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

a_{1},a_{2},a_{3},a_{4},a_{5}….a_{n}

#### 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 S_{n}.

S_{n}= a_{1}+a_{2}+a_{3}+…+a_{n}

General nth term of AP series:

a_{1}, a_{2}, a_{3}, a_{4}, ….., a_{n}

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

a_{n}= a+(n-1)d

So formula is to calculate nth term is

a_{n}= 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. a_{n}= a+ (n-1) d

a_{13}= 2+ (13-1) 2

a_{13}=26

#### Q2: If 11^{th}term is 47 and first term is 7. What is common difference between them?

**Solution:**

a=7 a_{11}=47 n=11 d=?

a_{11}= 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 S_{50}= 1+2+3+4+…. + 49+50 ……(1)

Write down in reverse order we will get

S_{50}=50+49+……+4+2+3+1……(2)

Now add equation 1 and 2

2 S_{50}= 51+51+……+51+51+51+51 (50 times)

2S_{50}= 50X51

S_{50}=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 S_{n}= a+(a+d)+(a+2d)+…….+ [a+(n-2)d] +[a+(n-1)d]

Write these in reverse order

S_{n}= [a+(n-1)d]+ [a+(n-2)d] + ……+ (a+d)+a

Now add them

2S_{n}=[2a+(n-1)d]+ [2a+(n-1)d]+……… [2a+(n-1)d]+ [2a+(n-1)d]…… (n terms)

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

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

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

So S_{n}=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 S_{n}= n/2(2a+(n-1)d)

S_{n}= 10/2(2X 11+ (10-1)6)

S_{n}= 5( 22+9X6)

S_{n}= 5(22+54)

S_{n}=5(76)

S_{n}= 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 s_{n}= n/2(a+l)

S_{n}= 19/2(10+100)

S_{n}=19X110/2

S_{n}=1045

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