Question

In an arithmetic sequence, a17 = -40 and

a28 = -73. Explain how to use this information to write a recursive formula for this sequence

Answer 1

The difference between the given terms is

–73 – (–40) = –33.

The difference between the term numbers is 28 – 17 = 11.

Dividing –33 / 11 = –3.

The common difference is –3.

The recursive formula is the previous term minus 3, or an = an – 1 - 3 where a17 = -40.

Answer 2

Tn = 2Tn-1 - Tn-2

Explanation:

Before we can generate the recursive sequence, we need to find the nth term of the given sequence.

nth term of an AP is given as:

Tn = a+(n-1)d

If a17 = -40

T17 = a+(17-1)d = -40

a+16d = -40 ...(1)

If a28 = -73

T28 = a+(28-1)d = -73

a+27d = -73 ...(2)

Solving both equations simultaneously using elimination method.

Subtracting 1 from 2 we have:

27d - 16d = -73-(-40)

11d = -73+40

11d = -33

d = -3

Substituting d = -3 into 1

a+16(-3) = -40

a - 48 = -40

a = -40+48

a = 8

Given a = 8, d = -3, the nth term of the sequence will be

Tn = 8+(n-1) (-3)

Tn = 8+(-3n+3)

Tn = 8-3n+3

Tn = 11-3n

Given Tn = 11-3n and d = -3

Tn-1 = Tn - d... (3)

Tn-1 = 11-3n +3

Tn-1 = 14-3n

Tn-2 = Tn-2d...(4)

Tn-2 = 11-3n-2(-3)

Tn-2 = 11-3n+6

Tn-2 = 17-3n

From 3, d = Tn - Tn-1

From 4, d = (Tn - Tn-2)/2

Equating both common difference

(Tn - Tn-2)/2 = Tn - Tn-1

Tn - Tn-2 = 2(Tn - Tn-1)

Tn - Tn-2 = 2Tn-2Tn-1

2Tn-Tn = 2Tn-1 - Tn-2

Tn = 2Tn-1 - Tn-2

The recursive formula will be

Tn = 2Tn-1 - Tn-2