Precalculus help dealing with series

Question

For the following series find the value of a(t₁)

 

A) Arithmetic series with the 4th term of 6 And the 8th term of 18

 

B) geometric series with the 3rd term of 12 and the 7th term of 192

 Thanks

 

Answer 1

I'll assume you're looking for the first terms of these, and that you mean sequences, not series? If so...

 

 

 

(A)

 

There are 8 - 4 = 4 steps between the 4th term and the 8th term. So, find the difference between the 8th term and the 4th term and divide by 4 to find the common difference:

 

(18 - 6)/4 = 12/4 = 3

 

There are 4 - 1 = 3 steps between the 1st term and the 4th term. Since the common difference is 3, that's a difference of 3(3) = 9.

 

So, the first term is 6 - 9 = -3.

 

 

 

(B)

 

There are 7 - 3 = 4 steps between the 3rd term and the 7th term. So, find the quotient of the terms and take the 4th root to find the common ratio:

 

_⁴√(192/12) = _⁴√(16) = 2

 

Divide by 2 twice to find the 1st term:

 

12/2/2 = 12/4 = 3.