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# Arithmetic Sequence Techniques

Extracts from this document...

Introduction

Diana Herwono D 0861 006

Portfolio

(Type II)

By Diana Herwono

## IB Student No: D 0861 006

May 2003

1. Deduce the formula Sn = n (a1 +an) for an arithmetic sequence.                                                                                2

Solution:

Let S = 1 + 2 + ... + 99 + 100

Written another way:

S = 100 + 99 + ... + 2 + 1

Let a1 = 1, a2 = 2, ... , an-1 = 99, an = 100

Notice that 1 + 100 = 2 + 99 = ... = 100 + 1

We can conclude that:

a1 + an = a2 + an-1 = ... = an + a1

Let Sn = a1 + a2 + ... + an-1 + an

Sn can also be written as:

Sn = an + an-1 + ... + a2 + a1

When we add the two sums together:

Sn  = a1 + a2 +                ...                             + an-1 + an

+) Sn  = an + an-1 +             ...                              + a2 + a1__

2Sn = (a1 + an) + (a2 +an-1) + ... + (an-1 + a2) + (an + a1)

= (a1 + an) + (a1 + an) + ... + (a1 + an) + (a1 + an)               a1 + an = a2 + an-1 = ... = an + a1

= n (a1 + an)                         n identical terms in the sum

2Sn = n (a1 +an)

Therefore:

Sn = n (a1 +an)

2

1. For an arithmetic sequence, a4 + an-3 = 8 and Sn = 32. Find n.

Solution:

From the previous question we know that:

a1 + an = a2 + an-1 = ... = an + a1

So if we extend the formula to:

a1 + an = a2 + an-1 = a3 + an-2 = a4 + an-3... = an + a1

Then:

a4 + an-3 = 8 = a1 + an

Sn = n (a1 +an)     = 32

2

= n (a4 + an-3) = 32

2

= n (8)            = 32

2

= 8n                = 64

Therefore:

n = 8

1. If a1 + a2 + a3 = 5, an-2 + an-1 + an = 10 and Sn

Middle

1 + (2n – 2)d

= 2 [a1 + (n – 1)d]

= 2an

Therefore:

2an = an-1 + an+1

1. 2 (a3 + a4) = (a1 + a2) + (a5 + a6)

So we get:

2 (an + an+1) = (an-2 + an-1) + (an+2 + an+3)

Proof:

2[a1 + (n – 1)d + a1 + nd] = a1 + (n – 3)d + a1 + (n – 2)d] + [a1 + (n + 1)d + a1

+ (n + 2)d]

2 (an + an+1)                      = (an-2 + an-1) + (an+2 + an+3)

Therefore:

2 (an + an+1) = (an-2 + an-1) + (an+2 + an+3)

1. 2 (a5 + a6 + a7) = (a2 + a3 + a4) + (a8 + a9 + a10)

So we get:

2 (an + an+1 + an+2) = (an-3 + an-2 + an-1) + (an+3 + an+4 + an+5)

Proof:

2[a1 + (n – 1)d + a1 + nd + a1 + (n + 1)d] = a1 + (n – 4)d + a1 + (n – 3)d + a1

+ (n – 2)d] + [a1 + (n + 2)d  + a1 + (n + 3)d + a1 + (n + 4)d]

6a1 + 6n = 6a1 + 6n

Therefore:

2 (an + an+1 + an+2) = (an-3 + an-2 + an-1) + (an+3 + an+4 + an+5)

General Conclusion:

2am = ap + aq

if m = p + q

Proof:

2[a1 + (m – 1)d] = a1 + (p – 1)d + a1 + (q – 1)d

= 2 a1 + (p + q – 2)d

= 2 a1 + (2m – 2)d                                                             2m = p + q

1. S5 = 60 and S10 = 80. Using the conclusions from question 4, find S15 and S20.

Solution:

 a1            a5 a6       a10 a11        a15

60                   20

80

From Question 4, we know that:

2am = ap + aq

if m = p + q

So:

2 (S10 – S5) = S5 + (S15 – S10)

2 (80 – 60)    = 60 + S15 – 80

2 (20)            = -20 + S15

S15               = 60

 a1              a5 a6             a10 a11           a15 a16           a20

60                        20                      -20

80

60

2 (S15 – S10)

Conclusion

1 + (q – 1)d

= 2a1 + (p + q – 2)d

= 2a1 + (m + n – 2)d                        m + n = p + q

= a1 + (m – 1)d + a1 + (n – 1)d

= am + an

Therefore:

am + an = ap + aq

if m + n = p + q

1. For an arithmetic sequence, a5 = 8 and a36 = 32. Using the conclusion from Question 7, find a12 + a29 and a2 + a11 + a30 + a39.

Solution:

From Question 7, we get:

am + an = ap + aq

if m + n = p + q

So:

a5 + a36 = 8 + 32 = a12 + a29 =

Because:

5 + 36 = 40 = 12 + 29

So:

a12 + a29 = 40

So:

a39 + a2 = a30 + a11 =a12 + a29 = 40

Because:

39 + 2 = 30 + 11 = 12 + 29 = 40

a2 + a11 + a30 + a39 = a39 + a2 + a30 + a11

= 2 (a39 + a2)                                             a39 + a2 = a30 + a11     = 2 (a30 + a11)

= 2 (a12 + a29)

= 2 (40)

= 80

Therefore:

a12 + a29 = 40

a2 + a11 + a30 + a39 = 80

1. a3 + a5 + a7 + a9 = 16. Using the conclusion from Question 7, find a6.

Solution:

From Question 7, we get:

am + an = ap + aq

if m + n = p + q

So:

a9 + a3 = a5 + a7

Because:

9 + 3 = 5 + 7

a3 + a5 + a7 + a9 = 16

2 (a9 + a3) = 16                                                                    a9 + a3 = a5 + a7

2 (a5 +a7) = 16

(a5 +a7)    = 8

From Question 4, we know:

2am = ap + aq

if m = p + q

So:

a5 + a7 = 2a6 = 8

a6 =  4

Therefore:

a6 =  4

1. a2 + a5 = 19 and S5 = 40. Using the conclusion from Question 7, find a1 and a6.

Solution:

From Question 7, we get:

am + an = ap + aq

if m + n = p + q

So:

a2 + a5 = a3 + a4 = a1 + a6 = 19

S5 = a1 + a2 + a3 + a4 + a5

40 = a1 + (a2 + a5) + (a3 + a4)

= a1 + 2 (19)

a1 = 2

a6 = 17                                                                                   a1 + a6 = 19

Therefore:

a1 = 2

a6 = 17

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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1. ## Math IB HL math portfolio type II. Deduce the formula Sn = ...

Solution: If we add the two equations together, we get: a1 + a2 + a3 = 5 +) an-2 + an-1 + an = 10 a1 + an-2 + a2 + a n-1 + a3 + an = 15 (a1 + an)

2. ## IB HL Math Portfolio - Arithmetic Sequence Techniques (Type II)

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