SUM OF N TERMS

Sum of n terms can be calculated in arithmetic progression as well as in Geometric Progression.In this article, we’ll focus on Arithmetic Progression(sum of ap).(image will be updated soon)

In this article, we’ll focus on Arithmetic Progression.


What Is a Sequence?

  • We can say that a sequence is in progression if the last term of the sequence can be represented with the help of a formula.

           The two types of sequence are: (image will be updated soon)

  • If the co-domain of the function is the set of complex numbers, the sequence is known as complex sequence.

  • On the other hand, if it is a set of real numbers, it can be called real sequence.

What is Arithmetic Progression?

  • In simple words, arithmetic progression can be defined as a sequence of numbers in which each differs from the preceding (previous) one by a number.

  • In Arithmetic Progression also known as AP, the difference between all the consecutive terms remains constant.

 Formula:

Sum of n terms of AP = [2a + (n - 1)d]

where, n= Number of terms

              d= Constant difference


How do we Find the Sum of N Terms of an AP?

  • The sum of n terms of an arithmetic progression can be defined as the sum of the first n terms of the sequence.

Examples

2,10,18,26……...        

↖️   First term

Here, the common difference(d)=8 which is positive.

50,45,40,35………

 ↖️  First term

Here, the common difference(d)= -5 which is negative.

1,4,9,16,25………...

 ↖️ First Term

Here, the series represents a sequence of squares of natural numbers.

Note: The common difference (d) can be positive, negative or a fraction.


LET’S DERIVE THE FORMULA TO FIND THE SUM :

Step 1: Let us take an example,

3,5,7, 9….

Step 2: Let us take the first term = a 

The second term = a+d and the Third term = a+2d

So, the nth term will be = a+(n-1) d

step 3: How do we find the sum of all terms?

In the terms of average,

We know that, Average = (Sum of all the terms)/n

step 4: Therefore, Sum of terms can be written as,

Sum of terms = n × Average --------> Equation 1

NOTE: The average of the evenly spaced numbers can be written as, 

Average = (First term + Second term) / 2 

Step 5: Now, substituting the value of average in Equation 1 we get,

Sum of terms = n×  (First term + Second term) / 2

By taking (n/2) common and replacing the last term, we get the sum of ap formula or the sum of ap series,

Sum of n terms of AP = [2a + (n - 1)d]

The sum of terms in a sequence is known as series.

Sum OF AP Formula 

The sum of AP formula can be written as,

                                      Sn= (n/2) [2a+(n-1) d] 

                                                  or

                                       Sn=  (n/2) (  a+an    )

LIST OF THE SUM OF NATURAL NUMBERS:

Here’s a table that lists down the sum of natural numbers.

Numbers

Sum

1-10

55

1-100

5050

1-1000

500500

1-10000

50005000

1-100000

5000050000

1-1000000

500000500000


FORMULAS YOU NEED TO KNOW:

Sum of terms when the first(a) and last term (l)is known and where n is the number of terms.

    (n/2) [a+l]

Sum of terms when last term is unknown, a and n are known.

    (n/2)[2a+(n-1) d]


To find the last term of the series( an) when d and n is known.

an = a1+(n-1) d



NOTATIONS:

Constant difference between the terms

d

First term of the series

a1

Total number of terms

n

Last term of the series

a


NOW LET’S SOLVE A FEW QUESTIONS

Question1) If a1= 5 and a20=62, find the sum of the first 20 terms of the arithmetic series.

Solution -> Let’s list down the given information,

 a1= 5, which is the first term of the series.

a20=62, which is the last term of the series.

We know the formula to find the sum of n terms in ap  is:

Sn= n(a1 + a2) / 2

S20= 20(5 + 62) / 2 = S20 = 670

Therefore, the sum of the 20 terms =670


Question 2) Find the sum of the first 40 terms of the A.P.

2,5,8,11, 14………….

Solution-> Let us first find the last term (the 40th term),

The given information, 

Difference (d) = 3

Number of terms(n)=40

a40 = a1+(n-1) d

     = 2+(40-1) ×3 = 119

Now, we will find the sum of the n terms in ap,

Sn= n(a1 + a2) / 2

S40= 40(2 +119) / 2 =2420

Therefore, the sum of ap series is 2420.


Question 3) A theatre has 50 rows of seats. There are 18 seats in the first row,20 seats in the second and there are 22 in the row and so on. Find how many seats will be there in the last row.

Solution -> Let’s list down the given information, 

Difference (d) = 2

Number of rows = 50

Using the formula,

a50 = a1+(n-1) d

     = 18+(50-1) ×2 = 116 seats

There are a total of 116 seats in the last row.


Question 4) The first and the last terms of an AP are 3 and 27. The common difference between the terms is 2. How many terms are there?

Solution -> Let’s list down the given information,

Difference (d) = 2

First term (a)=3 and last term(l)=27

Now, since we know the difference,

3,5,7………27

an= a1+(n-1) d

27 = 3+(n-1) ×2

3+ 2n-2 = 27

2n+1=27

2n = 26

n= 13

Therefore, the number of terms is equal to 13.


Question 5) Is the row 1,11,21,31 …… an arithmetic progression?

Solution-> Yes, the given sequence is an arithmetic progression, having the first term as 1 and the difference between the terms as 10, which remains constant.


Question 6) Show that the sum of an arithmetic progression having the first term as a and the second term as b and the last term of the series as c is = (a+c)(b+c-2a) /2(b-c)

Solution ->Let’s list the information given,

1st term = a

2nd term = b

Last term =c

Let the common difference cd be =d

d= b-a -->Equation 1

Last term of the sequence = an=c

Since we know the last term,

an= a1+(n-1) d

c= a1+(n-1) d

(n-1) d=c-a

(n-1) = (c-a) / d

Substituting the value of from equation 1,

(n-1) = (c-a) / (b-a)

n= (c-a) / (b-a)   + 1

Taking (b-a) as the L.C.M,

n= (c-a) + (b-a) / (b-a) 

n=  (c-a+b-a) / (b-a) 

n= (c+b-2a) / (b-a)

n= (b+c-2a) / (b-a)   --------> Equation 2

We know that the sum of n terms in ap is,

Using sum of ap formula,

Sum (Sn)=   [2a+(n-1) d]

         =   (n/2) [a+a+(n-1) d]

Sn= (n/2) [a+an]

Substituting the values of a, a

Substituting the value of n from Equation 2,

Sn= [((b+c-2a)/(b-a))/2] [a+c] 

Therefore,(sum of n terms in ap)  Sn =(a+c)(b+c-2a) / 2(b-c) proved.

FAQ (Frequently Asked Questions)

1) What is n in AP?

Ans: The full form of AP is Arithmetic Progression and n stands for the number of terms.

2) How do You Find the Sum of AP?

Ans: You can find the sum of AP by using the formula 2a+(n-1) d .

3) Give a Proof for the Sum of the First n Terms?

Ans: Let us take the first term = a 

The second term = a+d and the Third term = a+2d

So, the nth term will be = a+(n-1) d

How do we find the sum of all terms?

In the terms of average,

We know that, Average = (Sum of all the terms)/n

Therefore, Sum of  ap series can be written as,

Sum of terms = n × Average --------> Equation 1

NOTE: The average of the evenly spaced numbers can be written as, 

Average =  (First term + Second term) / 2

Now, substituting the value of average in Equation 1 we get,

Sum of terms = n×  (First term + Second term) / 2

By taking (n/2) common and replacing the last term, we get 

Sum of n terms of AP = [2a + (n - 1)d]

This is formula which denotes the sum of ap series.