Here you get the CBSE Class 10 Mathematics chapter 5 Arithmetic Progression Objective Questions. You will find only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is also provided with a detailed solution, which the students can refer to at a later stage.
All these questions are crucial as it helps the students to prepare for CBSE Class 10 Mathematics Board Examination. As per the latest modified exam pattern, the question papers will contain more of objective type questions. This topic is also very important from the standpoint of competitive exams. Find the CBSE Class 1o Maths Objective Questions below:
List of SubTopics Covered In This Chapter
5.1 General Term of AP (7 MCQs From The Topic)
5.2 Introduction to AP (6 MCQs Listed From The Topic)
5.3 Sum of Terms in AP (7MCQs From This Topic)
Download Free CBSE Class 10 Maths Chapter 5 Arithmetic Progression Objective Questions PDF
General Term of AP

 Find the number of terms in each of the following APs:(i)Â 7,13,19….,205
(ii)Â 18,31/2,13,…âˆ’47
 26, 35
 27, 34
 35, 26
 34, 27
 Find the number of terms in each of the following APs:(i)Â 7,13,19….,205
Answer: (D) 34, 27
Solution: (i) 7, 13, 19…., 205
First term,Â a=7
Common difference,Â d=13âˆ’7=6
a_{n}=205
Using formulaÂ an=a+ (nâˆ’1) dÂ to findÂ n^{th}Â term of arithmetic progression,Â we get
205=7+ (nâˆ’1)6
â‡’205=6n+1
â‡’204=6n
â‡’n=204/6=34
Therefore, there are 34 terms in the given arithmetic progression.
(ii)Â 18,31/2,13,…âˆ’47
First term,Â a=18
Common difference,Â d= (31/2) âˆ’18=âˆ’5/2
a_{n}Â = âˆ’ 47
Using formulaÂ a_{n}=a+ (nâˆ’1) dÂ to findÂ n^{thÂ }term of arithmetic progression, we get
âˆ’47=18 + (nâˆ’1) (âˆ’5/2)
â‡’âˆ’94=36âˆ’5n+5
â‡’5n=135
â‡’n=135/5=27
Therefore, there are 27 terms in the given arithmetic progression.

 Find theÂ 31stÂ term of an AP whoseÂ 11thÂ term is 38 andÂ 16thÂ term is 73.
 185
 210
 178
 150
 Find theÂ 31stÂ term of an AP whoseÂ 11thÂ term is 38 andÂ 16thÂ term is 73.
Answer: (C) 178
Solution: We are given that a_{11}=38Â andÂ a_{16}=73Â where, a_{11Â }is theÂ 11^{th}Â term andÂ Â a_{16}Â is theÂ 16^{th}Â term of an AP.
Using formulaÂ a_{n}=a+ (nâˆ’1) dÂ to findÂ nthÂ term of arithmetic progression, we get
38=a+10d….. (i)
73=a+15dÂ …… (ii)
equation (ii) – equation (i) gives,
35 = 5d
d = 7….. (iii)
Substituting (iii) in (i) we get, a = 32
a_{31}=âˆ’32+(31âˆ’1)(7)
â‡’âˆ’32+210=178
Therefore,Â 31^{st}Â term of AP is 178.

 If the third and the ninth terms of an AP are 4 and 8 respectively, which term of this AP is zero?
 5^{th}
 4^{th}
 3^{rd}
 6^{th}
 If the third and the ninth terms of an AP are 4 and 8 respectively, which term of this AP is zero?
Answer: (B) 4^{th}
Solution: It is given thatÂ 3rdÂ andÂ 9thÂ term of AP are 4 and 8 respectively.
It meansÂ a_{3}=4Â andÂ a_{9}=âˆ’8
Where,Â a_{3}Â andÂ a_{9}Â are third and ninth terms respectively.
Using formulaÂ a_{n}=a+(nâˆ’1)dÂ to findÂ n^{th}Â term of arithmetic progression, we get
4=a+(3âˆ’1)d
â‡’4Â =Â a+2d…(i)
âˆ’8Â =Â a+(9âˆ’1)d
â‡’âˆ’8=a+8d…(ii)
From equation (i)Â Â we haveÂ Â a=4âˆ’2d
Substituting in equation (ii) , we have
âˆ’8=4âˆ’2d+8d
â‡’âˆ’12Â =Â 6d
â‡’dÂ =âˆ’12/6Â =âˆ’2
Solving for (a) , we get
â‡’âˆ’8Â =Â aâˆ’16
â‡’aÂ =Â 8
Therefore, first termÂ a = 8Â and Common DifferenceÂ d =âˆ’2
We knowÂ Â a_{n}Â =Â a+ (nâˆ’1) dÂ (whereÂ a_{n}Â is theÂ n^{th}Â term)
Finding value of n whereÂ an=0
0=8+ (nâˆ’1) (âˆ’2)
â‡’0Â =Â 8âˆ’2n+2
â‡’0Â =Â 10âˆ’2n
â‡’2nÂ =Â 10
â‡’nÂ =Â 10/2=5
Therefore,Â 5thÂ term is equal toÂ 0.

