# Arithmetic Series Calculation Examples

Sum Formulas for Finite Arithmetic Series
If a1, a2, a3, ⋯ , an is a finite arithmetic sequence, then the corresponding series a1+a2+a3+ ⋯ +an 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, ⋯ , an and let Sn denote the sum of the series a1+a2+a3+ ⋯ +an.
Then

Sn=a1+(a1+d)+ ⋯ +[a1+(n-2)d]+[a1+(n-1)d]

Reversing the order of the sum, we obtain
Sn=[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
Sn=[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

Snn[2a1+(n-1)d]

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

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 Sn, using another formula:

Snn[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 Snn(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.

Snn[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
Snn[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 Snn(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 tn: {12, 25, 38, …}.
Steps:
(1) Write the formula for the sum of the first n terms in the arithmetic sequence.

Snn(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 Snn[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
Snn(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.

an=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
Snn[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, ⋯ , an, form an AP where an is defined as below:
(i) an=3+4n (ii) an=9-5n
Also find the sum of the first 15 terms in each case.
Solution:
(i) an=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:
a2a1=11-7=4
a3a2=15-11=4
a4a3=19-15=4

The difference between successive terms are same, hence a1, a2, ⋯ , an, is an AP The sum of n terms of an AP is given by
Snn[2a+(n-1)d]
S15=½⋅15⋅[2⋅7+(15-1)⋅4]
=½⋅15⋅(14+56)=15⋅½⋅70
=525

(ii) an=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: a2a1=-1-4=-5
a3a2=-6-(-1)=-5
a4a3=-11-(-6)=-S

The difference between successive terms are same, hence a1, a2, ⋯ , an, is an AP The sum of n terms of an AP is given by
Snn[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

Snn[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

Snn[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

Snn[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 Let’s read post “
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