Arithmetic Progression (A.P. or AP)
Let us recall some formulae and properties studied earlier.

A sequence a₁, a₂, a₃, …, a_n is called arithmetic sequence or arithmetic progression if a_(n+1)=a_n+d, n∈N, where a₁ is called the first term and the constant term d is called the common difference of the A.P.

Let us consider an A.P. (in its standard form) with first term a and common difference d, i.e., a, a+d, a+2d, ….
Then the nth term (general term) of the A.P. is a_n=a+(n-1)d.

We can verify the following simple properties of an A.P.:

(i) If a constant is added to each term of an A.P., the resulting sequence is also an A.P.
(ii) If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.
(iii) If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.
(iv) If each term of an A.P. is divided by a non-zero constant then the resulting sequence is also an A.P.
Here, we shall use the following notations for an arithmetic progression:
a= the first term, ℓ= the last term, d= common difference, n= the number of terms. S_n= the sum to n terms of A.P.
Let a, a+d, a+2d, a+(n-1)d be an A.P. Then ℓ=a+(n-1)d

S_n=½n[2a+(n-1)d]
We can also write, S_n=½n(a+ℓ)

Formula for Sum to n Terms Of an AP
Let

S_n=a+(a+d)+(a+2d)+…+(a+(n-2)d)+(a+(n-1)d) … (1)
Writing the expression in the reverse order, we get
S_n=(a+(n-1)d)+(a+(n-2)d)+…+(a+2d)+(a+d)+a … (2)
Adding (1) and (2) Vertically, we get
2S_n=[2a+(n-1)d]+[2a+(n-1)d]+…+[2a+(n-1)d] … (n expressions)
⇒S_n=½n[2a+(n-1)d]
Alternative form for the Sum Formula
S_n=½n[a+a+(n-1)d]
⇒S_n=½n[a+ℓ]
where ℓ=a+(n-1)d is the last term of the AP

(Proof 1). The first and last terms of an AP are a and ℓ respectively. Show that the sum of the nth term from the beginning and the nth term form the end is (a+ℓ).
Solution:
In the given AP, first term =a and last term =ℓ.
Let the common difference be d.
Then, nth term from the beginning is given by

a_n=a+(n-1)d … (1)
Similarly, nth term from the end is given by
a_n=ℓ-(n-1)d … (2)
Adding (1) and (2), we get
a+(n-1)d+{ℓ-(n-1)d}
=a+(n-1)d+ℓ-(n-1)d
=a+ℓ
Hence, the sum of the nth term from the beginning and the nth term from the end (a+ℓ).

(Proof 2). If the pth term of an AP is q and its qth term is p then show that its (p+q)th term is zero.
Solution:
In the given AP, let the first term be a and the common difference be d.
Then, a_n=a+(n-1)d

a_p=a+(p-1)d=q … (i)
a_q=a+(q-1)d=p … (ii)
On subtracting (i) from (ii), we get
(q–p)d=(p–q)
d=-1
Putting d=-1 in (i), we get
a=(p+q-1)
Thus, a=(p+q-1) and d=-1 Now, a_(p+q)=a+(p+q-1)d
=(p+q-1)+(p+q-1)(-1)
=(p+q-1)-(p+q-1)=0
Hence, the (p+q)th term is 0 (zero).

Example 1. Find the sum of each of the following arithmetic series:
(i) 7+10½+14+⋯ +84
(ii) 34+32+30+⋯ +10
(iii) (-5)+(-8)+(-11)+⋯ +(-230)
Solution:
(i) The given arithmetic series is 7+10½+14+⋯ +84.
Here, a=7, d=10½-7=3½ and ℓ=84.
Let the given series contains n terms. Then, a_n=84.

[a_n=a+(n-1)d]
7+(n-1)∙3½=84 … (×2)
14+(n-1)∙7=168
n-1=(168-14)÷7
n-1=22
n=23

S_n=½n[a+ℓ]

∴ Required sum S₂₃=½∙23∙(7+84)
=½∙23∙91
=½∙2030
=1046½

(ii) The given arithmetic series is 34+32+30+⋯ +10.
Here, a=34, d=32-34=-2 and ℓ=10.
Let the given series contain n terms. Then, a_n=10.

