ an = a + (n - 1)d
⇒ 15 + (n - 1 )(- 3)= 0
⇒ 15  - 3n +3 = 0
⇒ 18 - 3n = 0
⇒ - 3n = -18
⇒ n = 6

Q6: How many two digit numbers are divisible by 3?      [2019, 1 Mark]

Ans: Two-digit numbers which are divisible by 3 are 12, 15, 13..... 99. which forms an A.P. with first term (a) = 12, common difference (d) = 15 - 12 = 3 and last term (l) or nth term = 99
a + (n - 1)d = 99
⇒ 12 + (n - 1)3 = 99
⇒ 3n = 99 - 9
⇒ n = 90/3
⇒ n = 30

Q7: If the 9th term of an AR is zero, then show that its 29th term is double of its 19th term.      [2019, 2 Marks]

Ans:  Given, a9 = 0 . we have to show that a29 = 2a19
a + 8d = 0
⇒ a = - 8 d
Now, a19 = a + 18d = -8d + 18d = 10d
a29 = a + 28d = -8d + 28d = 20d = 2(10d ) = 2a19
Hence,  a29 = 2a19

Q8: Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21st term?    [2019, 2 Marks]

Ans:  We have, first term, a = 3, common difference,  d = 15 - 3 =12
nth term of an A.P. is given by an = a + (n - 1)d
∴ a21 =  3 + (20) x 12
= 3 + 240
= 243
Let the rth term of the AP. be 120 more than the 21st term.

Q9: If the 17th term of an A.P. exceeds its 10th term by 7, find the common difference.   [2019, 2 Marks]

Ans:  According to question, a17 - a10 = 7
i.e. a + 16d- (a + 9d) = 7
where a = first term d = common difference
⇒ 7d = 7
∴ d = 1

Q10: Which term of the Arithmetic Progression -7, -12, -17,  -22.._ will be -82 ? Is -100 any term of the A.P. ? Give reason for your answer.   [2019, 4,5,6 Marks]

Ans: Given, Arithmetic Progression -7, -12, -17,  -22..
and nth term of given AP. is -82.
∴ an = a + (n - 1)d

Hence, -100 is not any term of given A.P.

Q11: If Sn the sum of first it terms of an A.P. is given by Sn = 3n2 - 4n, find the nth term.    [2019, 2 Marks]

Ans: Given, Sn = 3n2 - 4n
We know that Sn - Sn-1 = an

Q12: If Sn the sum of first it terms of an A.P. is given by Sn = 2n2 + n, find the nth term.    [2019, 2 Marks]

Ans: We have Sn = 2n2 + n

Q13:  If mth term of an A.P. is 1/n and nth term is 1/m, then find the sum of its first mn terms.    [2019, 3 Marks]

Ans: Let a be first term and d be the common difference of an A.P.
Since, we have,

Subtracting (ii) from (i), we get

On Substituting (iii) in (i), we get

Sum of first mn terms is

=

The document Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

## Mathematics (Maths) Class 10

126 videos|477 docs|105 tests

## FAQs on Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions

 1. What is an arithmetic progression (AP)?
Ans. An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. For example, 2, 5, 8, 11, 14 is an arithmetic progression with a common difference of 3.
 2. How do you find the nth term of an arithmetic progression?
Ans. The nth term of an arithmetic progression can be found using the formula: $$a_n = a_1 + (n-1)d$$, where $$a_n$$ is the nth term, $$a_1$$ is the first term, $$n$$ is the position of the term, and $$d$$ is the common difference.
 3. What is the sum of the first n terms of an arithmetic progression?
Ans. The sum of the first n terms of an arithmetic progression can be calculated using the formula: $$S_n = \frac{n}{2} [2a_1 + (n-1)d]$$, where $$S_n$$ is the sum of the first n terms, $$a_1$$ is the first term, $$n$$ is the number of terms, and $$d$$ is the common difference.
 4. How do you determine if a given sequence is an arithmetic progression?
Ans. To determine if a given sequence is an arithmetic progression, check if the difference between consecutive terms is constant. If the difference remains the same throughout the sequence, then it is an arithmetic progression.
 5. Can an arithmetic progression have negative terms?
Ans. Yes, an arithmetic progression can have negative terms. As long as the difference between consecutive terms remains constant, the sequence can have negative terms.

## Mathematics (Maths) Class 10

126 videos|477 docs|105 tests

### Up next

 Explore Courses for Class 10 exam

### Top Courses for Class 10

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;