Math IB HL math portfolio type II. Deduce the formula Sn = n (a1 +an) for an arithmetic sequence.
. 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
...
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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
Proof:
From the general arithmetic sequence formula:
an = a1 + (n - 1)d:
We know:
2an = 2 [a1 + (n - 1)d]
an-1 = a1 + (n - 1 - 1)d]
an+1 = a1 + (n + 1 - 1)d]
an-1 + an+1 = 2a1 + (2n - 2)d
= 2 [a1 + (n - 1)d]
= 2an
Therefore:
2an = an-1 + an+1
II. 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)
III. 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
5. 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) = (S10-S5) + (S20 - S15)
2 (60 - 80) = (80 - 60) + (S20 - 60)
-40 = S20 - 40
S20 = 0
Therefore:
S15 = 60
S20 = 0
6. For an arithmetic sequence, Sk = p and S2k = q. Using the conclusions from Question 4, find S3k and S4k.
Solution:
S0 Sk S2k S3k S4k
p q - p
q
From Question 4, we know that:
2am = ap + aq
if m = p + q
So:
2(S2k - Sk) = Sk + (S3k - S2k)
2(q - p) = p + (S3k - q)
2q - 2p = p + S3k - q
S3k = 3q - 3p
S0 Sk S2k S3k S4k
p q - p
q
3q - 3p
2(S3k - S2k) = (S2k - Sk) + (S4k - S3k)
2[(3q-3p)-q] = (q - p) + [S4k - (3q - 3p)]
4q - 6p = q - p + S4k - 3q + 3p
S4k = 6q - 8p
Therefore:
S3k = 3q - 3p
S4k = 6q - 8p
7. Using the sequence 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 find the relationship between each pair of results.
I. (3 + 11) and (5 + 9)
II. (7 + 17) and (11 + 13)
III. (5 + 19) and (9 + 15)
Then for the general arithmetic sequence a1, a2, a3, ... an-1, an, state the general conclusion for I. II. and III.
Solution:
Let a1 = 3, a2 = 5, a3 = 7 ... a10 = 21
I. a1 + a5 = a2 + a4
II. a3 + a8 = a5 + a6
III. a2 + a9 = a4 + a7
So we get:
am + an = ap + aq
if m + n = p + q
Proof:
From the general arithmetic sequence formula:
an = a1 + (n - 1)d
We know:
am = a1 + (m - 1)d
ap = a1 + (p - 1)d
aq = a1 + (q - 1)d
am + an = ap + aq
= a1 + (p - 1)d + a1 + (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
8. 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
9. 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
0. 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
Diana Herwono D 0861 006
