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An arithmetic sequence is a set of numbers in which each term is the sum of the terms before it and a fixed number. This fixed number is known as a common difference. As a result, the differences between every two successive terms in an arithmetic series are the same. The total of all the digits in an arithmetic progression or series is calculated using the sum of the arithmetic sequence formula.
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Key Terms: Arithmetic Progression, Common Difference, Finite and Infinite AP, General Term, Sum
Arithmetic Progression
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Arithmetic progressions are progressions in which the difference between two consecutive terms is constant.
- Example: 2, 5, 8, 11, 14…. is an arithmetic progression.
Common Difference
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"Common difference"(d) is the difference between two consecutive terms in an AP (which is constant).
- In the progression: 2, 5, 8, 11, 14 …the common difference is 3.
If the common difference of any two consecutive terms, for any A.P, is as follows:
- Positive, the AP is increasing.
- Zero, the AP is constant.
- Negative, the A.P is decreasing.
Finite and Infinite Arithmetic Progression
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A finite AP is an A.P in which the number of terms is finite.
- Example the A.P: 2, 5, 8……32, 35, 38
An infinite A.P is an A.P in which the number of terms is infinite.
- Example 2, 5, 8, 11…..
While an infinite A.P does not have the last term, a finite A.P will.
General Term of Arithmetic Progression
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The nth term of an AP
The nth term of an A.P is given by Tn= a+(n−1)d, where a is the first term, d is a common difference and n is the number of terms.
The general form of an AP
The general form of an A.P is: (a, a+d,a+2d,a+3d……) where a is the first term and d is a common difference. Here, d=0, OR d>0, OR d<0
What Is Sum of Arithmetic Sequence Formula?
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The formula for calculating the total of all the terms in an arithmetic sequence is known as the sum of the arithmetic sequence formula. We know that the addition of the members leads to an arithmetic series of finite arithmetic progress, which is given by (a, a + d, a + 2d, …) where “a” = the first term and “d” = the common difference.
Formula for Sum of Arithmetic Sequence Formula
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We can find the sum of the arithmetic series in one of two methods. The following are the formulae for the sum of the arithmetic sequence:
Sum of Arithmetic Sequence Formula
When the Last Term is Given: S = n/2(a + L)
When the Last Term is Not Given: S = n/2 {2a + (n − 1) d}
Notations:
“S” is the sum of the arithmetic sequence,
“a” is the first term,
“d” is the common difference between the terms,
“n” is the total number of terms in the sequence and
“L” is the last term of the sequence.
Derivation of Sum of Arithmetic Series Formula
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Every term following the first is derived by adding a constant, referred to as the common difference (d), in an arithmetic sequence.
Step 1: The nth term of an arithmetic sequence,
an = a1 + (n – 1)d, where the first term is a1, the second term is a1 + d, the third term is a1+ 2d, etc and this gives the formula of the sum of the arithmetic series, Sn
Sn = a1 + (a1 + d) + (a1+ 2d) + … + [a1 + (n–1)d] _____ (1)
Step 2: We can also write it as,
Sn= an + (an – d) + (an– 2d) + … + [an– (n–1)d] _____ (2) that is, the nth term and successively subtracted the common difference
Step 3: Add the above two equations together, we get
2Sn= n (a1+ an) ⇒ Sn = n(a1 + an )/2. Thus, Sn = n/2(a1+ an).
Step 4: Substituting an = a1+ (n – 1)d,
Sn = n/2 [a1 + a1 + (n – 1)d]
Thus, Sn= n/2 [ 2a1 + (n – 1)d]
Things to Remember
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- The sum of the arithmetic sequence formula refers to the formula that gives the sum the total of all the terms present in an arithmetic sequence.
