sum of an arithmetic progression
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Sum of an Arithmetic Progression. Last Updated: October 11, 2005. Let a 1 = first term of an AP Let a n = last term of an AP And d = the common difference Hence, the A.P can be written as a 1 , a 1 + d, a 1 + 2d, …. a n And the SUM OF A.P is - PowerPoint PPT PresentationTRANSCRIPT
Sum of an Arithmetic Progression
Last Updated: October 11, 2005
Let a1 = first term of an AP
Let an = last term of an AP
And d = the common difference
Hence, the A.P can be written as
a1, a1 + d, a1 + 2d, …. an
And the SUM OF A.P is
Sn = a1 + (a1 + d) + (a1 + 2d) + …+ an
OR
Sn = an + (an - d) + (an - 2d) + …+ a1
Jeff Bivin -- LZHS
Summing it up
Sn = a1 + (a1 + d) + (a1 + 2d) + …+ an
Sn = an + (an - d) + (an - 2d) + …+ a1
1
2n
n
n(a a )S
)(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
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
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
Review -- Arithmetic
2 12nnS [ a (n )d ]
nth term Sum of n terms
1nT a d(n ) 2
)( 1 nn
aanS
Jeff Bivin -- LZHS
1n n nT S S
Problem solvingThe sum of the first n terms of a progression
is given by Sn = n2 + 3n. Find, in terms of nthe nth term.
Sn = n2 + 3n
Sn-1 = (n-1)2 + 3(n-1) = n2 – 2n + 1 + 3n – 3 = n2 + n – 2
Tn = Sn – Sn – 1
= n2 + 3n – (n2 + n – 2)
= 2n + 2 Jeff Bivin -- LZHS