arithmetic sequences & series [email protected] last updated: october 11, 2005
TRANSCRIPT
Arithmetic Progression
5, 8, 11, 14, 17, 20, … 3n+2, …
-4, 1, 6, 11, 16, … 5n – 9, . . .
11, 7, 3, -1, -5, … -4n + 15, . . .
Jeff Bivin -- LZHS
nth term
nth term
nth term of arithmetic sequence
Tn = a + d(n – 1)
Jeff Bivin -- LZHS
a = First termd = common differencen = number of terms.
Common difference = the difference between two consecutive terms in a sequence.d = Tn – Tn-1
8, 11, 14, 17, 20, …First term is 8
Common difference is 3
Tn = a + d(n – 1)
Tn = 8 + 3(n – 1)
Tn = 8 + 3n – 3
Tn = 3n + 5Jeff Bivin -- LZHS
Find the nth term of the following AP.
Finding the nth term
6, 1, 8, 15, 22, …First term is -6
common difference is 7
Tn = a + d(n – 1)
Tn = -6 + 7(n – 1)
Tn = -6 + 7n – 7
Tn = 7n - 13Jeff Bivin -- LZHS
Finding the nth term
23, 19, 15, 11, 7, …First term is 23
common difference is -4
Tn = a + d(n – 1)
Tn = 23 + -4(n – 1)
Tn = 23 -4n + 4
Tn = -4n + 27Jeff Bivin -- LZHS
Finding the 956th term
156, 140, 124, 108, . . .
Tn = a + d(n – 1)
T956 = 156 + -16(956 – 1)
T956 = 156 - 16(955)
T956 = 156 - 15280
T956 = -15124
a1 = 156
d = -16
n = 956
Jeff Bivin -- LZHS
Finding the 100th term5, 11, 17, 23, 29, . . .
Tn = a + d(n – 1)
T100 = 5 + 6(100 – 1)
T100 = 5 + 6(99)
T100 = 5 + 594
T100 = 599
a = 5
d = 6
n = 100
Jeff Bivin -- LZHS
Finding the number of terms in the AP10, 8, 6, 4, 2, . . .-24
Tn = a + d(n – 1)
-24 = 10 -2(n – 1)
-34 = -2(n – 1)
17 = n-1
n = 18
a = 10
d = -2
Tn = -24
Jeff Bivin -- LZHS
The 5th term of an AP is 13 and the 13th term is -19. Find the first term & the common difference.
T5 = a + 4d = 13……..(1)T13= a + 12d = -19……….(2)(2) – (1): 8d = -19 - 13 8d = - 32 d = -4Substitute d = -4 into (1): a + 4(-4) = 13 a – 16 = 13 a = 29
Jeff Bivin -- LZHS
Problem solving
In a race, a swimmer takes 35 seconds to
swim the first 100 m, 39 seconds to swim
the second 100m and 43 seconds to swim
the third 100 m. If he continues to swim in
this manner, how long does he take to finish
the 10th lap of 100m.
Summing it up
Sn = a1 + (a1 + d) + (a1 + 2d) + …+ an
Sn = an + (an - d) + (an - 2d) + …+ a1
2
)( 1
1
nn
iin
aanaS
)(2 1 nn aanS
)(...)()()(2 1111 nnnnn aaaaaaaaS
Jeff Bivin -- LZHS
1 + 4 + 7 + 10 + 13 + 16 + 19
a1 = 1
an = 19
n = 7
2
)( 1 nn
aanS
2
)191(7 nS
2
)20(7nS
70nSJeff Bivin -- LZHS
4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24
a1 = 4
an = 24
n = 11
2
)( 1 nn
aanS
2
)244(11 nS
2
)28(11nS
154nSJeff Bivin -- LZHS
Find the sum of the integers from 1 to 100
a1 = 1
an = 100
n = 100
2
)( 1 nn
aanS
2
)1001(100 nS
2
)101(100nS
5050nSJeff Bivin -- LZHS
Find the sum of the multiples of 3 between 9 and 1344
a1 = 9
an = 1344
d = 3 2
)( 1 nn
aanS
2
)13449( n
Sn
2
)1353(446nS
301719nS
)1(1 ndaan)1(391344 n3391344 n
631344 nn31338 n446
Sn = 9 + 12 + 15 + . . . + 1344
Jeff Bivin -- LZHS
Find the sum of the multiples of 7 between 25 and 989
a1 = 28
an = 987
d = 7 2
)( 1 nn
aanS
2
)98728( n
Sn
2
)1015(138nS
70035nS
)1(1 ndaan)1(728987 n7728987 n
217987 nn7966 n138
Sn = 28 + 35 + 42 + . . . + 987
Jeff Bivin -- LZHS
Evaluate
a1 = 16
an = 82
d = 3
n = 232
)( 1 nn
aanS
2
)8216(23 nS
2
)98(23nS
1127nS
Sn = 16 + 19 + 22 + . . . + 82
25
3
)73(i
i
Jeff Bivin -- LZHS
Evaluate
a1 = -29
an = -199
d = -2
n = 862
)( 1 nn
aanS
2
)19929(86 nS
2
)228(86 nS
9804nS
Sn = -29 - 31 - 33 + . . . - 199
100
15
)12(k
k
Jeff Bivin -- LZHS
Find the sum of the multiples of 11 that are 4 digits in length
a1 = 1001
an = 9999
d = 11 2
)( 1 nn
aanS
2
)99991001( n
Sn
2
)11000(819nS
4504500nS
)1(1 ndaan)1(1110019999 n111110019999 n
990119999 nn119009 n819
Sn = 10 01+ 1012 + 1023 + ... + 9999
Jeff Bivin -- LZHS
Review -- Arithmetic
nth term Sum of n terms
)1(1 ndaan2
)( 1 nn
aanS
Jeff Bivin -- LZHS