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
a = 29© iTutor. 2000-2013. All Rights Reserved

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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