**How to find the specified term of an arithmetic sequence ?**

Using the general term formula, we can easily find any term of the sequence. nth term means any term in the sequence like 20th or 15th or 28 like that.

The formula to find nth term is

**an = a + (n - 1)d**

**Question 1 :**

Find the common difference and 15th term of an A.P 125 , 120 ,115 , 110 , ……….….

**Solution :**

First term (a) = 125

Common difference (d) = a2 – a1 ==> 120 – 125 ==> -25

General term of an A.P (an) = a + (n - 1) d

= 125 + (15 - 1) (-25) ==> 125 + 14 (-25) ==> 125 – 350

a₁₅ = -225

Therefore 15th term of A.P is -225

Let us see the next example on "How to find the specified term of an arithmetic sequence".

**Question 2 :**

Which term of the arithmetic sequence is 24 , 23 ¼ ,22 ½ , 21 ¾ , ………. Is 3?

**Solution :**

First term (a) = 24

Common difference = a_{2} – a_{1} ==> 23 ¼ – 24 ==> (93/4) – 24

d = -3/4

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

Let us consider 3 as nth term

an = 3

3 = 24 + (n-1) (-3/4)

3 – 24 = (n-1) (-3/4)

(-21 x 4)/(-3) = n -1 ==> 84/3 = n -1 ==> 28 = n – 1 ==> n=29

Hence, 3 is the 29th term of A.P.

Let us see the next example on "How to find the specified term of an arithmetic sequence".

**Question 3 :**

Find the 12th term of the A.P √2 , 3 √2 , 5 √2 , …………

**Solution :**

First term (a) = √2,

Common difference = 3 √2 - √2 ==> 2 √2

n = 12

General term of an A.P

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

a_{₁₂} = √2 + (12 - 1) (2√2)

= √2 + 11 (2√2)

= √2 + 22 √2

= 23 √2

Hence, 12th of A.P is 23 √2

**Question 4 :**

Find the 17th term of the A.P 4 , 9 , 14 ,…………

**Solution :**

First term (a) = 4

Common difference (d) = 9 - 4 ==> 5

n = 17

General term of an A.P

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

a_{n} = 4 + (17 - 1) (5) ==> 4 + 16 (5) ==> 84

Hence, 17th of A.P is 84

**Question 5 :**

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

**Solution :**

10th term = 41 ==> a + 9 d = 41 ------- (1)

18th term = 73 ==> a + 17 d = 73 ------- (2)

Subtracting the second equation from first equation

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

a + 17d - a - 9d = 32 ==> 8d = 32 ==> d = 4

Substitute d = 4 in the first equation

a + 9 (4) = 41 ==> a + 36 = 41 ==> a = 5

Now, we have to find 27th term

an = a + (n - 1) d

here n = 27

= 5 + (27-1) 4 ==> 5 + 26 (4) ==> 5 + 104 ==> 109

Hence, 27th term of the sequence is 109.

**Question 6 :**

Find n so that the nth 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

nth term of the first sequence

a = 1 d = t₂-t₁ ==> 7-1 ==> d = 6

a_{n} = 1 + (n-1) 6 ==> 1 + 6 n – 6 ==> 6 n – 5 -----(1)

nth term of the second sequence

a = 100 d = a₂-a₁ ==> 95 - 100 ==> -5

a_{n} = 100 + (n-1) (-5) ==> 100 - 5 n + 5 ==> 105 - 5 n -----(2)

(1) = (2)

6 n – 5 = 105 – 5 n

6 n + 5 n = 105 + 5

11 n = 110 ==> 110/11 ==> 11

Hence, 11th terms of the given sequence are equal

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