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Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions

Previous Year Questions 2024

Q1:  In the given figure, if PT is a tangent to a circle with centre O and ∠TPO = 35º, then the measure of ∠x is      (2024)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions(a) 110º
(b) 115º
(c) 120º
(d) 125º

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: (d)
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions∠OTP = 90º [Line from centre is ⊥ to tangent at point of contact] 
∠x = ∠TPO + ∠OTP [Exterior Angle Prop.] 
x = 35º + 90º 
= 125º


Q2: The ratio of the 10th term to its 30th term of an A.P. is 1 : 3 and the sum of its first six terms is 42. FInd the first term and the common difference of A.P.       (2024)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans:
Let the AP be a, a + d, a + 2d,.............
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
⇒ 3a + 27d= a + 29d
⇒ 2a – 2d = 0
⇒ 2a + 2d
∴ a = d ...(i)      
Now,  S6 = 42
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
⇒ 3[2a + 5d] = 42
⇒ 2a + 5d =14
⇒ 2a + 5a = 14 [From eqn (i)]
⇒ 7a = 14
⇒ a = 14/7
∴ a = 2
So First term = 2
Common difference = 2.


Q3: If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of its first 20 terms.   (2024)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans:
S7 = 49
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
⇒ 2a + 6d = 14
⇒ a + 3d = 7 ...(i)
S17 = 289
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
⇒ 2(a + 8d) = 34
⇒ a + 8d = 17 ...(ii)  
From eqn (i) and (ii).    
–5d = –10
∴  d =2  
Put the value of d in eqn.(i),
∴ a + 3 × 2 =7
⇒ a + 6 = 7
⇒ a = 7 - 6
⇒ a = 1
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions


Previous Year Questions 2023

Q4: If a, b,  form an A.P. with common difference d. then the value of a- 2b-c is equal to  (2023)
(a) 2a + 4d
(b) 0
(c) -2a- 4d 
(d) -2a - 3d    

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: (c)
Sol: 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


Q5: If k + 2, 4k - 6. and 3k - 2 are three consecutive terms of an A.P. then the value of k is  (2023)
(a) 3
(b) -3
(c) 4
(d) -4  

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: (a)
Sol: 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


Q6: 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)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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.


Q7: Which term of the A.P. : 65, 61, 57, 53, _____ is the first negative term ?   (2023)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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.


Q8: 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   (2023)
(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.

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: (b)
Sol: 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 = n/2[2a + (n -1)d]
= n/2 [2 + (n - 1)2] = n2
Hence, assertion and reason are true but reason is not the correct expiation of assertion.


Q9: The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.    (2023)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: Here, a = 15 and S15 = 750
∵ Sn = n/2[2a + (n -1)d]
∴ S15 = 15/2 [2 x 15 + (15 -1)d] = 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


Q10: 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)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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
Sum of 30 instalment, S30 = 30/2[a + 1]
⇒ 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.


Q11: The ratio of the 11th term to the 18th term of an AP is 2:3. Find the ratio of the 5th term to the 21st term and also the ratio of the sum of the first five terms to the sum of the first 21 terms.   (2023)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: Let a and d be the first term and common difference of an AP.
Given that, a11 : a18 = 2 : 3
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
⇒ 3a + 30d = 2a + 34d
⇒ a = 4d ...(i)
Now, a5 = a + 4d = 4d + 4d = 8d [from Eq.(i)]
And a21 = a + 20d = 4d + 20d = 24d [from Eq. (i)]
a5 : a21 = 8d : 24d = 1 : 3
Now, sum of the first five terms, S5 = 5/2 [2a + (5−1)d]
= 5/2 [2(4d) + 4d]   [from Eq.(i)]
= 5/2 (8d + 4d) = 5/2 × 12d = 30d
And, sum of the first 21 terms, S21 = 21/2 [2a + (21−1)d]
= 21/2 [2(4d) + 20d]= 21/2 × 28 d = 294 d from Eq..(i)]
So, ratio of the sum of the first five terms to the sum of the first 21 terms is,
S5 : S21 = 30d : 294d = 5 : 49


Q12: 250 logs are stacked in the following manner: 22 logs in the bottom 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?
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: Let there be n rows to pile of 250 logs
Here, the bottom row has 22 logs and in next row, 1 log reduces
It means, we get an AP 22, 21, 20, 19, ..................... n with first term or a = 22 and d = -1
Now, we know that total logs are 250 or we can say that,
Sn =250
Since sum of n terms of an A.P. Sn = n/2 (2a + (n-1) d)
= 250 Therefore, n/2 (2 x 22 + (n-1) x (-1))
or 500 = n (44 - (n-1))
500 = n (45- n)
n2 - 45 n + 500 = 0
By solving this, we get (n-20) (n-25) = 0
Since, there are 22 logs in first row and in next row, 1 log reduces, then we can not have more than 22 terms
∴ n ≠ 25
and n = 20
Means, 20th row is the top row of the pile
Now let's find out number of logs in 20th row
We know that value of nth term of an A.P. = a + (n-1) d
N20 = [22 + (20-1) (-1)]
=(22-19) = 3
Therefore, there are 3 logs in the top row.


Previous Year Questions 2022

Q13: Find a and b so that the numbers a, 7, b,  23 are in A.P.      (2022)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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.


Q14: Find the number of terms of the A.P. : 293, 235, 27,....., 53      (2022)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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


Q15: Determine the A.P. whose third term is 5 and seventh term is 9.     (2022)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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, .....


