# Write down the sequence whose nth term is (1/3)*n + 1/6. Verify if the sequence is an A.P.

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To write out the sequence we can just find the first few terms:

n = 1 --> (1/3) + (1/6) = 1/2

n = 2 --> (2/3) + (1/6) = 5/6

n = 3 --> 1 + (1/6) = 7/6

n = 4 --> (4/3) + (1/6) = 3/2

So the sequence is:

1/2, 5/6, 7/6, 3/2,…,(1/3)n+(1/6)

To prove it is an arithmetic series (which it is) we can show that the differences between terms are the same:

i.e. (5/6) - (1/2) = (7/6)- (5/6) = (3/2) - (7/6)

(5/6) - (1/2) = 1/3

(7/6) - (5/6) = 2/6 = 1/3

(3/2) - (7/6) = 2/6 = 1/3

Since a2-a1 = a3-a2 = a4-a3 this is an arithmetic series.

a1,a2,....,an is an arthimatical progression.

an = (1/3)*n + 1/6

a1= 1/3 + 1/6 = 3/6 = 1/2

a2= 2/3 + 1/6 = 5/6

a3= 3/3 + 1/6 = 7/6

.....

We notice that a1,a2,a3,...,an are parts of an arthimaltical progression with constant difference r= 2/6 = 1/3 between terms.

We'll start to write down each term of the sequence, substituting n by the values 1, 2, 3,....

a1 = (1/3)*1 + 1/6

a1 = 1/3 + 1/6

a2 = (1/3)*2 + 1/6

a2 = (2/3) + 1/6

a3 = (1/3)*3+ 1/6

a3 = 1 + 1/6

a4 =4/3 + 1/6

........................

To verify if the sequence is an a.p., we'll have to calculate the difference between each 2 consecutive terms of the sequence and to check if the result is the same:

a2 - a1 = a3 - a2 = a4 - a3 = ............

a2 - a1 = 2/3 + 1/6 - 1/3 - 1/6

We'll eliminate like terms:

a2 - a1 = 1/3

a3 - a2 = 1 + 1/6 - 2/3 - 1/6

We'll eliminate like terms:

a3 - a2 = 1/3

We notice that the value of the difference between 2 consecutive terms is the same, so, the sequence is an a.p., where a1 = 1/2 and c.d. = 1/3.

The nth term Tn = (1/3)*n+1/6)

Put n =1.

T1 = (1/3)*1+1/6

Put n=2

T2 = (1/3)(2)+1/6

Put n = 3

T3 = (1/3)*3+1/6

Clearly T3 -T2 = (1/3)(3-2) = 1/3 and

T2-T1 = (1/3)(2-1) = 1/3

Tr+1 -Tr = (1/3)(r+1-r) = 1/3.

Therefore the sequence is an AP with common difference (1/3) and the starting term T1 = 1/3+1/6 = 1/2.

The sequence is {Tn} = {(n/3)+(1/6)} = {(2n+1)/6} for n =1,2,3....

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