 Which term of the AP: 3, 15, 27, 39, â€¦ will be 132 more than itsÂ 54^{th}Â term?
 70
 65
 80
 55
 Which term of the AP: 3, 15, 27, 39, â€¦ will be 132 more than itsÂ 54^{th}Â term?
Answer: (B) 65
Solution: Letâ€™s first calculateÂ 54^{th}Â of the given AP.
First term = a = 3
Common difference = d =15 – 3 = 12
Using formulaÂ a_{n}=a+ (nâˆ’1) d, Â Â to findÂ n^{thÂ Â }term of arithmetic progression,Â we get
a_{54}=a+ (54âˆ’1) d
a_{54}Â = 3 + 53 (12) =3 + 636=639
We want to find which term is 132 more than itsÂ 54^{th}Â term. Letâ€™s suppose it isÂ n^{th}Â term which is 132 more thanÂ 54^{th}Â term.
Therefore, we can say that
a_{n}Â =Â a_{54}+132
a_{n}=a+ (nâˆ’1) d=3+ (nâˆ’1) (12)}
â‡’3+ (nâˆ’1)12=639+132
â‡’3+12nâˆ’12=771
â‡’12nâˆ’9=771
â‡’12n=780
â‡’n=Â 780/12=65
Therefore,Â 65^{th}Â term is 132 more than itsÂ 54^{th}Â term.

 Two AP’s have the same common difference. The difference between theirÂ 100^{th}Â terms 100, what is the difference between theirÂ 1000^{th}Â terms.
 200
 150
 100
 55
 Two AP’s have the same common difference. The difference between theirÂ 100^{th}Â terms 100, what is the difference between theirÂ 1000^{th}Â terms.
Answer: (C) 100
Solution: Let first term of first AP =Â a
Let first term of 2nd AP =Â aâ€²
It is given that their common difference is same. Let their common difference beÂ d.
It is given that difference between theirÂ 100^{th}Â terms is 100.Â Using formulaÂ a_{n}Â = a + (nâˆ’1) d, Â to findÂ n^{th}Â term ofÂ arithmetic progression, we can say that
a+ (100âˆ’1) dâˆ’ (aâ€² +(100âˆ’1)d)=a+99dâˆ’aâ€²âˆ’99d=100
â‡’aâˆ’aâ€²=100Â Â Â ……..Â (1)
We want to find difference between theirÂ 1000^{th}Â terms which means we want to calculate:
a+ (1000âˆ’1) dâˆ’(aâ€²+(1000âˆ’1)d)=a+999dâˆ’aâ€²âˆ’999d=aâˆ’aâ€²
Putting equationÂ (1)Â in the above equation we get,
a+ (1000âˆ’1) dâˆ’(aâ€²+(1000âˆ’1)d)=a+999dâˆ’aâ€²999d=aâˆ’aâ€²=100
Therefore, difference between theirÂ 1000^{th}Â terms would be equal to 100

 How many threedigit numbers are divisible by 7?
 112
 114
 128
 110
 How many threedigit numbers are divisible by 7?
Answer: (C) 128
Solution: We have an APÂ starting atÂ 105Â because it is the first three digit number divisible byÂ 7.
AP will end atÂ 994Â because it is the last three digit number divisible byÂ 7.
Therefore, we have an AP 105,112,119…994
First term, a = 105
Common difference,Â dÂ = 112 – 105 = 7
Using formulaÂ a_{n}=a+ (nâˆ’1) d, to findÂ Â n^{th}Â term of arithmetic progression,Â we can say that
994=105+ (nâˆ’1) (7)
â‡’994=105+ (nâˆ’1) (7)
â‡’889=7(nâˆ’1)
â‡’nâˆ’1=889/7â‡’n=127+1=128
994Â is theÂ 128^{th}Â term of AP. Therefore, there areÂ 128Â terms in AP. In other words, we can also say that there are 128 three digit numbers divisible by 7.

