**Derivation**

To find the nth term (general term) of an Arithmetic Progression, First of all, We know the general form of an Arithmetic Progression which is given as

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

Here, we can see that the first term is a.

To find the second term we are adding common difference d to first term a or we can say that we are multiplying common difference (d) with (2 – 1) and then adding it to the first term a.

a_{2} = a + d = a + (2 – 1)d

To find the third term, same as above we are multiplying d with (3 – 1) and adding to the first term a.

a_{3 }= a + 2d = a + (3 – 1)d

Similarly, to find the nth term (general term) of an Arithmetic Progression we will multiply d with (n – 1) and then add to the first term as written in general form also.

**a _{n} = a + (n – 1)d**

Here, a_{n} = n^{th} term or general term

a = First term

n = Number of terms

d = Common difference

If there are finite terms in an AP then a_{n} represents the last term which is also denoted by *l*.

**Some examples are given below**

**Example 1) **find the 9^{th }term of the AP 6, 12, 18, 24, 30…………..

**Solution – **Here first term (a) = 6, common difference (d) = 12 – 6 = 6

Number of term (n) = 9, 9^{th }term (a_{9}) = ?

By the formula of n^{th }term, a_{n} = a + (n – 1)d

a_{9} = 6 + (9 – 1)6

a_{9} = 6 + (8)6 = 6 + 48

a_{9} = 54

So, the 9^{th} term of the given AP is 54. **Ans.**

**Example 2) **Which term of an AP 4, 8, 12, 16……….. is 464 ?

**Solution – **First term (a) = 4, common difference (d) = 8 – 4 = 4

n^{th} term (a_{n}) = 464, Number of term (n) = ?

By the formula, a_{n} = a + (n – 1)d

464 = 4 + (n – 1)4

464 – 4 = 4n – 4

460 = 4n – 4

460 + 4 = 4n

464 = 4n

464/4 = n

**n = 116**

Therefore, the 116^{th} term of the given AP is 464. **Ans.**

**Example 3) **Find the n^{th} term (general term) of AP 7, 11, 15, 19……..

**Solution – **First term (a) = 7, common difference (d) = 11 – 7 = 4

n^{th} term (a_{n}) =?

By the formula of n^{th} term, a_{n} = a + (n – 1)d

a_{n} = 7 + (n – 1)4

a_{n} = 7 + 4n – 4

a_{n} = 3 + 4n

So, the n^{th} term of the AP is a_{n} = 3 + 4n **Ans.**

**Example 4) **Find the number of terms of AP 1, 8, 15, 22……………..,204.

**Solution – **Here First term (a) = 1, common difference (d) = 8 – 1 = 7

n^{th} term (a_{n}) = 204, Number of terms (n) = ?

By formula, a_{n} = a + (n – 1)d

204 = 1 + (n – 1)7

204 – 1 = 7n – 7

203 = 7n – 7

203 + 7 = 7n

210/7 = n

**n = 30**

So, there are 30 terms in the given AP. **Ans.**

**Example 5) **For AP 10, 7, 4, 1, -2, -5, -8, -11….check whether -31 is the term of it or not.

**Solution – **First term (a) = 10, common difference (d) = 7 – 10 = -3

If -31 is the term of given AP then number of terms(n) for it will be a natural number.

So, n^{th} term (a_{n}) = -31 then number of term (n) = ?

n^{th} term formula, a_{n} = a + (n – 1)d

-31 = 10 + (n – 1)(-3)

-31 – 10 = -3n + 3

-41 = -3n + 3

3n = 41 + 3

**n = 44/3 = 14.67**

Since value of n is the decimal number. n is always natural number for AP.

Therefore, -31 is not the term of the given AP. **Ans.**

**Example 6) **Determine the AP whose 5^{th} term is 9 and 9^{th} term is 17.

**Solution – **By the formula of n^{th} term, a_{n} = a + (n – 1)d

5^{th} term a_{5} = a + (5 – 1)d 9^{th} term a_{9} = a + (9 – 1)d

9 = a + 4d 17 = a + 8d

Solving both the equation,

a + 4d = 9 ……………..(1) a + 8d = 17 ………………(2)

