Download - Arithmetic Progression
Arithmetic Progressiona) 5, 8, 11, 14, 17, 20, … 3n+2, …b) -4, 1, 6, 11, 16, … 5n – 9, . . .
c) 11, 7, 3, -1, -5, … -4n + 15, . . .
In all the lists above, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an Arithmetic Progression ( AP ).
So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero
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nth term of arithmetic sequenceTn = a + d(n – 1)a = First termd = common differencen = number of terms.
Common difference = the difference between two consecutive terms in a sequence. d = Tn – Tn-1
ExampleFind the nth term of the following AP.
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Finding the 956th term56, 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
Example
Finding the number of terms in the AP 10, 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
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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
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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
Sum of First terms of an AP
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1 + 4 + 7 + 10 + 13 + 16 + 19a1 = 1
an = 19
n = 7
2
)( 1 nn
aanS
2
)191(7 nS
2
)20(7nS
70nS
Example
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
5050nS
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Find the sum of the multiples of 3 between 9 and 1344
a1 = 9
an = 1344
d = 32
)( 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
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Find the sum of the multiples of 7 between 25 and 989
a1 = 28
an = 987
d = 72
)( 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
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Evaluate
a1 = 16
an = 82
d = 3n = 23
2
)( 1 nn
aanS
2
)8216(23 nS
2
)98(23nS
1127nS
Sn = 16 + 19 + 22 + . . . + 82
25
3
)73(i
i
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Review -- Arithmetic
nth term Sum of n terms
)1(1 ndaan2
)( 1 nn
aanS
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