Question

The explicit rule for a sequence is an=7(−4)n−1

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What is the recursive rule for the sequence?


an=−4(an−1),a1=7
an=−7(an−1),a1=4
an=−7(an+1),a1=4
an=−4(an+1),a1=7

Answer 1

Option (a) is correct.

The recursive rule for the given sequence
`a_n=7(-4)^(n-1)` is
`a_n = (-4)\cdot a_(n-1)` with
`a_1=7`

Explanation:

The explicit sequence of the geometric sequence is given by:

`a_n = a_1r^(n-1)`

where,

`a_1` is the first term

r is the common ratio

n is the number of terms

For the given explicit rule,
`a_n=7(-4)^(n-1)`

Comparing with above sequence , we have,

`a_1=7` and r = -4

Recursive formula for the geometric sequence having
`a_1=7` and r = -4 is given by:

`a_n=r\cdot a_(n-1) \ for\ n\geq 2`

Putting values, we get,

`a_n = (-4)\cdot a_(n-1)`

Hence, the recursive rule for the given sequence
`a_n=7(-4)^(n-1)` is
`a_n = (-4)\cdot a_(n-1)` with
`a_1=7`

hence, option (a) is correct.

Answer 2

The correct answer is


`a_n=-4(a_(n-1)), a_1=7`

The explicit formula is of the form

`a_n=a_1(r)^(n-1)`,
where a₁ is the first term and r is the common ratio. Comparing this to our explicit form, we see that r = -4 and a₁ = 7.

The recursive formula is of the form

`a_n=a_(n-1)*r`; using r=-4 gives us

`a_n=-4(a_(n-1)), a_1=7`