**General Form of an A.P: **

Let us consider a series with n number of terms:

a_{1 }+ a_{2 }+ a_{3 }+ a_{4 }+ a_{5 }+ a_{6 }+ a_{7 }…………………………….. _{ }= a_{n}

The above series of numbers is said to form an Arithmetic Progression (A.P) if,

a_{2 }– a_{1 }= a_{3 }– a_{2 }= a_{4 }– a_{3 }= …………………………. a_{n }– a_{n-1 }= d

Here, d = common difference of the A.P and it can be positive negative or zero.

Thus the general form of an A.P can be written as:

**a, a+ d, a+ 2d, a+ 3d, a+ 4d, a+ 5d …………………………. a + (n – 1)d**

Where **a = First Term**

And, ** d = Common Difference**

**Q-1: From the given A.P write the first term (a) and common difference (d). **

**(i) 2, 4, 6, 8, 10…..**

**Sol.**

Here, **a = ****2**

Since d = a_{2 }– a_{1}

Therefore, **d** = 4 – 2 = **2**

**Q-2: Check whether the following list of numbers form an A.P? If they form an A.P then write next three terms. **

**(i) 5, 25, 45, 65………**

**Sol**.

Here, **a = 5**

Since for an A.P: a_{2} – a_{1 }= a_{3 }– a_{2 }= a_{4 }– a_{3}

Therefore, 25 – 5 = 45 – 25 = 65 – 45 = 20

Hence the given number forms an A.P with **d = 20**

Therefore next three terms of this A.P are: 65 + 20 = **85**

85 + 20 =**105 **

105 + 20 = **125**

**(ii) −12,−14,−18……**

**Sol.**

Here, **a = −12**

Since for an A.P: a_{2} – a_{1 }= a_{3 }– a_{2 }= a_{4 }– a_{3}

Therefore,

Hence the given numbers does not form an A.P

**The n ^{th} term of an A.P is given by:**

** a _{n }= a + (n – 1)×d**

**Here, a = first term of an A.P and d = common difference**

**Example-1: Let us consider an A.P 2, 7, 12, 17……. Find its 9 ^{th} term.**

**Sol. **

Here, **a = 2**

d = 7 – 2 = 17 – 12 = 5

**n = 9**

Therefore, a_{9} = 2+ (9 – 1)5

a_{9 }= 42

**Therefore, 9 ^{th} term of this A.P = 4**

**Example-2 Fourth term of an A.P is 12 and Sixth term of an A.P is 18. Determine the A.P**

**Sol.**

Given, a_{4} = 12 and a_{6} = 18

Let **a** be the first term and **d** be the common difference of an A.P

Then 12 = a + (4 – 1)d

Therefore, 12 = a + 3d . . . . (1)

And 18 = a + (6 – 1)d

Therefore 18 = a + 5d . . . . . . . (2)

Now, **equation (2) – equation (1)**

18 – 12 = 5d – 3d **Therefore d = 3**

Putting d = 3 in equation (2) we get

18 = a + 5×(3)

**Therefore a = 3**

**Hence, the required A.P is 3, 6, 9, 12, 15, 18, 21………**

** **

**Example -3 Determine which term of the following A.P : 14, 9, 4 . . . . . . is -96 ?**

**Sol.**

From the given A.P,

a = 14,

d = -5,

a_{n} = -96,

n =??

Since, a_{n }= a + (n – 1)d

Therefore -96 = 14 + (n – 1)×-5

(or) 5n = -96 -14 -5

Therefore n = -23 (neglecting –ve sign)

**Therefore, 23 ^{rd} term of this A.P is -96.**

** **

**Example-4 Check whether 328 is the term in the following A.P: 76, 97, 118, 139………**

**Sol.**

From the given A.P:

a = 76,

d = 97 – 76 = 21,

a_{n} = 328

Since, a_{n }= a + (n – 1)d

Therefore 328 = 76 + (n – 1)×21

(or) 21n = 328 + 21 – 76

(or) n = 13

Since n is a +ve integer

**Therefore ****328 is the 13 ^{th} term of this given A.P**

** **

**Example -5 Find how many two digit numbers are divisible by 5.?**

**Sol.**

Two digit numbers that are divisible by 5 are: 10, 15, 20, 25…………………95

Now, from the given A.P

a = 10

d = 5

a_{n }= 95

n = ??

Since, a_{n }= a + (n – 1)d

Therefore 95 = 10 + (n – 1)×5

(or) 5n = 95 – 10 + 5

**n = 18**

**Therefore there are 18 two digit numbers that are divisible by 5. **

**Example-6** **From the given A.P: 15, 8, 1, -6………….. -111. Find 8 ^{th} term from last term.**

**Sol. **

On reversing the given A.P:

**a = -111** (in this case, last term will now become its first term)

**d = 7** (If we go from first term to last term then **d** = -7 (8 – 15), so if we reverse this A.P **d** will become +7)

**n = 8**

a_{8} = ??

Since, a_{n }= a + (n – 1)d

Therefore, a_{8} = -111 + (8 – 1)×7 = -111 + 49

a_{8} = -62

**Therefore, 8 ^{th} term from last term = -62**