Previous Year Questions 2020

Q16: If -5/7, a, 2 are consecutive terms in an Arithmetic  Progression, then the value of a' is    (2020)
(a) 9/7
(b) 9/14
(c) 19/7
(d) 19/14     

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions


Q17: Which of the following is not an A.P?     (2020)
(a) -1.2, 0.8.2.8, ....
(b) 3, 3+√2, 3+2√2,3 + 3√2,...
(c) 4/3, 7/3, 9/3, 12/3, ...

(d) -1/5, -2/5, -3/5,..

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: (c)
Sol: In option (c), We have
Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions
As a2 - a1 ≠  a3- a2 the given list of numbers does not form an A.P.


Q18: The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an A.P, is  (2020)
(a) 6
(b) -6
(c) 18
(d) -18

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: (a)
Sol: 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


Q19: Show that (a - b)2, (a2 + b2) and (a + b)2 are in A.P.    (2020)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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.


Q20: 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)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions

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.


Q21: Find the sum of the first 100 natural numbers.       (2020)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: First 100 natural numbers are 1 , 2 , 3 ...... 100 which form an A.P. with a = 1, d = 1.
Sum of n terms =Sn = n/2 [2a + (n - 1)d] 

 = 100/2 [2 x 1 + (100 - 1) x 1] = 50 [2 + 99] = 50 x 101 = 5050


Q22: Find the sum of first 16 terms of an Arithmetic Progression whose 4th and 9th terms are - 15 and - 30 respectively.   (2020)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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, Sn = n/2 [2a + (n - 1)d]
⇒ S16 = 16/2 [2(-6) + (16 - 1) (-3)]
= 8 [2(-6) + (15) (-3)] = 8 [-12 - 45] = -456


Q23:  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)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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,
Sn = n/2 [2a+(n-1)d]
= 19/2 [2 × 54 +(19 - 1)(-3)]
= 19/2 [108 - 54] = 19/2 × 54 = 513


Q24: Find the Sum (−5) + (−8)+ (−11) + ... + (−230).      (2020)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans: (−5) + (−8)+ (−11) + ... + (−230) .
Common difference of the A.P. (d) = a- a
= -8-(-5)
= -8+5
= -3
So here,
First term (a) = −5
Last term (l) = −230
Common difference (d) = −3
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
an =  a + (n-1) d
So, for the last term,
- 230 = -5 + ( n-1) (-3)
- 230 = -5-3n + 3
-23 +2 = -3n
- 228/-3 = n
n = 76
Now, using the formula for the sum of n terms, we get
Sn = 76/2 [2(-5) + (76-1)(-3)]
= 38 [-10 + (75)(-3)]
=38 (-10-225)
= 38(-235)
= -8930
Therefore, the sum of the A.P is  Sn = -8930 


Previous Year Questions 2019

Q25: Write the common difference of A.P.  (2019)
√3, √12, √27, √48, ......

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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


Q26: Which term of the A.P. 10, 7, 4, ...is -41 ?  (2019)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

Ans:  Let nth term of A.P. 10, 7, 4, ...is -41
∴ an = a + (n - 1)d
⇒ - 41 = 10+(n-1)(-3)     [∵ d = 7 - 10 = -3]
⇒ - 41 = 10 - 3n + 3
⇒ -41 = 13 - 3n
⇒ 3n = 54=n= 18
∴ 18th term of given A.P. is - 41 


Q27: If in an A.P. a = 15, d = - 3 and an = 0, then find the value of n.  (2019)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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


Q28: How many two digit numbers are divisible by 3?  (2019)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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


Q29: If the 9th term of an AR is zero, then show that its 29th term is double of its 19th term.      (2019, 2 Marks)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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


Q30: Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21st term?    (2019)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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.
⇒ a + (r - 1) d = 243 + 120
⇒ 3 + (r - 1) 12 = 363
⇒ (r - 1) 12 = 360 ⇒ r - 1= 30 ⇒ r = 31


Q31: If the 17th term of an A.P. exceeds its 10th term by 7, find the common difference.   (2019)

Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions  View Answer

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

The document Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 10 Maths Chapter 5 Previous Year Questions - Arithmetic Progressions

1. What is an arithmetic progression (AP) and how is it defined?
Ans.An arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is called the "common difference." An AP can be expressed in the form: a, a+d, a+2d, ..., where 'a' is the first term and 'd' is the common difference.
2. How can I find the nth term of an arithmetic progression?
Ans.The nth term of an arithmetic progression can be found using the formula: Tn = a + (n-1)d, where Tn is the nth term, 'a' is the first term, 'd' is the common difference, and 'n' is the term number.
3. What is the formula for the sum of the first n terms of an arithmetic progression?
Ans.The sum of the first n terms (Sn) of an arithmetic progression can be calculated using the formula: Sn = n/2 × (2a + (n-1)d) or Sn = n/2 × (first term + last term), where 'n' is the number of terms, 'a' is the first term, 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, calculate the differences between consecutive terms. If the differences are constant, then the sequence is an AP. If the differences vary, it is not an AP.
5. Can an arithmetic progression have a negative common difference?
Ans.Yes, an arithmetic progression can have a negative common difference. In this case, the terms of the sequence will decrease as you move along the sequence, indicating that the terms will get smaller. For example, in the sequence 10, 7, 4, 1, the common difference is -3.
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