- The sum of first n terms of an arithmetic sequence where nth the term is not known: Sn=n/2[2a+(n−1)d]
- The sum of first n terms of the arithmetic sequence where the nth term, an is known: Sn=n/2[a1+an]
Sample Question
Ques. Find the sum of the first 30 terms of the sequence 1, 3, 5, 7, 9 …… (2 Marks)
Ans. Given,
a = 1
d = 2
n = 30
Using the formula: S = n/2 {2a + (n − 1)d}
S = 30/2 {2(1) + (30 − 1)2}
= 900
Ques. Find the sum of arithmetic sequence -4, -1, 2, 5, ... up to 10 terms. (2 Marks)
Ans. Here, a1=−4
a1=−4 and n=10
Using the sum of arithmetic sequence formula,
Sn=n/2[2a1+(n−1)d]=10/2[2(−4)+(10−1)3]=5×(−8+27)=95
Ques. Find the sum of 7 terms of an arithmetic sequence whose first and last terms are 10 and 40 respectively. (2 Marks)
Ans. Here,
a1=10 and a7=40
Using the sum of arithmetic sequence formula,
Sn=n/2[a1+an]=7/2[10+40]=7×50/2=7×25=175
Ques. Find the sum of the first 22 terms of an AP in which d = 7 and the 22nd term is 149. (3 Marks)
Ans.Given,
Common difference, d = 7
22nd term, a22 = 149
Sum of first 22 terms, S22 =?
By the formula of nth term,
an = a+(n−1)d
a22 = a+(22−1)d
149 = a+21×7
149 = a+147
a = 2 = First term
Sum of n terms,
Sn = n/2(a+an)
S22 = 22/2 (2+149)
= 11×151
= 1661
Ques. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. (3 Marks)
Ans.Given that,
Second term, a2 = 14
Third term, a3 = 18
Common difference, d = a3−a2 = 18−14 = 4
a2 = a+d
14 = a+4
a = 10 = First term
Sum of n terms;
Sn = n/2 [2a + (n – 1)d]
S51 = 51/2 [2×10 (51-1) 4]
= 51/2 [20+(50)×4]
= 51 × 220/2
= 51 × 110
= 5610
Ques. If the sum of the first n terms of an AP is 4n − n^2, what is the first term (that is S1)? What is the sum of the first two terms? What is the second term? Similarly find the 3rd, the the10th and the nth terms. (4 Marks)
Ans. Given that,
Sn = 4n−n^2
First term, a = S1 = 4(1) − (1)^2 = 4−1 = 3
Sum of first two terms = S2= 4(2)−(2)^2 = 8−4 = 4
Second term, a2 = S2 − S1 = 4−3 = 1
Common difference, d = a2−a = 1−3 = −2
Nth term, an = a+(n−1)d
= 3+(n −1)(−2)
= 3−2n +2
= 5−2n
Therefore, a3 = 5−2(3) = 5-6 = −1
a10 = 5−2(10) = 5−20 = −15
Hence, the sum of the first two terms is 4. The second term is 1.
The 3rd, the 10th, and the nth terms are −1, −15, and 5 − 2n respectively.
Ques. Find the sum of the first 40 positive integers divisible by 6. (4 Marks)
Ans. The positive integers that are divisible by 6 are 6, 12, 18, 24 ….
We can see here that this series forms an A.P. whose first term is 6 and the common difference is 6.
a = 6
d = 6
S40 = ?
By the formula of the sum of n terms, we know,
Sn = n/2 [2a +(n – 1)d]
Therefore, putting n = 40, we get,
S40 = 40/2 [2(6)+(40-1)6]
= 20[12+(39)(6)]
= 20(12+234)
= 20×246
= 4920
Ques. Find the sum of the first 15 multiples of 8. (4 Marks)
Ans. The multiples of 8 are 8, 16, 24, 32…
The series is in the form of AP, having the first term as 8 and the common difference as 8.
Therefore, a = 8
d = 8
S15 =?
By the formula of sum of nth term, we know,
Sn = n/2 [2a+(n-1)d]
S15 = 15/2 [2(8) + (15-1)8]
= 15/2[16 +(14)(8)]
= 15/2[16 +112]
= 15(128)/2
= 15 × 64
= 960
Ques. Find the sum of the odd numbers between 0 and 50. (4 Marks)
Ans. The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.
