Respuesta :

Answer:

a(n) = 19 - 4(n-1)

Step-by-step explanation:

The general formula for an arith. seq. is

a(n) = a(1) + (n-1)d, where d is the "term to term rule" or the "common difference," a(1) is the first term and a(n) is the nth term.

Here, a(n) = a(1) - (n-1)(4).

If the 3rd term is 11, then:

a(3) = a(1) - (3-1)(4) = 11, or a(3) = a(1) - 8 = 11

Adding 8 to both sides yields a(1) = 19.

Then the specific expression for the nth term is:

a(n) = 19 - 4(n-1).

The nth term of the sequence can be expressed as:

[tex]T_n = -4n + 23[/tex]

Given that:

  • The term to term rule is "take away 4"
  • The third term of a sequence = 11

Calculations of first term and of nth term:

Since term to term rule is to take away 4, thus when we go back in sequence, the rule will be to add 4 term to term.

Thus:

Second term = 4 + third term = 4 + 11 = 15

First term = 4 + second term = 15 + 4 = 19

Since the given sequence has a constant difference of -4 between each adjacent terms, thus it is an arithmetic progression with d = -4

The nth term of an arithmetic progression with difference d is given by:

[tex]T_n = T_1 + (n-1) \times d[/tex]

Since d = -4 and first term is 19, thus we have:

[tex]T_n = 19 + (n-1) \times (-4) = -4n + 23\\T_n = -4n + 23[/tex]

Thus, the nth term of the sequence can be expressed as:

[tex]T_n = -4n + 23[/tex]

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