 For what value of n, are theÂ n^{th}Â terms of two APâ€™s: 63, 65, 67 …. and 3, 10, 17,.. equal?
 11
 14
 12
 13
 For what value of n, are theÂ n^{th}Â terms of two APâ€™s: 63, 65, 67 …. and 3, 10, 17,.. equal?
Answer: (D) 13
Solution: Letâ€™s first consider AP:Â 63, 65, 67…..
First termÂ =a=63
Common differenceÂ =d=65âˆ’63=2
Using formulaÂ a_{n}=a+ (nâˆ’1) d, Â to findÂ Â n^{th}Â term of arithmetic progression, we can say that
a_{n}=63+ (nâˆ’1) (2)Â Â Â Â Â Â Â Â Â Â Â Â (1)
Now, consider second APÂ 3, 10, 17…
First term = a = 3
Common difference = d = 10 3 = 7
Using formulaÂ a_{n}=a+ (nâˆ’1) d, Â to findÂ Â n^{th}Â term of arithmetic progression, we can say that
a_{n}=3+(nâˆ’1)(7)Â Â Â Â Â Â Â Â Â Â Â (2)
According to the given condition, we can write
(1) = (2)
â‡’63+ (nâˆ’1) (2) =3+ (nâˆ’1) (7)
â‡’63+2nâˆ’2=3+7nâˆ’7
â‡’65=5n
â‡’n=65/5=13
Therefore,Â 13thÂ terms of both the AP’s are equal.
Introduction to AP

 IfÂ (x+1),3xÂ andÂ (4x+2)Â are first three terms of an AP, then itsÂ 5thÂ term is :
 14
 19
 24
 28
 IfÂ (x+1),3xÂ andÂ (4x+2)Â are first three terms of an AP, then itsÂ 5thÂ term is :
Answer: (D) 28
Solution: Given,Â (x+1), 3x, (4x+2)Â are in AP.
Hence, the difference of two consecutive terms will be same.
Hence,Â 3xâ€“(x+1) = (4x+2)â€“3x
â‡’2xâˆ’1=x+2
â‡’x=3
So, the first term,
a=(x+1) =4.
TheÂ common difference,
d=9â€“4=5.
TheÂ n^{th}Â term of an AP is given by
t_{n }=a+ (nâˆ’1) d
â‡’t_{5}=4+4(5) =24.

 The sum of first ten terms of an A.P. is four times the sum of its first five terms, then ratio of the first term and common difference is
 Â½
 Â¼
 4
 1
 The sum of first ten terms of an A.P. is four times the sum of its first five terms, then ratio of the first term and common difference is
Answer: (A) Â½
Solution: LetÂ S_{10}Â be the sum of first 10 terms andÂ S5Â be theÂ sum of first 5 terms.
We know,
S_{n = (n/2) [2a + (n1) d] }
Given,
S_{10}=4S_{5}
â‡’ (10/2) [2a + (101) d] = 4 Ã— (5/2) [2a + (51) d]
â‡’ (10/2) [2a+ 9d] = 4 Ã— (5/2) [2a+ 4d]
â‡’ 2a+ 9d = 4a+ 8d
â‡’ a/d= Â½

 Find the common difference (d) in the following APs respectively.
(i) Â Â 20,Â 40, 60, 80, 100,..
(ii) Â Â 5, 0, 5, 10, 15, …
(iii) Â 2, 2, 2, 2, 2..


