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Arithmetic Progression and How to solve Arithmetic Progression (AP)

In nature many things follow the pattern such as the hole of honeycomb, Petals of Rose Flower. As like that Arithmetic Progression is type of number pattern. In this number are arranged in a pattern.
Sequence: It is a set of numbers which are arranged in a particular order. Sequence is
a1,a2,a3,a4,a5….an

For example Odd number sequence

1, 3, 5, 7……..

Series: Series is some of terms in a sequence. If there are n terms in a sequence then sum of n term is denoted by Sn.

Sn= a1+a2+a3+…+an

General nth term of AP series:

a1, a2, a3, a4, ….., an

a, a+d, a+d+d, a+d+d+d, ……..

a1=a=a(1-1)d

a2=a+d= a(2-1)d

a3= a+2d=a(3-1)d

an= a+(n-1)d

So formula is to calculate nth term is

an= First term + ( term number- 1) common difference

Q1: find the 13 the term of AP series

2, 4, 6, 8, 10…………

Solution:

First term is a= 2 Common difference (d) = 4-2= 2=6-4

So apply formula I.e. an= a+ (n-1) d

a13= 2+ (13-1) 2

a13=26

Q2: If 11thterm is 47 and first term is 7. What is common difference between them?

Solution:

a=7 a11=47 n=11 d=?

a11= a + (n-1) d

47=7 + (11-1) d

47-7=10d

40=10d

d=4

Common difference (d) = 4.

Sum of first n terms of an AP series:

Suppose that this is AP series 1, 2, 3, 4, …… , 49, 50

So sum of these terms is S50= 1+2+3+4+…. + 49+50 ……(1)

Write down in reverse order we will get

S50=50+49+……+4+2+3+1……(2)

Now add equation 1 and 2

2 S50= 51+51+……+51+51+51+51 (50 times)

2S50= 50X51

S50=50X51/2

Now for n terms of an AP

First n terms of AP series

a, a+d, a+2d,………. a+ (n-2)d, a+(n-1)d

so  Sn= a+(a+d)+(a+2d)+…….+ [a+(n-2)d] +[a+(n-1)d]

Write these in reverse order

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

Now add them

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

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

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

Sn= n/2{ a+ an}; where an= a+ (n-1)d= l (Last term)

So Sn=n/2{a+l)

Q4:  Find out the sum of first 10 terms

11,17, 23, 29,35,…………

Solution: From equation a= 11 d= 6 n=10

So we can use the formula Sn= n/2(2a+(n-1)d)

Sn= 10/2(2X 11+ (10-1)6)

Sn= 5( 22+9X6)

Sn= 5(22+54)

Sn=5(76)

Sn= 380

Q5: Find out the sum of this sequence…..

10,15,20,25,30,…………..,100

Solution: From equation a=10; l=100  d= 5

L= a+ (n-1)d

100= 10+(n-1)5

90= (n-1)5

90=5n-5

90+5=5n

95/5=n

n=19

Now we can use sn= n/2(a+l)

Sn= 19/2(10+100)

Sn=19X110/2

Sn=1045

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Scarlett

About Scarlett

Bachelor in Mathematics, MSc in physics from the university of Calgary, Canada. Scarlett has been teaching mathematics to high school students from last 5 years.