To find n^{th} term of the arithmetic sequence, we use the formula
a_{n} = a_{1} + (n – 1)d
Here a_{1} is the first term, n represents the position of the term and d represents the common difference.
a) Find
the common difference of each arithmetic sequence.
b) Write the n^{th} term of each sequence for the given value of n.
Problem 1 :
3, 6, 9, 12, …, n = 8
Solution :
a) Common difference :
a_{1 }= 3, a_{2 }= 6, a_{3 }= 9
d = a_{2} – a_{1} = 6 - 3 = 3
So, the common difference is 3.
b) Finding the 8^{th} term :
a_{n} = a_{1} + (n – 1)d
a_{8} = 3 + (8 – 1)3
= 3 + 7(3)
= 3 + 21
a_{8 }= 24
So, the 8^{th} term of the sequence is a_{8} = 24.
Problem 2 :
2, 7, 12, 17, …, n = 12
Solution :
a) Common difference :
a_{1 }= 2, a_{2 }= 7, a_{3 }= 12
d = a_{2} – a_{1} = 7 - 2 = 5
So, the common difference is 5.
b) n^{th} term of sequence :
a_{n} = a_{1} + (n – 1)d
Finding 12th term :
a_{12} = 2 + (12 – 1)5
= 2 + 11(5)
= 2 + 55
a_{12 }= 57
So, the nth term of the sequence is a_{12} = 57.
Problem 3 :
18, 16, 14, 12, …, n = 10
Solution :
a) Common difference :
a_{1 }= 18, a_{2 }= 16, a_{3 }= 14
d = a_{2} – a_{1} = 16 - 18 =
-2
So, the
common difference is -2.
b) 10^{th} term of sequence :
a_{n} = a_{1} + (n – 1)d
a_{10} = 18 + (10 – 1)(-2)
= 18 + 9(-2)
= 18 - 18
a_{10 }= 0
So, the 10^{th} term of the sequence is a_{10} = 0.
Problem 4 :
1/2, 1, 3/2, 2, …, n = 7
Solution :
a) Common difference :
a_{1 }= 1/2, a_{2 }= 1, a_{3 }= 3/2
d = a_{2} – a_{1} = 1 – 1/2
= (2 – 1)/2
= 1/2
So, the
common difference is 1/2.
b) 7^{th} term of sequence :
a_{n} = a_{1} + (n – 1)d
a_{7} = 1/2 + (7 – 1)(1/2)
= 1/2 + 6(1/2)
= 1/2 + 6/2
a_{7 }= 7/2
So, the nth term of the sequence is a_{7} = 7/2.
Problem 5 :
-1, -3, -5, -7, …, n = 10
Solution :
a) Common difference :
a_{1 }= -1, a_{2 }= -3, a_{3 }= -5
d = a_{2} – a_{1} = -3 + 1 =
-2
So, the
common difference is -2.
b) 10^{th} term of sequence :
a_{n} = a_{1} + (n – 1)d
a_{10} = -1 + (10 – 1)(-2)
= -1 + 9(-2)
= -1 - 18
a_{10 }= -19
So, the nth term of the sequence is a_{10} = -19.
Problem 6 :
2.1, 2.2, 2.3, 2.4, …, n = 20
Solution :
a) Common difference :
a_{1 }= 2.1, a_{2 }= 2.2, a_{3 }= 2.3
d = a_{2} – a_{1} = 2.2 – 2.1
= 0.1
So, the
common difference is 0.1.
b) 20^{th} term of sequence :
a_{n} = a_{1} + (n – 1)d
a_{20} = 2.1 + (20 – 1)(0.1)
= 2.1 + 19(0.1)
= 2.1 + 1.9
a_{20 }= 4
So, the nth term of the sequence is a_{20} = 4.
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