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3 years ago

Given a2 = 15 and a5= -3,240 of a geometric sequence, what is the recursive equation for the nth term

Mathematics

1 answer:

Assoli18 [71]3 years ago

6 0

Answer:

a[1] = -5/2

a[n] = -6·a[n-1]

Step-by-step explanation:

The explicit formula for the n-th term is ...

an = a1(r^(n-1))

We can use this to find the common ratio, r.

a2 = 15 = a1·r^(2-1) = a1·r

a5 = -3240 = a1·r(5-1) = a1·r^4

Dividing the second equation by the first gives ...

-3240/15 = (a1·r^4)/(a1·r)

-216 = r^3 . . . . simplify

-6 = r . . . . . take the cube root

The first term is ...

15 = a1·(-6)

a1 = 15/-6 = -5/2

The entire recursive formula for the sequence is ...

a[1] = -5/2 . . . . . . . . . . the initial condition

a[n] = -6·a[n-1] . . . . . . the recursion relation

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tresset_1 [31]

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