[a_n=a+(n-1)d]
34+(n-1)⋅(-2)=10
-2n+36=10
-2n=10-36=-26
n=13.

S_n=½n[a+ℓ]

∴ Required sum S₁₃=½∙13∙(34+10)
=½∙13∙44
=½∙286

(iii) The given arithmetic series is (-5)+(-8)+(-11)+⋯ +(-230).
Here, a=-5, d=-8-(-5)=-8+5=-3 and ℓ=230.
Let the given series contain n terms. Then, a_n=-230.

[a_n=a+(n-1)d]
-5+(n-1)⋅(-3)=-230
-3n-2=-230
-3n=-230+2=-228
n=76

S_n=½n[a+ℓ]

∴ Required sum S₇₆=½∙76∙(-5-230)
=38∙(-235)
=-8930

Find the sum of each arithmetic series Ex2 to Ex11.
Ex2. 4+8+12+…+200
Solution:

8-4=4
12-8=4
The common difference is 4.
a₁=4, a_n=200, d=4
a_n=a₁+(n-1)d
200=4+(n-1)4
4n=200
n=50
Find the sum of the series.
S_n=½⋅n⋅(a₁+a_n )
S₅₀=½⋅50⋅(4+200)
=5100
Answer:
5100

Ex3. -18+(-15)+(-12)+…+66
Solution:

-15-(-18)=3
-12-(-15)=3
The common difference is 3.
a₁=-18, a_n=66
Find the value of n.
a_n=a₁+(n-1)d
66=-18+(n-1)3
3n=87
n=29
Find the sum.
S_n=½n(a₁+a_n )
S₂₉=½⋅29⋅(-18+66)
=696
Answer: 696

Ex4. -24+(-18)+(-12)+…+72
Solution:

-18-(-24)=6
-12-(-18)=6
The common difference is 6.
a₁=-24, a_n=72
Find the value of n.
a_n=a₁+(n-1)d
72=-24+(n-1)6
6n=102
n=17
Find the sum.
S_n=½n(a₁+a_n )
S₁₇=½⋅17⋅(-24+72)
=408
Answer: 408

Ex5. a₁=12, a_n=188, d=4
Solution:

a_n=a₁+(n-1)d
188=12+(n-1)4
4n=180
n=45
Find the sum of the series.
S_n=½⋅n⋅(a₁+a_n )
S₄₅=½⋅45⋅(12+188)
=4500
Answer:
4500

Ex6. a_n=145, d=5, n=21
Solution:

a_n=a₁+(n-1)d
145=a₁+(21-1)5
a₁=45
Find the sum of the series.
S_n=½⋅n⋅(a₁+a_n )
S₂₁=½⋅21⋅(45+145)
=1995
Answer:
1995

Ex7. the first 50 natural numbers
Solution:

S₅₀=½⋅50⋅(1+50)
=½⋅50⋅51
=1275
Answer: 1275

Ex8. the first 100 odd natural numbers
Solution: Here a₁=1, a₁₀₀=199 and n=100
Find the sum.

S_n=½n(a₁+a_n )
S₁₀₀=½⋅100⋅(1+199)
=10,000
Answer: 10,000

Ex9. the first 200 odd natural numbers
Solution:
Here a₁=1 and a₂₀₀=399.
Find the sum.

S_n=½n(a₁+a_n )
S₂₀₀=½⋅200⋅(1+399)
=40,000
Answer: 40,000

Ex10. the first 100 even natural numbers
Solution: Here a₁=2 and a₁₀₀=200.
Find the sum.

S_n=½n(a₁+a_n )
S₁₀₀=½⋅100⋅(2+200)
=10,100
Answer: 10,100

Ex11. the first 300 even natural numbers
Solution:
Here a₁=2, a₃₀₀=300 and n=300.
Find the sum.

