Sum Formulas for Finite Arithmetic Series
If a₁, a₂, a₃, ⋯ , a_n is a finite arithmetic sequence, then the corresponding series a₁+a₂+a₃+ ⋯ +a_n is called an arithmetic series. We will derive two sim- ple and very useful formulas for the sum of an arithmetic series. Let d be the common difference of the arithmetic sequence a₁, a₂, a₃, ⋯ , a_n and let S_n denote the sum of the series a₁+a₂+a₃+ ⋯ +a_n.
Then

S_n=a₁+(a₁+d)+ ⋯ +[a₁+(n-2)d]+[a₁+(n-1)d]
Reversing the order of the sum, we obtain
S_n=[a₁+(n-1)d]+[a₁+(n-2)d]+ ⋯ +(a₁+d)+a₁
Adding the left sides of these two equations and corresponding elements of the right sides, we see that
S_n=[2a₁+(n-1)d]+[2a₁+(n-1)d]+ ⋯ +[2a₁+(n-1)d]
This can be restated as in Theorem 1.

Theorem 1: Sum of an Arithmetic Series — First Form

S_n=½n[2a₁+(n-1)d]
By replacing a₁+(n-1)d with a_n, we obtain a second useful formula for the sum.
Theorem 2: Sum of an Arithmetic Series — Second Form
S_n=½n(a₁+a_n)

An arithmetic series is an arithmetic progression with plus signs between the terms instead of commas. We can find the sum of the first n terms, which we will denote by S_n, using another formula:

S_n=½n[2a+(n-1)d]

Example 1: Determining the Sum, Given the First Term and Common Difference
An arithmetic series has t₁=5.5 and d=-2.5; determine S₄₀.
solution:
Use S_n=½n(2t₁+d(n-1))
Substitute: n=40, t₁=5.5, d=-2.5

S₄₀=½⋅40⋅(2⋅5.5-2.5⋅(40-1))
S₄₀=20⋅(11-97.5)
S₄₀=-1730

Example 2: Finding the Sum of an Arithmetic Series
Find the sum of the first 26 terms of an arithmetic series if the first term is -7 and d=3.
Solution:
Let n=26, a₁=-7, d=3, and use Theorem 1.

S_n=½n[2a₁+(n-1)d]
S₂₆=½⋅26⋅[2(-7)+(26-1)3]=793

Example 3. Determine the sum of the first 20 terms of the series: 3+7+11+15+ ⋯
Solution:

a=3, n=20, d=4
S_n=½n[2a+(n-1)d]
S₂₀=½⋅20⋅[2⋅3+19⋅4]
=10⋅(6+76)
S₂₀=820
The sum of the first 20 terms is 820.

Example 4. If the first 3 terms in an arithmetic progression are 3, 7, 11 then what is the sum of the first 10 terms? Note that a=3, d=4 and n=10.

S₁₀=½⋅10⋅[2⋅3+(10-1)⋅4]
=5(6+36)=210
Alternatively, but more tediously, We add the first 10 terms together:
S₁₀=3+7+11+15+19+23+27+31+35+39=210
This method would have drawbacks if we had to add 100 terms together!

Example 5. If the first 3 terms in an arithmetic progression are 8, 5, 2 then what is the sum of the first 16 terms?

S₁₆=½⋅16⋅[2⋅8+(16-1)⋅(-3)]
=8(16-45)=-232

Example 6. Find the sum of 4+7+10+13+ ⋯ to 50 terms.
solution:
The series is arithmetic with u₁=4, d=3 and n=50.
Now S_n=½n(2u₁+(n-1)d)

∴S₅₀=½⋅50⋅(2⋅4+(50-1)⋅3)
=25⋅(8+147)=25⋅155
=3875

Example 7. Find the sum of the first 20 terms in the sequence t_n: {12, 25, 38, …}.
Steps:
(1) Write the formula for the sum of the first n terms in the arithmetic sequence.

