This document defines and provides examples of arithmetic progressions. An arithmetic progression is a sequence of numbers where each term is calculated by adding a fixed number, called the common difference, to the preceding term. The document provides formulas for calculating the nth term and sum of terms in an arithmetic progression. Examples are given for finding specific terms, number of terms, common differences, and sums of arithmetic progressions.

1. Arithmetic
Progression
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2. Arithmetic Progression
a) 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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3. nth
term of arithmetic sequence
Tn = a + d(n – 1)
a = First term
d = common difference
n = number of terms.
Common difference = the difference between two consecutive
terms in a sequence. d = Tn – Tn-1
Example
Find the nth
term of the following AP.
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4. Finding the 956th
term
56, 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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5. 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 = -4
Substitute d = -4 into (1):
a + 4(-4) = 13
a – 16 = 13
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6. Sn = a1 + (a1 + d) + (a1 + 2d) + …+ an
Sn = an + (an - d) + (an - 2d) + …+ a1
2
)( 1
1
n
n
i
in
aan
aS
+
== ∑=
)(2 1 nn aanS +=
)(...)()()(2 1111 nnnnn aaaaaaaaS ++++++++=
Sum of First terms of an AP
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7. 1 + 4 + 7 + 10 + 13 + 16 + 19
a1 = 1
an = 19
n = 7
2
)( 1 n
n
aan
S
+
= 2
)191(7 +
=nS
2
)20(7
=nS 70=nS
Example
Find the sum of the integers from 1 to 100
a1 = 1
an = 100
n = 100
2
)( 1 n
n
aan
S
+
=
2
)1001(100 +
=nS
2
)101(100
=nS 5050=nS
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8. Find the sum of the multiples of 3
between 9 and 1344
a1 = 9
an = 1344
d = 3
2
)( 1 n
n
aan
S
+
=
2
)13449( +
=
n
Sn
2
)1353(446
=nS
301719=nS
)1(1 −+= ndaan
)1(391344 −+= n
3391344 −+= n
631344 += n
n31338 =
n=446
Sn = 9 + 12 + 15 + . . . + 1344
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9. Find the sum of the multiples of 7
between 25 and 989
a1 = 28
an = 987
d = 7
2
)( 1 n
n
aan
S
+
=
2
)98728( +
=
n
Sn
2
)1015(138
=nS
70035=nS
)1(1 −+= ndaan
)1(728987 −+= n
7728987 −+= n
217987 += n
n7966 =
n=138
Sn = 28 + 35 + 42 + . . . + 987
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10. Evaluate
a1 = 16
an = 82
d = 3
n = 23
2
)( 1 n
n
aan
S
+
=
2
)8216(23 +
=nS
2
)98(23
=nS
1127=nS
Sn = 16 + 19 + 22 + . . . + 82
∑=
+
25
3
)73(
i
i
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11. Review -- Arithmetic
nth
term Sum of n terms
)1(1 −+= ndaan
2
)( 1 n
n
aan
S
+
=
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12. The End
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