S_n=½n(a₁+a_n )
S₃₀₀=½⋅300⋅(2+600)
=90,300
Answer: 90,300

Ex12. Determine the sum of the series: 19+22+25+…+121
solution:

a=19 and d=3
a_n=3n+16=121=ℓ
3n=105
n=35
S_n=½n(a+ℓ)
S₃₅=½⋅35⋅(19+121)=35⋅½⋅140=35⋅70=2450

Ex13. Find the sum of the series 1+3.5+6+8.5+…+101.
Solution:
This is an arithmetic series, because the difference between the terms is a constant value, 2.5. We also know that the first term is 1, and the last term is 101. But we do not know how many terms are in the series. So we will need to use the formula for the last term of an arithmetic progression,

ℓ=a+(n-1)d
to give us 101=1+(n-1)×2.5.
Now this is just an equation for n, the number of terms in the series, and we can solve it. If we subtract 1 from each side we get
100=(n-1)×2.5
and then dividing both sides by 2.5 gives us
40=n-1
so that n=41. Now we can use the formula for the sum of an arithmetic progression, in the version using ℓ, to give us
S_n=½n(a+ℓ)
S₄₁=½×41×(1+101)
=½×41×102
=41×51
2091

Ex14. Find the sum of -6+1+8+15+…+141.
solution:
The series is arithmetic with u₁=-6, d=7 and u_n=141. First we need to find n.

u₁+(n-1)d=141
-6+7(n-1)=141
7(n-1)=147
n-1=21
n=22
Using S_n=½n(u₁+u_n )
S₂₂=½⋅22⋅(-6+141)
=11⋅135=1485

Ex15. Find the number of terms of the AP -12, -9, -6, …, 21.
If 1 is added to each term of this AP then the sum of all terms of the AP thus obtained.
Solution:
The given AP is -12, -9, -6, …, 21.
Here, a=-12, d=-9-(-12)=-9+12=3 and ℓ=21.
Suppose there are n terms in the AP.

∴ ℓ=a_n=21
[a_n=a+(n-1)d]

-12+(n-1)⋅3=21
3n-15=21
3n=21+15=36
n=12.

Thus, there are 12 terms in the AP.
If 1 is added to each term of the AP, then the new AP so obtained is -11, -8, -5, …, 22.
Here, first term, a=-11, last term, ℓ=22 and n=12.
Get the sum of the terrns of this AP.
S_n=½n(a+ℓ)
∴ S₁₂=½⋅12⋅(-11+22)
=6⋅11=66
Hence, the required sum is 66.

Ex16. The sum of the first 20 odd natural numbers is
(a) 100 (b) 210 (c) 400 (d) 420
Solution:
The first 20 odd natural numbers are 1, 3, 5, …, 39.
These numbers are in AP.
Here a=1, ℓ=39 and n=20.
∴ Sum of first 20 odd tural number

S_n=½n(a+ℓ)
S₂₀=½⋅20⋅(1+19)
=10⋅40
=400.
Answer: (c) 400

Ex17. Find the sum of all odd numbers between 0 and 50.
Solution:
All odd numbers between 0 and 50 are 1, 3, 5, 7, …, 49.
This is an AP in which a=1, d=(3-1)=2 and ℓ=49.
Let the number of terms be n.
Then, a_n=49

a+(n-1)d=49
1+(n-1)⋅2=49
2n=50
n=25
∴ Required sum =½n(a+ℓ)
=½⋅25⋅(1+49)
=25⋅½⋅50
=25⋅25=625
Hence, the required sum is 625.

Ex18: Finding the Sum of an Arithmetic Series
Find the sum of all the odd numbers between 51 and 99, inclusive.
Solution:
First, use a₁=51, a_n=99 to find n:

a_n=a₁+(n-1)d
99=51+(n-1)2
n=25
Now, find S₂₅.
S_n=½n(a₁+a_n )
S₂₅=½⋅25⋅(51+99)
=1,875

Ex19. Find the sum of all integers between 100 and 1000 which are divisible by 9.
Solution: The first integer greater than 100 and divisible by 9 is 108 and the integer just smaller than 1000 and divisible by 9 is 999. Thus, we have to find the sum of the series.

