**How to find the total number of terms in an arithmetic sequence?**

By using the formula** **

**n = [(L- a)/d] + 1**

we can find the total number of terms of an arithmetic sequence.

L - Last term

a = first term

d = common difference (a_{2} - a_{1})

**Question 1 : **

How many terms are there in the following Arithmetic progressions ?

-1,-5/6,-2/3,……………10/3

**Solution :**

First term (a) = -1

Common difference (d) = a_{2} – a_{1}

d = (-5/6) – (-1)

d = 1/6

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

L = 10/3

n = 27

Hence, 27 terms are in the given A.P

**Question 2 : **

How many terms are there in the following Arithmetic progressions ?

7, 13, 19,……………205

**Solution :**

First term (a) = 7

Common difference (d) = a_{2} – a_{1}

d = 13 – 7

d = 6

L = 205

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

n = [(205 - 7)/6] + 1

n = (198/6) + 1

n = 34

Hence, 34 terms are in the given A.P

**Question 3 :**

The 10^{th} and 18^{th} terms of an A.P are 41 and 73 respectively. Find the 27^{th} term

**Solution :**

10th term = 41

a + 9 d = 41 ------- (1)

18th term = 73

a + 17 d = 73 ------- (2)

Subtracting (2) from (1)

a + 17 d - (a + 9 d) = 73 - 41

a + 17d - a - 9d = 32

8d = 32

d = 4

By substituting the value of d in (1)

a + 9 (4) = 41

a + 36 = 41

a = 5

Now, we have to find 27^{th} term

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

here n = 27

a_{27} = 5 + (27-1) 4

a_{27 }= 5 + 26 (4)

a_{27 }= 5 + 104

a_{27 } = 109

Hence, 27^{th} term of the sequence is 109.

**Question 4 :**

Find n so that the n^{th} terms of the following two A.P’s are the same

1, 7, 13, 19, ………………. and 100, 95, 90 ,………..

**Solution :**

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

n^{th} term of the first sequence

a = 1 d = t₂-t₁

d = 7-1

d = 6

a_{n} = 1 + (n-1) 6

a_{n} = 1 + 6n – 6

a_{n} = 6n – 5 -----(1)

n^{th} term of the second sequence

a = 100 d = t₂ - t₁

d = 95 - 100

d = -5

a_{n} = 100 + (n-1)(-5)

a_{n} = 100 - 5n + 5

a_{n} = 105 - 5n -----(2)

n^{th} terms of the above sequence are equal.

(1) = (2)

6n – 5 = 105 – 5n

6n + 5n = 105 + 5

11n = 110

n = 110/11

n = 11

Hence 11th terms of the given sequence are equal.

**Question 5 :**

How many two digit numbers are divisible by 13?

**Solution :**

Two digit numbers are

10, 11, 12,………… 99

Now we need to find how many terms from this sequence are divisible by 13

By writing the numbers divisible by 13 as sequence, we get

13, 26, 39, …………….. 91

Now, we need to find how many terms are there in this sequence for that let us use formula for n.

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

a = 13, d = 26 – 13==> d = 13, L = 91

n = [(91 - 13)/13] + 1

n = (78/13) + 1

n = 7

7 two digit numbers are divisible by 13.

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