[Maths Class Notes] on Sum of N Terms Pdf for Exam

An arithmetic progression is a sequence of numbers or variables wherein the difference between every two consecutive terms is the same. This difference is called the common difference. To find the sum of n terms of an AP, we use a formula involving the first term of the AP, the common difference, and the number of terms in the AP. 

The 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 (also, we will study the sum of terms in an AP).

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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 sequences are: 

• 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 a real sequence.

 

Sum of N Terms of an Arithmetic Progression Formula

The sum of n terms of an AP can be easily calculated using a simple formula that states that if the first term of the AP is a and the common difference is d, the sum of n terms of the AP is:

[Sn=frac{n}{2}(2a+(n-1)d)]

[Sn=frac{infty }{2}(2(2)+(infty -1)3)]

[Sn=infty ]

We observed the sum of infinite AP to be ∞ whilst d > 0. In an equal manner, the sum of endless AP is −∞ while, d < 0.

Thus, 

 Sum of [infty ],Ap ={ [infty ], if d > 0 – [infty ],if d < 0 

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

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

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=(frac{n}{2})2a+(n-1)d2a+(n-1)d2a+(n-1)d]  or

[Sn=(frac{n}{2})(a+an)]            

                            

List of the Sum of Natural Numbers:

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

Formulas You Need to Know:

Notations:

AP Tips and Tricks Sum of n Terms:

• [Sn=frac{n}{2}(2a+(n-1)d)] and [Sn=frac{n}{2}(a1+an)] are two formulas for calculating the sum of arithmetic progressions with the first term a and the common difference d.

• AP of n natural numbers is equal to n(n+1)/2.

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 theater 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]

        [=(frac{n}{2})a+a+(n-1)d]

[Sn=(frac{n}{2})[a+an]]

Substituting the values of a, an 

Substituting the value of n from Equation 2,

[Sn=frac{((b+c-2a)/(b-a))}{2}]

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