108+117+126+…+999.
Here t=a=108, d=9 and ℓ=999
Let n be the total number of terms in the series be n. Then
999=108+9(n-1) … (÷9)
111=12+(n-1)
n=100
Hence, the required sum
S_n=½n(a+ℓ)
= ½⋅100⋅(108+999)
=50(1107)=55350.
Lets read post Numbers which are divisible by or multiples of an integer in Arithmetic Sequence
Question 1. (5+13+21+…+181)=?
(a) 2476 (b) 2337 (c) 2219 (d) 2139
Solution:
Here, a=5, d=(13-5)=8 and ℓ=181.
Let the number of terms be n. Then a_n=181
a+(n-1)d=181
5+(n-1)⋅8=181
8n=184
n=23
∴ Required sum =½n(a+ℓ)
=23⋅½⋅(5+181)=23⋅93=2139.
Hence, the required sum is 2139.
Answer: (d) 2139

Q2. A local theater has 30 seats in the first row and 50 rows in all. Each successive row contains two additional seats. How many seats are in the theater?
solution:
I need to know how many seats are in the 50th row. I will find a formula for an to get started.

a_n=2n+x
a₁=2(1)+x
30=2+x
28=x
formula for sequence: a_n=2n+28.
Computation of number of seats in 50th row: a₅₀=2(50)+28=128.
I need to add these numbers to get my answer: 30+32+34+…..+128.
I will use the formula S_n=½(a₁+a_n )
S₅₀=½⋅50⋅(30+128)=25(158)=3950
Answer: There are 3950 seats.

Q3. An outdoor amphitheater has 40 seats in the first row, 41 seats in the second row, 42 seats in the third row. The pattern continues. The amphitheater has 70 rows.

How many seats does the amphitheater contain?
solution:
I need to know how many seats are in the 70th row. I will find a formula for an to get started.
a_n=1n+x
a₁=1(1)+x
40=1+x
39=x
formula for sequence: a_n=n+39.
Computation of number of seats in 70th row: a₇₀=70+39=109.
I need to add these numbers to get my answer: 40+41+42+…..+109.
I will use the formula S_n=½(a₁+a_n ).
S₇₀=½⋅70⋅(40+109)=35(149)=5215
Answer: There are 5215 seats

Q4. 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?
Solution:
Suppose there are n terms in the AP.
Here, a=17, d=9 and ℓ=350.

a_n=350
[a_n=a+(n-1)d]
17+(n-1)⋅9=350
9n+8=350
9n=350-8=342
n=38.
Thus, there are 38 terms in the P.
S_n=½n(a+ℓ)
S₃₈=½⋅38⋅(17+350)
=19⋅367
=6973
Hence, the required sum is 6973.

Q5. (Contests) The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?
Solution:
Given, a₁=150, d=50 and n=11.
Find the value of a₁₁.

a_n=a₁+(n-1)d
a₁₁=a₁+(11-1)d
=150+10⋅50
=650
Find the sum.
S_n=½⋅n⋅(a₁+a_n )
=11⋅½⋅(150+650)
=4400
A cash prizes totaled $4400 for the eleven week contest.
Answer: $4400

Q6. (Drama) Laura has a drama performance in 12 days. She plans to practice her lines each night. On the first night she rehearses her lines 2 times. The next night she rehearses her lines 4 times. The third night she rehearses her lines 6 times. On the eleventh night, how many times has she rehearsed her lines?
Solution:
The arithmetic sequence that represents the situation is 2, 4, 6, ….
Substitute 2 for a₁, 2 for d, and 11 for n in the formula for the nth term and find a₁₁.

a_n=a₁+(n-1)d
a₁₁=2+(11-1)2
=2+20
=22
Substitute 2 for a₁, 22 for a_n, 11 for n in the Sum formula
S_n=½n(a₁+a_n )
=11⋅½⋅(2+22)
=11⋅12
=132
Answer: 132