 20, 5, 0
 20, 5, 0
 20, 5, 0
 20, 5, 0

Answer: (D) 20, 5, 0
Solution: Common difference is
(i) Â 20, 40, 60, 80, 100,
d= 4020 =20
(ii) 5, 0, 5, 10, 15, …
d= 05 = 5
(iii) 2, 2, 2, 2, 2….
d= 22 =0

 Which of the following sequences form an AP?
(i) 2, 4, 8, 16â€¦â€¦..
(ii) 2, 3, 5, 7, 11â€¦â€¦.
(iii) 1, 1.25, 1.5, 1.75â€¦â€¦
(iv) 1, 1, 3, 5, 7â€¦â€¦â€¦â€¦
Answer: (iii) 1, 1.25, 1.5, 1.75â€¦â€¦ and
(iv) 1, 1, 3, 5, 7â€¦â€¦â€¦â€¦
Solution: Consider each list of numbers:
(i) 2, 4, 8, 16â€¦â€¦..
Difference between the first two terms = 4 – 2 = 2
Difference between the third and second termÂ = 8 – 4 = 4
SinceÂ a_{2}â€“a_{1}â‰ a_{3}â€“a_{2}, this sequence is not an AP
(ii) 2, 3, 5, 7, 11â€¦â€¦..
Difference between the first two terms = 3 – 2 = 1
Difference between the third and second termÂ = 5 – 3 = 2
SinceÂ a_{2}â€“a_{1}â‰ a_{3}â€“a_{2}, this sequence is not an AP
(iii) 1, 1.25, 1.5, 1.75â€¦â€¦.
Difference between the first two terms = 1.25 â€“ (1) = 0.25
Difference between the third and second termÂ = 1.5 â€“ (1.25) = 0.25
Difference between the fourth and third termÂ = 1.75 â€“ (1.5) = 0.25
SinceÂ a_{2}â€“a_{1}=a_{3}âˆ’a_{2}=a_{4}âˆ’a_{3}, this sequence is an AP
(iv) 1, 1, 3, 5, 7â€¦â€¦â€¦â€¦
Difference between the first two terms = 1 â€“ 1 = 2
Difference between the third and second termÂ = 3 â€“ (1) = 2
Difference between the fourth and third termÂ = 5 â€“ (3) = 2
SinceÂ a_{2}â€“a_{1}=a_{3}âˆ’a_{2}=a_{4}âˆ’a_{3}, this sequence is an AP

 What are the conditions for a sequenceÂ to be an AP?
 The sum of two consecutive terms should be constant.
 The product of two consecutive numbers should be constant.
 The difference between two consecutive terms should be constant.
 The ratio of two consecutive terms should be constant.
 What are the conditions for a sequenceÂ to be an AP?
Answer: (C)The difference between two consecutive terms should be constant.
Solution: Let a1,Â a_{2},Â a_{3},Â a_{4},_{Â }a_{5},Â a_{6,}Â a_{7,}Â a_{8}â€¦ be a sequence.
For this sequence to be an AP, the difference between any two consecutive terms should be constant.
a_{2}â€“a_{1}=a_{3}â€“a_{2}=a_{4}â€“a_{3}=d
This difference is called the common difference of the AP and is denoted by d.
So an AP can also be represented in this form as wellÂ a, a+d, a+2d…

 Which of the following list of numbers forms an AP?
(i) 4, 4 +Â âˆš3Â , 4 + 2âˆš3, 4 + 3âˆš3, 4 + 4âˆš3
(ii) 0.3, 0.33, 0.333, 0.3333, 0.33333
(iii)Â 3/5, 6/5,Â 9/5,Â 12/5, 3
(iv) 1/5, 1/5, 1/5, 1/5, 1/5
Answer: (i), (iii) and (iv)
Solution: Consider each series (i) 4, 4 +Â âˆš3, 4 + 2Â âˆš3, 4 + 3Â âˆš3, 4 + 4Â âˆš3
Difference between first two consecutive terms = 4 +Â âˆš3Â – 4 =Â âˆš3
Difference between third and second consecutive terms = 4 + 2Â âˆš3Â – 4 +Â âˆš3Â =Â Â âˆš3
Difference between fourth and third consecutive terms = 4 + 3Â âˆš3â€“ 4 + 2Â âˆš3Â =Â Â âˆš3
SinceÂ a_{2}â€“a_{1}Â =Â a_{3}âˆ’a_{2}Â =Â a_{4}âˆ’a_{3}
This series is an AP
(ii) 0.3, 0.33, 0.333, 0.3333, 0.33333
Difference between first two consecutive terms = 0.330.3 = 0.03
Difference between third and second consecutive terms = 0.333 – 0.33 = 0.003
SinceÂ Â a_{2}â€“a_{1}Â â‰ Â a_{3}âˆ’a_{2}
This series is not an AP
(iii)Â 3/5,Â 6/5,Â 9/5,Â 12/5, 3
Difference between first two consecutive terms =Â (6/5)â€“(3/5)Â =Â 3/5
Difference between third and second consecutive terms =Â (9/5)â€“(6/5)Â =Â 3/5
Difference between fourth and third consecutive terms =Â (12/5)Â â€“Â (9/5)Â =Â 3/5
Since Â a_{2}â€“a_{1}Â =Â a_{3}â€“a_{2}Â =Â a_{4}â€“a_{3}
This series is an AP
(iv) 1/5, 1/5, 1/5, 1/5, 1/5
Difference between first two consecutive terms = 1/5Â â€“ (Â 1/5) = 0
Difference between third and second consecutive terms = 1/5Â â€“ (1/5) = 0
Difference between fourth and third consecutive terms = 1/5Â â€“ (1/5) = 0
SinceÂ a_{2}â€“a_{1}Â =Â a_{3}â€“a_{2}Â =Â a_{4}â€“a_{3}
This series is an AP
Sum of Terms in AP

