Question

the frist term of the arithmetic sequence is 1/2 and the its common difference is 1/6 write the algebric form of the sequence At what position does the an integer appears as a term in the sequence first time what is the integer term frist appear in this sequence Is this sequence contain all natural number as the terms​

Answer 1

Answer:

4th term,    1    ,  Contains all natural numbers

Step-by-step explanation:

Let the first integer be 'k' and 'nth' term be the required term.

           Using aₙ = a + (n - 1)d

⇒ k = a + (n - 1)d

⇒ k = (1/2) + (n - 1)(1/6)

⇒ k = [3 + (n - 1)] / 6

⇒ k = (n + 2)/6

⇒ 6k = (n + 2)

      Note that: for k to be an integer,  (n + 2) must be divisible by 6.

For k = 1(smallest possible no. for +ve value of n):

          ⇒ 6 = n + 2    

          ⇒ 4 = n          

Integer term = a₄ = (1/2) + (4 - 1)(1/6)

                            = 1/2 + 1/2

                            = 1

  To contain all the natural numbers:

⇒ General term = (n + 2)/6

⇒ Tₙ = (n + 2)/6

⇒ 6Tₙ = n + 2

⇒ 6T - 2 = n

       n will always be a natural number(as required), for any value natural value of T.          [based on 'n can't be fractional, 0 or -ve]

Hence, AP contains all natural numbers.

Answer 2

Given :-

The first term of the arithmetic sequence is 1/2 and its common difference is 1/6

To Find :-

Write the algebraic form of the sequence At what position does the integer appears as a term in the sequence first time what is the integer term first appear in this sequence Is this sequence contain all-natural number as the terms​

Solution :-

We know that

(i) Algebraic form

aₙ = a + (n - 1)d

aₙ = (1/2) + (n - 1)1/6

aₙ = 1/2 + (n - 1)/6

aₙ = 3 + n - 1/6

aₙ = 2 + n/6 (i)

(ii) At what position does an integer appears as a term in the sequence first time

Let aₙ = x

x = 2 + n/6

6(x) = 2 + n

6x = 2 + n

Putting x as 1

6(1) = 2 + n

6 = 2 + n

6 - 2 = n

4 = n

Putting x as -1

6(-1) = 2 + n

-6 = 2 + n

-6 - 2 = n

-8 = n

No. of terms can't be negative. So, 4 = n

a₄ = a + (4 - 1)d

a₄ = a + 3d

a₄ = 1/2 + 3 × 1/6

a₄ = 1/2 + 3/6

a₄ = 3 + 3/6

a₄ = 6/6

a₄ = 1

(iii)  Is this sequence contain all-natural number as the terms​

Using 1

aₙ = a + (n - 1)d

aₙ = 2 + n/6

6(aₙ) = 2 + n

6aₙ = 2 + n

6aₙ - 2 = n

It contains all-natural number because it can't be negative number and zero. Only positive numbers are natural number. Hence, the answer is Yes.