Therefore, we can see that these odd numbers are in the form of A.P.
Hence,
First-term, a = 1
Common difference, d = 2
Last term, l = 49
By the formula of last term, we know,
l = a+(n−1) d
49 = 1+(n−1)2
48 = 2(n − 1)
n − 1 = 24
n = 25 = Number of terms
By the formula of sum of nth term, we know,
Sn = n/2(a +l)
S25 = 25/2 (1+49)
= 25(50)/2
=(25)(25)
= 625
Ques. A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay a penalty if he has delayed the work by 30 days. (4 Marks)
Ans.
We can see, that the given penalties are in the form of A.P. having the first term as 200 and common difference as 50.
Therefore, a = 200 and d = 50
The penalty that has to be paid if the contractor has delayed the work by 30 days = S30
By the formula of the sum of nth term, we know,
Sn = n/2[2a+(n -1)d]
Therefore,
S30= 30/2[2(200)+(30 – 1)50]
= 15[400+1450]
= 15(1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as a penalty.
Ques. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students? (4 Marks)
Ans. It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2−1 = 1
Sn = n/2 [2a +(n-1)d]
S12 = 12/2 [2(1)+(12-1)(1)]
= 6(2+11)
= 6(13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3×78 = 234
Therefore, 234 trees will be planted by the students.
Ques. An AP consists of 37 terms. The sum of the three middlemost terms is 225 and the sum of the last three is 429. Find the AP. (4 Marks)
Ans. We know that,
First term of an AP = a
Common difference of AP = d
nth term of an AP, an = a + (n – 1)d
Since, n = 37 (odd),
Middle term will be (n+1)/2 = 19th term
Thus, the three middle most terms will be,
18th, 19th and 20th terms
According to the question,
a18 + a19 + a20 = 225
Using an = a + (n – 1)d
a + 17d + a + 18d + a + 19d = 225
3a + 54d = 225
3a = 225 – 54d
a = 75 – 18d … (1)
Now, we know that the last three terms will be 35th, 36th and 37th terms.
According to the question,
a35 + a36 + a37 = 429
a + 34d + a + 35d + a + 36d = 429
3a + 105d = 429
a + 35d = 143
Substituting a = 75 – 18d from equation 1,
75 – 18d + 35d = 143 [ using eqn1]
17d = 68
d = 4
Then,
a = 75 – 18(4)
a = 3
Therefore, the AP is a, a + d, a + 2d….
i.e. 3, 7, 11….
Ques. Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5. (5 Marks)
Ans. We know that,
Multiples of 2 as well as of 5 = LCM of (2, 5) = 10
Multiples of 2 as well as of 5 from 1 and 500 = 10, 20, 30…, 500.
Hence,
We can conclude that 10, 20, 30…, 500 is an AP with common difference, d = 10
First term, a = 10
Let the number of terms in this AP = n
Using nth term formula,
an = a + (n – 1)d
500 = 10 + (n – 1)10
490 = (n – 1)10
n – 1 = 49
n = 50
Sum of an AP,
Sn = (n/2) [ a + an], here an is the last term, which is given]
= (50/2) ×[10+500]
= 25× [10 + 500]
= 25(510)
= 12750
Therefore, sum of those integers from 1 to 500 which are multiples of 2 as well as of 5= 12750
Ques. Find the sum of the integers between 100 and 200 that are divisible by 9. (5 Marks)
Ans. The number between 100 and 200 which is divisible by 9 = 108, 117, 126, …198
Let the number of terms between 100 and 200 which is divisible by 9 = n
an = a + (n – 1)d
198 = 108 + (n – 1)9
90 = (n – 1)9
n – 1 = 10
n = 11
Sum of an AP = Sn = (n/2) [ a + an]
Sn = (11/2) × [108 + 198]
= (11/2) × 306
= 11(153)
= 1683
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