ARITHMETIC PROGRESSION (AP)

In mathematics, an arithmetic progression  (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means second minus(- )the first term for instance, the sequence 1,3,5,7,9, . . . is an arithmetic progression with common difference of 2.
Arithmetic Progression, AP
Definition:
Arithmetic Progression (also called arithmetic sequence), is a sequence of numbers such that the difference between any two consecutive terms is constant. Each term therefore in an arithmetic progression will increase or decrease at a constant value called the common difference, d.

Examples of arithmetic progression are:
2, 5, 8, 11,... common difference = 3
23, 19, 15, 11,... common difference = -4
If the initial term of an arithmetic progression is 'a' and the common difference of successive members is 'd', then the nth term of the sequence is given by:
T(n)=a+(n-1).d
Derivation of Formulas:
Let
d = common difference
a1 = first term
a2 = second term
a3 = third term
am = mth term or any term before anan
an = nth term or last term

A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.

The behavior of the arithmetic progression depends on the common difference d. If the common difference is:

positive, then the members (terms) will grow towards positive infinity;
negative, then the members (terms) will grow towards negative infinity.

Some important questions on terms of A.P:

1.Find the Tn for an Arithmetic Progression where T3=22 ,T17=−20.

Solution:
Let a be the first term and d be the common difference of given AP.
Hence given that T3​=22 and T17=−20
⇒a+2d=22...(1) and a+16d=−20...(2)
subtracting (1) from (2), we get
14d=−42
⇒d=−3
putting d in (1), we get
a+2×−3=22
⇒a=28
Nowt
n
​
 =a+(n−1)d
⇒t
n
​
 =28+(n−1)×−3
⇒Tn=28−3n+3
⇒Tn=−3n+31
Suppose if the question were asked T10 then, we just use here 10 at the place of n
i.e,n=3;
The answer will be,
T10=-3n+31=>-3(10)+31=>1 {used n=10}
Sum of A.P:
We will learn how to find the sum of first n terms of an Arithmetic Progression.

Prove that the sum Snn of n terms of an Arithmetic Progress (A.P.) whose first term ‘a’ and common difference ‘d’ is

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

Or, S = n/2[a + l], where l = last term = a + (n - 1)d

Proof:

Suppose, a11, a22, a33, ……….. be ann  Arithmetic Progression whose first term is a and common difference is d.



Then,

a11 = a

a22 = a + d

a33 = a + 2d

a44 = a + 3d

………..

………..

ann = a + (n - 1)d

Now,

S = a11 + a22 + a33 + ………….. + an−1n−1 + ann

S = a + (a + d) + (a + 2d) + (a + 3d) + ……….. + {a + (n - 2)d} + {a + (n - 1)d} ……………….. (i)

By writing the terms of S in the reverse order, we get,

S = {a + (n - 1)d} + {a + (n - 2)d} + {a + (n - 3)d} + ……….. + (a + 3d) + (a + 2d) + (a + d) + a

Adding the corresponding terms of (i) and (ii), we get

2S = {2a + (n - 1)d} + {2a + (n - 1)d} + {2a + (n - 1)d} + ………. + {a + (n - 2)d}

2S = n[2a + (n -1)d

⇒ S = n/2[2a + (n - 1)d]

Now, l = last term = nth term = a + (n - 1)d

Therefore, S = n/2[2a + (n - 1)d] = n2n2[a {a + (n - 1)d}] = n/2[a + l].



We can also find find the sum of first n terms of ann Arithmetic Progression according to the process below.

Suppose, S denote the sum of the first n terms of the Arithmetic Progression {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d ……………...}.

Now nth term of the given Arithmetic Progression is a + (n - 1)d

Let the nth term of the given Arithmetic Progression = l

Therefore, a + (n - 1)d = l

Hence, the term preceding the last term is l – d.

The term preceding the term (l - d) is l - 2d and so on.

Therefore, S = a + (a + d) + (a + 2d) + (a + 3d) + …………………….. to n tems

Or, S = a + (a + d) + (a + 2d) + (a + 3d) + …………………….. + (l - 2d) + (l - d) + l ……………… (i)

Writing the above series in reverse order, we get

S = l + (l - d) + (l - 2d) + ……………. + (a + 2d) + (a + d) + a………………(ii)

Adding the corresponding terms of (i) and (ii), we get

2S = (a + l) + (a + l) + (a + l) + ……………………. to n terms

⇒ 2S = n(a + l)

⇒ S = n/2(a + l)

⇒ S = Numberofterms2Numberofterms2 × (First term + Last term) …………(iii)

⇒ S = n/2[a + a + (n - 1)d], Since last term l = a + (n - 1)d

⇒ S = n/2[2a + (n - 1)d]
Solved examples to find the sum of first n terms of an Arithmetic Progression:

1. Find the sum of the following Arithmetic series:

1 + 8 + 15 + 22 + 29 + 36 + ………………… to 17 terms

Solution:

First term of the given arithmetic series = 1

Second term of the given arithmetic series = 8

Third term of the given arithmetic series = 15

Fourth term of the given arithmetic series = 22

Fifth term of the given arithmetic series = 29

Now, Second term - First term = 8 - 1 = 7

Third term - Second term = 15 - 8 = 7

Fourth term - Third term = 22 - 15 = 7

Therefore, common difference of the given arithmetic series is 7.

The number of terms of the given A. P. series (n) = 17

We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is

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

Therefore, the required sum of first 20 terms of the series = 172172[2 ∙ 1 + (17 - 1) ∙ 7]

= 172172[2 + 16 ∙ 7]

= 172172[2 + 112]

= 172172 × 114

= 17 × 57

= 969



2. Find the sum of the series: 7 + 15 + 23 + 31 + 39 + 47 + ……….. + 255

Solution:

First term of the given arithmetic series = 7

Second term of the given arithmetic series = 15

Third term of the given arithmetic series = 23

Fourth term of the given arithmetic series = 31

Fifth term of the given arithmetic series = 39

Now, Second term - First term = 15 - 7 = 8

Third term - Second term = 23 - 15 = 8

Fourth term - Third term = 31 - 23 = 8

Therefore, the given sequence is ann arithmetic series with the common difference 8.

Let there be n terms in the given arithmetic series. Then

ann = 255

⇒ a + (n - 1)d = 255

⇒ 7 + (n - 1) × 8 = 255

⇒ 7 + 8n - 8 = 255

⇒ 8n - 1 = 255

⇒ 8n = 256

⇒ n = 32

Therefore, the required sum of the series = 322322[2 ∙ 7 + (32 - 1) ∙ 8]

= 16 [14 + 31 ∙ 8]

= 16 [14 + 248]

= 16 × 262

= 4192



KEY TO REMEMBER:

1. We know the formula to find the sum of first n terms of ann Arithmetic Progression is S = n2n2[2a + (n - 1)d]. In the formula there are four quantities. They are S, a, n and d. If any three quantities  are known, the fourth quantity can be determined.

Suppose when two quantities are given then, the remaining two quantities are provided by some other relation.

2. When the sum Snn of n terms of an Arithmetic Progression is given, then the nth term a_n of the Arithmetic Progression cann be determined by the formula ann = Snn - Sn−1n−1.