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

Sagot :

sqdancefan

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

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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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