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Example 1: Find the 10 th term and the n th term for the sequence 7, 10, 13, …. Solution:. U n =. U 10 =. ( a - d )( a )( a + d )=480. ( a - d ) + a + ( a + d ) = 24. Example 2. Find the three numbers in an arithmetic progression whose sum is 24 and whose product is 480. Solution. - PowerPoint PPT Presentation
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1
Example 1:
Find the 10th term and the nth term for the sequence 7, 10, 13, … .
Solution:7a 3d
U10 = 34
31107
Un= 43
317
nn
2
Example 2
Find the three numbers in an arithmetic progression whose sum is 24 and whose product is 480.
Solution
Let the three numbers be (a-d), a, (a+d) where d is the common difference.
(a-d) +a + (a+d) = 243a = 24
a = 8
(a-d)(a)(a+d)=480Substituting a =8, we get
2d
3
When d = 2, required numbers are 6, 8, 10When d = -2 , required numbers are 10, 8, 6
4
Example 3
The sum of the first eighteen terms of an arithmetic series is -45 and the eighteenth term is also -45.
Find the common difference and the sum of the first hundred terms.
Solution
])118(2[2
184518 daS
4515318 da --------- (1)
122
dnan
Sn
4518 U
5
4518 U
45118 da4517 da ------------ (2)
Solving equation 1 and 2, we get
5
40
d
a
6
Example 4
Find the sum of the positive integers which are less than 500 and are not multiples of 11.
Solution
Required sum=Sum of integers from 1 to 499(SI)
Sum of multiples of 11(SII)
499............4321 IS
49912
499
124750
7
495.......44332211 IIS 45........432111
451
2
4511
11385
lan
Sn 2
Required sum = SI-SII
=124750-11385
=113365
8
Example 5
The sum of the first 10 terms of an A.P. is 145 and the sum of the next 6 terms is 231
Find (i) the 31st term , and
(ii) the least number of terms required for the sum to exceed 2000.
daU 3031
14510 S
231161514131211 UUUUUU10S 145
16S da 1522
16
9
Solution
2311451522
1616 daS
3761528 da47152 da (1)
14510 S
145922
10 da
(2)2992 da
10
(1)(2): 186 d3d
From (2):
12
3929
a
9133013031 daU
(ii)Find the least number of terms required for the sum to exceed 2000.
Let n be the least number of terms required for 2000nS
11
2000122
dnan
200031122
nn
400013 nn
040003 2 nn
Consider 040003 2 nn
3.36or 7.366
40003411
n
12
3.36 7.36++
7.36or 3.36 nn
Since n must be a positive integer, n > 36.7
Hence, the least number of terms required is 37.
040003 2 nnFor
13
Example 6
Given that the fifth term of a geometric
progression is and the third term is 9.
Find the first term and the common ratio if all the terms in the G.P. are positive.
4
81
4
814 ar
92 ar
(1)
(2)
14
(1)(2):9
1
4
812
4
ar
ar
4
92 r
Since all the terms in the G.P. are positive, r > 0
2
3r
From (2): 92
3 2
a
4a
15
Example 7
Three consecutive terms of a geometric
progression are and 81. Find the value of x. If 81 is the fifth term of the geometric progression, find the seventh term.
13 ,3 xx
x
x
r3
3 1
13
81
x
1
41
3
3
3
3
xx
x
16
x 333x31
2xGiven 81 is the fifth term, find the seventh term:
U5= ar4 = 81
33
3
3
813
4
1 x
r
813 4 a
13
814a
U7 = ar6 729)3(1 6
17
Example 8
Find the sum of the first eight terms of the series
Soln:.........
9
8
3
423
G.P. with ,3a3
2r
3
2 ,
3
2 ,
3
2
3
4
2
3
1
2 U
U
U
U
U
U
r
raS
1
1 8
8
3
21
3
213
8
18
3
21
3
213
8
36561
25613
729
6305
19
Example 9
A geometric series has first term 1 and the common ratio r, where r 1, is positive. The sum of the first five terms is twice the sum of the terms from the 6th to 15th inclusive. Prove that
).13(2
15 r
1a
5S
1576 ..... UUU 157654321 ........ UUUUUUUU
515 SS 54321 UUUUU
20
SolutionGiven 5S 5152 SS
155 23 SS
1
)1(2
1
)1(3
155
r
ra
r
ra
Since and a = 1, we have 1r
)1(2)1(3 155 rr
0132 515 rr
013)(2 535 rr
21
Let .5rx So the equation becomes 0132 3 xx
Let f(x) = 132 3 xxSince f(1) = 2 – 3 + 1 = 0
(x –1) is a factor of f(x).
132 3 xx )12)(1( 2 kxxx
Comparing coefficient of x:
2
13
k
k
132 3 xx )122)(1( 2 xxx
22
Hence, 0)122)(1( 2 xxx
01221 2 xxorx
)2(2
)1)(2(442 x
)13(2
1)13(
2
1 or
4
322
4
1221 55
rorr
Since 1r
1r
and r > 0, )13(2
15 r
(r<0)
23
SSum to infinity, S (or ) = r
a
1
The sum to infinity exists (the series converges or series is convergent) provided 1r
Example 10
Determine whether the series given below converge. If they do, give their sum to infinity.
........16842 (a)G.P. with >12r
Does not converge
24
(b) ...........64
1
16
1
4
11 G.P. with
4
1r
Series converges4
1r <1
r
aS
1
4
11
1
5
4
25
(c) ....................4
27
2
93 G.P. with
2
3r
2
3r >1
Does not converge
26
A geometric series has first term a and the common ratio .
Show that the sum to infinity of the geometric progression is .
Example 11
2
1
)22( a
Solutionr
aS
1
2
11
a
12
2
a
12
12
12
)22(
a )22( a (shown)
27
Example 12
A geometric series has first term a and common ratio r. S is the sum to infinity of the series, T is the sum to infinity of the even-numbered terms (i.e. ) of the series. Given that S is four times the value of T, find the value of r.
....642 UUU
r
aS
1arT 3ar .........5 ar
1
ar2r
Given S = 4T, we have r
a
1 214
r
ar
common ratio r 2
28
)1)(1(
4
1
1
rr
r
r
r
r
1
41
rr 41
3
1r