**Sum Formulas for Finite Arithmetic Series**

If *a _{1}*,

*a*,

_{2}*a*, ⋯ ,

_{3}*a*is a finite arithmetic sequence, then the corresponding series

_{n}*a*+

_{1}*a*+

_{2}*a*+ ⋯ +

_{3}*a*is called an arithmetic series. We will derive two sim- ple and very useful formulas for the sum of an arithmetic series. Let

_{n}*d*be the common difference of the arithmetic sequence

*a*,

_{1}*a*,

_{2}*a*, ⋯ ,

_{3}*a*and let

_{n}*S*denote the sum of the series

_{n}*a*+

_{1}*a*+

_{2}*a*+ ⋯ +

_{3}*a*.

_{n}Then

*S*=

_{n}*a*+(

_{1}*a*+

_{1}*d*)+ ⋯ +[

*a*+(

_{1}*n*-2)

*d*]+[

*a*+(

_{1}*n*-1)

*d*]

Reversing the order of the sum, we obtain

*S*=[

_{n}*a*+(

_{1}*n*-1)

*d*]+[

*a*+(

_{1}*n*-2)

*d*]+ ⋯ +(

*a*+

_{1}*d*)+

*a*

_{1}Adding the left sides of these two equations and corresponding elements of the right sides, we see that

*S*=[2

_{n}*a*+(

_{1}*n*-1)

*d*]+[2

*a*+(

_{1}*n*-1)

*d*]+ ⋯ +[2

*a*+(

_{1}*n*-1)

*d*]

This can be restated as in

__Theorem 1__.

Theorem 1:

Sum of an Arithmetic Series — First Form

S=½_{n}n[2a+(_{1}n-1)d]

By replacinga+(_{1}n-1)dwitha, we obtain a second useful formula for the sum._{n}

Theorem 2:Sum of an Arithmetic Series — Second Form

S=½_{n}n(a+_{1}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*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

Example 1: **Determining the Sum, Given the First Term and Common Difference**

An arithmetic series has *t*_{1}=5.5 and *d*=-2.5; determine *S*_{40}.

solution:

Use *S _{n}*=½

*n*(2

*t*

_{1}+

*d*(

*n*-1))

Substitute:

*n*=40,

*t*

_{1}=5.5,

*d*=-2.5

*S*

_{40}=½⋅40⋅(2⋅5.5-2.5⋅(40-1))

*S*

_{40}=20⋅(11-97.5)

*S*

_{40}=-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, *a _{1}*=-7,

*d*=3, and use

__Theorem 1__.

*S*=½

_{n}*n*[2

*a*+(

_{1}*n*-1)

*d*]

*S*

_{26}=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*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{20}=½⋅20⋅[2⋅3+19⋅4]

=10⋅(6+76)

*S*

_{20}=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.

*S*

_{10}=½⋅10⋅[2⋅3+(10-1)⋅4]

=5(6+36)=210

Alternatively, but more tediously, We add the first 10 terms together:

*S*

_{10}=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?

*S*

_{16}=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 *u*_{1}=4, *d*=3 and *n*=50.

Now *S _{n}*=½

*n*(2

*u*

_{1}+(

*n*-1)

*d*)

*S*

_{50}=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*=½

_{n}*n*(2

*a*+(

*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.

*S*

_{20}=½⋅20⋅(2⋅12+19⋅13)

*S*

_{20}=10⋅(24+247)=10⋅271

*S*

_{20}=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}*=½

*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{16}=½⋅16⋅[2⋅3+(16-1)⋅8]

=8⋅[6+15⋅8]

=8(6+120)

*S*

_{16}=8×126=1008

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

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*=½

_{n}*n*(2

*a*+(

*n*-1)

*d*)

*S*

_{50}=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, *a _{2}*=14,

*a*=18 and

_{3}*n*=51.

*a*=

_{n}*a*+(

*n*-1)

*d*

*a*=

_{2}*a*+(2-1)

*d*

14=

*a*+

*d*

*a*=14-

*d*⋯ (1)

and

*a*=

_{3}*a*+(3-1)

*d*

*a*+2

*d*

Putting the value of

*a*from equatio

*n*(1), we get

*d*+2

*d*

*d*=4

Putting the value of

*d*in equatio

*n*(1), we get

*a*=14-4=10

The sum of

*n*terms of an AP is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{51}=½⋅51⋅[2⋅10+(51-1)⋅4]

=½⋅51⋅[20+200]

=51⋅½⋅220

=51⋅110=5610

Example 11. Show that *a _{1}*,

*a*, ⋯ ,

_{2}*a*, form an AP where

_{n}*a*is defined as below:

_{n}(i)

*a*=3+4

_{n}*n*(ii)

*a*=9-5

_{n}*n*

Also find the sum of the first 15 terms in each case.

Solution:

(i)

*a*=3+4

_{n}*n*

Putting

*n*=1, we get

*a*=3+4(1)=7

_{1}Putting

*n*=2, we get

*a*=3+4(2)=11

_{2}Similarly,

*a*=3+4(3)=15

_{3}*a*=3+4(4)=19

_{4}Difference between the successive terms:

*a*–

_{2}*a*=11-7=4

_{1}*a*–

_{3}*a*=15-11=4

_{2}*a*–

_{4}*a*=19-15=4

_{3}The difference between successive terms are same, hence

*a*,

_{1}*a*, ⋯ ,

_{2}*a*, is an AP The sum of

_{n}*n*terms of an AP is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{15}=½⋅15⋅[2⋅7+(15-1)⋅4]

=½⋅15⋅(14+56)=15⋅½⋅70

=525

(ii)

*a*=9-5

_{n}*n*

Putting

*n*=1, we get

*a*=9-5(1)=4

_{1}Putting

*n*=2, we get

*a*=9-5(2) =-1

_{2}Similarly,

*a*=9-5(3)=-6

_{3}*a*=9- 5(4)=-11

_{4}Difference between the successive terms:

*a*–

_{2}*a*=-1-4=-5

_{1}*a*–

_{3}*a*=-6-(-1)=-5

_{2}*a*–

_{4}*a*=-11-(-6)=-S

_{3}The difference between successive terms are same, hence

*a*,

_{1}*a*, ⋯ ,

_{2}*a*, is an AP The sum of

_{n}*n*terms of an AP is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{15}=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*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

⇒

*S*

_{10}=½⋅10⋅[2(2)+(10-1)(5)]

⇒

*S*

_{10}=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*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

⇒

*S*

_{12}=½⋅12⋅[2(-37)+(12-1)(4)]

⇒

*S*

_{12}=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*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

⇒

*S*

_{100}=½⋅100⋅ [2(0.6)+(100-1)(1.1)]

⇒

*S*

_{100}=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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