# FIND FIRST FIVE TERMS OF SEQUENCE WHEN nth TERM IS GIVEN

To find the first 5 terms of the sequence, we have apply n = 1, 2, 3, .......in the general rule.

Write the first five terms of the sequence. Determine whether or not the sequence is arithmetic. If it is, find the common difference.

Problem 1 :

an = 8 +13n

Solution :

an = 8 +13n

 a1 = 8 +13(1)= 8 + 13a1 = 21 a2 = 8 +13(2)= 8 + 26a2 = 34 a3 = 8 +13(3)= 8 + 39a3 = 47 a4 = 8 +13(4)= 8 + 52a4 = 60

a5 = 8 +13(5)

= 8 + 65

a= 73

So, first five terms of the sequence is 21, 34, 47, 60, 73.

To find the common difference :

d = a2 – a1 = 34 – 21 = 13

a3 – a2 = 47 – 34 = 13

The sequence is arithmetic because common difference is 13.

Problem 2 :

an = (1/n) +1

Solution :

an = (1/n) +1

 a1 = (1/1) +1a1 = 2 a2 = (1/2) +1a2 = 3/2 a3 = (1/3) +1)a3 = 4/3 a4 = (1/4) + 1a4 = 5/4

a5 = (1/5) +1)

a5 = 6/5

So, first five terms of the sequence is 2, 3/2, 4/3, 5/4, 6/5.

To find the common difference :

d = a2 – a1 = (3/2) - 2

= (3 - 4)/2

d = -1/2

Problem 3 :

an = 2n + n

Solution :

an = 2n + n

 a1 = 21 + 1= 2 + 1a1 = 3 a2 = 22 + 2= 4 + 2a2 = 6 a3 = 23 + 3= 8 + 3a3 = 11 a4 = 24 + 4= 16 + 4a4 = 20 a5 = 25 + 5= 32 + 5a5 = 37

So, first five terms of the sequence is 3, 6, 11, 20, 37.

To find the common difference :

d = a2 – a1 = 6 – 3 = 3

a3 – a2 = 11 – 6 = 5

Since the common difference is not same, it is not arithmetic sequence.

Find the first four terms of the sequence given the explicit formula.

Problem 1 :

an = n2 + 3

Solution :

an = n2 + 3

 a1 = 12 + 3= 1 + 3a1 = 4 a2 = 22 + 3= 4 + 3a2 = 7 a3 = 32 + 3= 9 + 3a3 = 12

a4 = 42 + 3

= 16 + 3

a= 17

So, first four terms of the sequence is 4, 7, 12, 17.

Problem 2 :

an = 15/(n + 3)

Solution :

an = 15/(n + 3)

 a1 = 15/(1 + 3)a1 = 15/4 a2 = 15/(2 + 3)a2 = 15/5a2 = 3 a3 = 15/(3 + 3)a3 = 15/6a3 = 5/2

a4 = 15/(4 + 3)

a4 = 15/7

a4 = 15/7

So, first four terms of the sequence is 15/4, 3, 5/2, 15/7.

Problem 3 :

an = 2n - 1

Solution :

an = 2n - 1

 a1 = 2(1) – 1= 2 – 1a1 = 1 a2 = 2(2) – 1= 4 – 1a2 = 3 a3 = 2(3) – 1= 6 – 1a3 = 5

a4 = 2(4) – 1

= 8 – 1

a= 7

So, first four terms of the sequence is 1, 3, 5, 7.

Problem 4 :

an = n/(n +1)

Solution :

an = n/(n +1)

 a1 = 1/(1 +1)a1 = 1/2 a2 = 2/(2 +1)a2 = 2/3 a3 = 3/(3 +1)a3 = 3/4

a4 = 4/(4 +1)

a= 4/5

So, first four terms of the sequence is 1/2, 2/3, 3/4, 4/5.

Problem 5 :

an = 12 - 3n

Solution :

an = 12 - 3n

 a1 = 12 – 3(1)= 12 – 3a1 = 9 a2 = 12 – 3(2)= 12 – 6a2 = 6 a3 = 12 – 3(3)= 12 – 9a3 = 3

a4 = 12 – 3(4)

= 12 – 12

a= 0

So, first four terms of the sequence is 9, 6, 3, 0.

Problem 6 :

an = 4n/3

Solution :

an = 4n/3

 a1 = 4(1)/3a1 = 4/3 a2 = 4(2)/3a2 = 8/3 a3 = 4(3)/3a3 = 12/3 a3 = 4

a4 = 4(4)/3

a4 = 16/3

So, first four terms of the sequence is 4/3, 8/3, 4, 16/3.

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