Question

135,114,93,72,51….

Write a recursive and an exiplicit rule for each arithmetic sequence. Then find the 15th term.

Answer 1

`\textsf{Recursive Rule:} \quad \begin{cases}a_1=135\\a_n=a_(n-1)-21\end{cases}`

`\textsf{Explicit Rule:} \quad a_n=156-21n`

`\textsf{15th term:} \quad a_(15)=-159`

Explanation:

An arithmetic sequence is a sequence of numbers in which the difference between any two consecutive terms is constant.

`\hrulefill`

Recursive Rule

A recursive rule for an arithmetic sequence is a formula that generates each term in the sequence in terms of the preceding term.

The general formula for a recursive arithmetic sequence is:

`\large\boxed{a_n=a_(n-1)+d}`

where:

• 
  `a_n` is the nth term in the sequence.
• 
  `a_(n-1)` is the previous term.
• 
  `d` is the common difference.

To find the common difference (d) of the given sequence, subtract one term from the next term:

`d=a_2-a_1\\\\d=114-135\\\\d=-21`

Therefore, the recursive rule is:

`a_n=a_(n-1)-21`

When providing a recursive rule for an arithmetic sequence, it is important to define the first term of the sequence, as it is not part of the recursive formula itself. Therefore, the full recursive rule for the given sequence is:

`\Large\boxed{\boxed{\begin{cases}a_1=135\\a_n=a_(n-1)-21\end{cases}}}`

`\hrulefill`

Explicit Rule

An explicit rule for an arithmetic sequence is a formula that directly calculates any term in the sequence based on its position in the sequence without referring to previous terms.

The general formula for a recursive arithmetic sequence is:

`\large\boxed{a_n=a_1+(n-1)d}`

where:

• 
  `a_n` is the nth term of the sequence.
• 
  `a_1` is the first term of the sequence.
• 
  `n` is the position of the term in the sequence.
• 
  `d` is the common difference between consecutive terms.

The first term of the given sequence is a₁ = 135.

We have already calculated the common difference as d = -21.

So, the explicit rule for the given sequence is:

`a_n=135+(n-1)(-21)`

Simplify:

`a_n=135-21(n-1)`

`a_n=135-21n+21`

`a_n=156-21n`

Therefore, the explicit rule for the given sequence is:

`\Large\boxed{\boxed{a_n=156-21n}}`

`\hrulefill`

15th term

To find the value of the 15th term, we can substitute n = 15 into the explicit formula:

`a_(15)=156-21(15)`

`a_(15)=156-315`

`a_(15)=-159`

Therefore, the value of the 15th term is -159.

`\Large\boxed{\boxed{a_(15)=-159}}`

Answer 2

Recursive Rule:
`\sf a_n = a_(n-1) - 21`

Explicit Rule:
`\sf a_n = 135 - 21(n - 1)`

15th term: - 159

Explanation:

Arithmetic sequence:

An arithmetic sequence is a sequence of numbers in which each term is equal to the previous term plus a constant value, called the common difference.

Recursive formula for an arithmetic sequence:

The recursive formula for an arithmetic sequence is as follows:

`\sf a_1 =\textsf{ first term}`

`\sf a_n = a_(n-1 )+ d`

where d is the common difference.

Explicit formula for an arithmetic sequence:

The explicit formula for an arithmetic sequence is as follows:

`\sf a_n = a_1 + d(n - 1)`

where n is the term number.

Now,

Recursive rule for the arithmetic sequence 135, 114, 93, 72, 51…

The first term of the sequence is 135 and the common difference is 114 - 135 = -21.

Therefore, the recursive rule for the sequence is as follows:

`\sf a_1 = 135`

`\sf a_n = a_(n-1) - 21`

Explicit rule for the arithmetic sequence 135, 114, 93, 72, 51…

To find the explicit rule for the sequence, we can substitute the values of the first term and the common difference into the explicit formula for an arithmetic sequence:

`\sf a_n = a_1 + d(n - 1)`

`\sf a_n = 135 - 21(n - 1)`

Finding the 15th term of the arithmetic sequence by:

Using Recursive Rule:

`\sf a_(15) = a_(14) - 21`

`\sf a_(14)= a_(13) - 21`

.......

`\sf a_2 = a_(1) - 21`

Now, let's calculate the terms one by one:

`\sf a_2 = 135 - 21 = 114`

`\sf a_3 = 114 - 21 = 93`

`\sf a_4 = 93 - 21 = 72`

`\sf a_(14) = 135 - 21(14-1) = 135 - 21(13) = 135 - 273 = -138`

Now, we can find the 15th term:

`\sf a_(15) = a_(14) - 21 = -138 - 21 = -159`

Using Explicit Rule:

To find the 15th term of the sequence, we can substitute the value of n = 15 into the explicit formula:

`\sf a_n = 135 - 21(n - 1)`

`\sf a_(15) = 135 - 21(15 - 1)`

`\sf a_(15) = 135 - 21(14)`

`\sf a_(15) = 135 -294`

`\sf a_(15) = - 159`

Therefore, the 15th term of the arithmetic sequence is -159.