By substitution method,

a + 4d = 9 9 – 4d + 8d = 17

a = 9 – 4d[putting the value in equation (2)] 4d = 17 – 9

a = 9 – 4d ……………..(3) d = 8/4

a = 9 – 4⨯2 **d = 2** (Common difference)[putting the value in equation (3)]

a = 9 – 8

**a = 1** (First term)

So, the required AP is 1, 3, 5, 7, 9, 11, 13, 15, 17, 19……….. **Ans.**

**Example 7) **If the 6th term of an AP is 22 and the 30^{th} term is 94. Find its 25^{th} term.

**Solution – **By n^{th} term formula, a_{n} = a + (n – 1)d

6^{th }term, a_{6} = 22 30^{th} term, a_{30} = 94

a + (6 – 1)d = 22 a + (30 – 1)d = 94

a + 5d = 22 a + 29d = 94

Solving by elimination method,

-24d = – 72

d = -72/-24

Common difference d = 3

Putting the value d = 3 in any equation,

a + 5d = 22

a + 5⨯3 = 22

a = 22 – 15

First term **a = 7**

Now 25^{th} term, a_{25} = a + (25 – 1)d

a_{25} = 7 + 24⨯3

a_{25} = 7 + 72

**a _{25} = 79**

Hence the 25^{th} term of the given AP is 79. **Ans.**

**Example 8) **How many two digits numbers are divisible by 9?

**Solution – **Two digits numbers = 10 to 99

Numbers which are divisible by 9, 18, 27, 36…………………..99

This sequence will be AP because all numbers are divisible by 9.

First-term (a) = 18, common difference (d) = 27 – 18 = 9

n^{th} term (a_{n}) = 99, number of terms (n) = ?

By the formula, a_{n} = a + (n – 1)d

99 = 18 + (n – 1)9

99 – 18 = 9n – 9

81 + 9 = 9n

90 = 9n

n = 90/9

**n = 10**

So, there are 10 two digits numbers divisible by 9. **Ans.**

**n**^{th} Term From the Last Term

^{th}Term From the Last Term

If we take the last term ‘*l *’ as the first term and reverse the common difference as – d then the n^{th }term from last can be written as follows

n^{th} term from last a_{n} = ^{ }*l* + (n – 1)(-d)

**a _{n} = l – (n – 1)d**

**Example 9) **Find the 16^{th} term from the last term of the AP 21, 18, 15, 12…………….-81.

**Solution – **Here last term ( *l* ) = -81, common difference (d) = 18 – 21 = -3, number of terms (n) = 16

By the formula of n^{th} term from last term, a_{n} = *l* – (n – 1)d

a_{16 }= -81 – (16 – 1)(-3)

a_{16} = -81 – (-48 + 3)

a_{16} = -81 – (-45)

a_{16} = -81 + 45

**a _{16} = – 36**

So, the 16^{th} term from the last term is -36. **Ans.**

**How to Select Terms of Arithmetic Progression if not given –**

If there is a question in which neither AP is given nor any term is given and we have to find the AP then we have to select the terms of AP in a different order. Selection of the terms of AP in different order helps us to find the AP easily. Let’s see how to select the terms.

If we have to select **three terms** then the terms are – a – d, a, a + d

If we have to select **four terms** then the terms are – a – 3d, a – d, a + d, a + 3d

If we have to select** five terms** then the terms are – a – 2d, a – d, a, a + d, a + 2d

If we have to select **six terms** then the terms are – a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d

**Example 10) **Three numbers are in AP if their sum is 15 and product is -55 then find the numbers.

**Solution – **Let the three numbers which are in AP a – d, a, a + d

Then according to question sum is 15 so a – d + a + a + d = 15

3a = 15

a = 15/3

a = 5

And product is -55 (a – d)⨯a⨯(a + d) = -55

(a^{2} – d^{2})⨯a = -55

∵ a = 5

{(5)^{2} – d^{2}}⨯5 = -55

{25 – d^{2}}⨯5 = -55

25 – 5d^{2} = -55

125 + 55 = 5d^{2}

180 = 5d^{2}

180/5 = d^{2}

d^{2 }= 36

d = ±√36

**d = ±6**

Putting the value of a and d. if a = 5 and d = +6 then numbers 5 – 6, 5, 5 + 6 ⇒ **-1, 5, 11**

If a = 5 and d = -6 then numbers 5 – (-6), 5, 5 + (-6) ⇒ **11, 5, -1** **Ans.**

**nth Term (General Term) of an Arithmetic Progression Class 10 ^{th} in Hindi**

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