 The first term of an AP is 5, the last term is 50 and the sum is 440. Find the number of terms and the common difference
 17; 3
 16; 2
 17; 2
 16; 3
 The first term of an AP is 5, the last term is 50 and the sum is 440. Find the number of terms and the common difference
Answer: (D)16; 3
Solution: First term,Â a=5
Last term,Â l=50
S_{n}=440
Applying formula,Â S_{n}= n/2 (a+l)Â to findÂ sum of n terms of AP,Â we get
440= n/2 (5+50)
â‡’440/55=n/2
â‡’8=n/ 2
â‡’n=16
Applying formula,Â S_{n}=n/2(2a+ (nâˆ’1) d)Â to findÂ sum ofÂ nÂ terms of APÂ and putting value ofÂ n,Â we get
440=16/2(2(5) + (16âˆ’1) d)
â‡’440=8(10+15d)
â‡’10+15d=55
â‡’15d=45
â‡’d=45/15=3

 The first and the last terms of an AP are 17 and 350 respectively. If, the common difference is 9, how many terms are there and what is their sum?
 38; 7114
 32; 7114
 38; 6973
 32; 6973
 The first and the last terms of an AP are 17 and 350 respectively. If, the common difference is 9, how many terms are there and what is their sum?
Answer: (C) 38; 6973
Solution: First term =Â a = 17
Last term = lÂ = 350
Common difference =Â d = 9
Using formulaÂ a_{nÂ }= a + (nâˆ’1) d,Â to findÂ nth term of arithmetic progression,Â we can say that
350=17+ (nâˆ’1) (9)
â‡’350=17+9nâˆ’9
â‡’342=9n
â‡’n=342/9=38
Applying formula,Â S_{n}=n/2(2a+ (nâˆ’1) d)Â to findÂ sum of n terms of APÂ and putting value of n, we get
S_{38}=38/2(34+ (38âˆ’1) (9))
â‡’S_{38}=19(34+333) =6973
Therefore, there are 38 terms and their sum is equal to 6973.

 Find the sum of first 22 terms of an AP in which d = 7 andÂ theÂ 22^{nd}Â term is 149.
 1623
 1712
 1542
 1661
 Find the sum of first 22 terms of an AP in which d = 7 andÂ theÂ 22^{nd}Â term is 149.
Answer: (D) 1661
Solution: It is given thatÂ 22ndÂ term is equal to 149.
It meansÂ a_{22}=149
Using formulaÂ a_{n}=a+ (nâˆ’1) d, Â to findÂ n^{th}Â term of AP,Â we can say that
149=a+ (22âˆ’1) (7)
â‡’149=a+147
â‡’a=2
Applying formula,Â S_{n}Â = n/2(2a+ (nâˆ’1) d)Â to find Sum of n terms of AP and putting value of a,Â we get
S_{22}=Â 22/2(4+ (22âˆ’1) (7))
â‡’S_{22}=11(4+147)
â‡’S_{22}=1661
Therefore, sum of first 22 terms of AP is equal toÂ 1661.

 Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
 5610
 5840
 5320
 5000
 Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer: (A) 5610
Solution: It is given that second and third terms of AP are 14 and 18 respectively.
Using formulaÂ a_{n}=a+ (nâˆ’1) d, Â to findÂ n^{th} term of arithmetic progression,Â we can say that
14=a+ (2âˆ’1) d
â‡’14=a+ dÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (1)
And, Â Â 18=a+ (3âˆ’1) d
â‡’18=a+2dÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (2)
These are equations consisting of two variables. We can solve them by the method of substitution.
Using equationÂ (1),Â we can say thatÂ a=14âˆ’d
Putting value ofÂ aÂ in equationÂ (2),Â we can say that
18=14âˆ’d+2d
â‡’d=4
Therefore, common differenceÂ d=4
Putting value ofÂ dÂ in equation numberÂ (1), we can say that
18=a+2(4)
â‡’a=10
Applying formula,Â S_{n}= n/2(2a+ (nâˆ’1) d)Â to findÂ sum of n terms of AP,Â we get
S_{15}=51/2(20+ (51âˆ’1) 4) =51/2(20+200) =51Ã—110 =5610
Therefore, sum of first 51 terms of an AP is equal to 5610.

 The sum of fourÂ consecutive numbers inÂ an A.P. with d > 0 is 20. Sum of their square is 120, then the middle terms are
 8, 10
 6,8
 4,6
 2,4
 The sum of fourÂ consecutive numbers inÂ an A.P. with d > 0 is 20. Sum of their square is 120, then the middle terms are
Answer: (C) 4, 6
Solution: Let the numbers areÂ aâˆ’3d,aâˆ’d,a+d,a+3d
Given:Â aâˆ’3d+aâ€”d+a+d+a+3d=20
4a=20
a=5Â and
(aâˆ’3d)^{2}+Â (aâˆ’d)^{2}Â +Â (a+3d)^{2}Â +Â (a+3d)^{2}=120
(4)(a)^{2}Â + 20Â d^{2}=120
(4)(5)^{2}+20Â d^{2}=120
d^{2}=1
d=+1Â orâˆ’1
HenceÂ numbersÂ areÂ 2,4,6,andÂ 8

 Find the sum of all the nonnegative terms of the following sequence:Â 100, 97, 94, â€¦ .
 1717
 1719
 1721
 1723
 Find the sum of all the nonnegative terms of the following sequence:Â 100, 97, 94, â€¦ .
Answer: (A) 1717
Solution: Here,Â t_{1}Â = 100, common difference, d =Â t_{2}Â â€“Â t_{1}Â = 97 â€“ 100 = 3
t_{n}Â =Â t_{1Â }+ (n1) d â‡’100 + (n1)(3) = 100 â€“ 3(n1) = 103 â€“ 3n.
LetÂ t_{n}Â be the first negative term. i.e,
t_{n}Â < 0Â â‡’Â 103 – 3n < 0Â â‡’Â n >Â 103/3Â > 34
That is, theÂ 35^{th}Â term will be negative.
Sum of the firstÂ 34Â termsÂ = (34/2) (first term + last term)
=Â 17(100+100+ (34âˆ’1) (âˆ’3)) =1717

 What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3?
 149700
 164749
 164850
 897
 What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3?
Answer: (C) 164850
Solution: The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.
The next number that will leave a remainder of 2 when divided by 3 is 104, 107, ….
The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.
So, it is an AP with the first term being 101 and the last term being 998 and common difference being 3.
Sum of an AP =Â Â n/2(a+l)
we know that in an A.P., the nth termÂ a_{n}=a_{1}+ (nâˆ’1)Ã—d
In this case, therefore,
998 = 101 + (n – 1)Â Ã—Â 3
â‡’897= (nâˆ’1) Ã—3
â‡’299= (nâˆ’1)
â‡’n=300
Sum of the AP will therefore, be (101+198)/ 2 Ã—300= 164, 850
In this chapter, concepts discussed include pattern in succeeding term obtained by adding a fixed number to the preceding terms. Students will also see how to find nth terms and the sum of n consecutive terms. Students will learn arithmetic progression effectively when they solve daily life problems.
This chapter has Arithmetic Progression Derivation of the nth term and sum of the first n terms of an A.P. and their application in solving daily life problems.
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