Question

Find the third, fourth, and fifth terms of the sequence defined by

Answer 1

By the recursive definition,

$a_3=2a_2-a_1=2\cdot9-4=14$

$a_4=2a_3-a_2=2\cdot14-9=19$

$a_5=2a_4-a_3=2\cdot19-14=24$

$a_6=2a_5-a_4=2\cdot24-19=29$

Answer 2

Answer:

  a3=14, a4=19, a5=24

Step-by-step explanation:

Put the numbers where the symbols are and do the arithmetic.

  a3 = 2(a2) -(a1) = 2(9) -4 = 14

  a4 = 2(a3) -(a2) = 2(14) -9 = 19

  a5 = 2(a4) -(a3) = 2(19) -14 = 24

  a6 = 2(a5) -(a4) = 2(24) -19 = 29

_____

With a little work, you can show that this is an arithmetic sequence with a common difference of a2-a1 = 5.

Let d = a[2] -a[1]

Of course, the second term is that difference added to the first:

  a[2] = a[1] + (a[2] -a[1]) = a[1] +d

The third term is ...

  a[3] = 2a[2] -a[1] = a[2] +(a[2] -a[1]) = a[2] +d

  a[4] = 2a[3] -a[2] = a[3] +(a[3] -a[2]) = a[3] -(a[2] +d) -a[2] = a[3] +d