S_n=½n(2a+(n-1)d)
(2) Identify the variables.
a=12, d=25-12=13, n=20
(3) Substitute values of a, d and n into the formula and evaluate.
S₂₀=½⋅20⋅(2⋅12+19⋅13)
S₂₀=10⋅(24+247)=10⋅271
S₂₀=2710

Example 8. Find the sum of the series 3+11+19+ ⋯ to 16 terms.
Solution:
Here a=3, d= 11-3=8, n=16
Using formula S_n=½n[2a+(n-1)d]

S₁₆=½⋅16⋅[2⋅3+(16-1)⋅8]
=8⋅[6+15⋅8]
=8(6+120)
S₁₆=8×126=1008

Example 9. Find the sum of the first 50 terms of the sequence

1, 3, 5, 7, 9, ⋯ .
Solution:
This is an arithmetic progression, and we can write down
a=1, d=2, n=50.
We now use the formula, so that
S_n=½n(2a+(n-1)d)
S₅₀=½×50×(2×1+(50-1)×2)
=25×(2+98)
=2500

Example 10. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Here, a₂=14, a₃=18 and n=51.

a_n=a+(n-1)d
a₂=a+(2-1)d
14=a+d
a=14-d ⋯ (1)
and a₃=a+(3-1)d
18=a+2d
Putting the value of a from equation (1), we get
18=14-d+2d
d=4
Putting the value of d in equation (1), we get
a=14-4=10
The sum of n terms of an AP is given by
S_n=½n[2a+(n-1)d]
S₅₁=½⋅51⋅[2⋅10+(51-1)⋅4]
=½⋅51⋅[20+200]
=51⋅½⋅220
=51⋅110=5610

Example 11. Show that a₁, a₂, ⋯ , a_n, form an AP where a_n is defined as below:
(i) a_n=3+4n (ii) a_n=9-5n
Also find the sum of the first 15 terms in each case.
Solution:
(i) a_n=3+4n
Putting n=1, we get

a₁=3+4(1)=7
Putting n=2, we get
a₂=3+4(2)=11
Similarly, a₃=3+4(3)=15
a₄=3+4(4)=19
Difference between the successive terms:
a₂–a₁=11-7=4
a₃–a₂=15-11=4
a₄–a₃=19-15=4
The difference between successive terms are same, hence a₁, a₂, ⋯ , a_n, is an AP The sum of n terms of an AP is given by
S_n=½n[2a+(n-1)d]
S₁₅=½⋅15⋅[2⋅7+(15-1)⋅4]
=½⋅15⋅(14+56)=15⋅½⋅70
=525
(ii) a_n=9-5n
Putting n=1, we get
a₁=9-5(1)=4
Putting n=2, we get
a₂=9-5(2) =-1
Similarly, a₃=9-5(3)=-6
a₄=9- 5(4)=-11
Difference between the successive terms: a₂–a₁=-1-4=-5
a₃–a₂=-6-(-1)=-5
a₄–a₃=-11-(-6)=-S
The difference between successive terms are same, hence a₁, a₂, ⋯ , a_n, is an AP The sum of n terms of an AP is given by
S_n=½n[2a+(n-1)d]
S₁₅=½⋅15⋅[2⋅4+(15-1)⋅(-5)]
=½⋅15⋅(8-70)=15⋅½⋅(-62)
=15⋅(-31)=-465

Example 12. Find the sum of the following APs:
(i) 2, 7, 12, …, to 10 terms (ii) -37, -33, – 29, …, to 12 terms
(iii) 0.6, 1.7, 2.8, …, to 100 terms (iv) 1/15, 1/12, 1/10, …, to 11 terms
Solution:
(i) AP: 2, 7, 12, …
Here, a=2 and d=7-2=5.
The sum of n terms of an AP is given by

S_n=½n[2a+(n-1)d]
⇒S₁₀=½⋅10⋅[2(2)+(10-1)(5)]
⇒S₁₀=5[4+45]=245

(ii) AP: -37, -33, -29, ⋯
Here, a=-37 and d=-33-(-37)=4.
The sum of n terms of an AP is given by

S_n=½n[2a+(n-1)d]
⇒S₁₂=½⋅12⋅[2(-37)+(12-1)(4)]
⇒S₁₂=6[-74+44]=-180

(iii) AP: 0.6, 1.7, 2.8, ⋯
Here, a=0.6 and d=1.7-0.6=1.1.
The sum of n terms of an AP is given by

S_n=½n[2a+(n-1)d]
⇒S₁₀₀=½⋅100⋅ [2(0.6)+(100-1)(1.1)]
⇒S₁₀₀=50[1.2+99×1.1]=50[110.1]=5505

(iv) AP: 1/15, 1/12, 1/10, …
Here, a=1/15 and d=1/12-1/15=1/60.
The sum of n terms of an AP is given by

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