TAREA arithmetic

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

2. 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

3. If a1 + a2 + a3 = 5, an-2 + an-1 + an = 10 and Sn = 20, what is n?

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) + (a2 + a n-1) + (a3 + a n-2) = 15

3 (a1 + an) = 15 a1 + an = a2 + an-1 = ... = an + a1

(a1 + an) = 5

Sn = n (a1 +an) = 20

2

= n (5) = 20

2

= 5n = 40

Therefore:

n = 8

4. Using the sequence 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 find the relationship between each pair of results.

I. 2 (7) and (5 + 9); 2(11) and (9 + 13); 2 (13) and (11 + 15)

II. 2 (7 + 9) and (3 + 5) + (11 + 13)

III. 2 (11 + 13 + 15) and (5 + 7 + 9) + (17 + 19 + 21)

For the general arithmetic sequence a1, a2, a3, ... an-1, an, state the general conclusions for I. II. and III.

Then summarize what you have found from I. II. and III.

Solution:

Let a1 = 3, a2 = 5, a3 = 7 ... a10 = 21

I. 2a3 = a2 + a4

2a5 = a4 + a6

2a6 = a5 + a7

So we get:

2an = an-1 + an+1

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