example 13
DESCRIPTION
and. Example 13. The first two terms of a geometric progression are 3 and –2. Find the least value of n for which the difference between sum of the first n terms and sum to infinity is within 2% of the sum to infinity. Since. Hence the least value of n is 10. (ans). - PowerPoint PPT PresentationTRANSCRIPT
1
Example 13
The first two terms of a geometric progression are 3 and –2. Find the least value of n for which the difference between sum of the first n terms and sum to infinity is within 2% of the sum to infinity.
3a3
2rand nSS S
100
2
)1
(100
2
1
)1(
1 r
a
r
ra
r
a n
2
32
1
3
100
2
32
1
)1(3
32
1
3 nr
100
2
3
2
n
n
3
2
:Noten
3
2
n
3
2
5
9
100
21
5
9
5
9 nr
5
9
100
2
5
9 nr
3
02.032
n
02.0lg3
2lg
n
65.9n
Hence the least value of n is 10. (ans)
Since
4
Example 14
The sum of the first n terms of a series is
Find the fifth term. Prove that the series is arithmetic and state its common difference.
.2 2 nn
SolutionGiven .nnSn 22
Therefore, 455 SSU
]4)4(2[]5)5(2[ 22 19
5
To prove that the series is an A.P., prove
tconsUU nn tan1
Given ,nnSn 22
112 21 nnSn
1 nnn SSU
)]1()1(2[)2( 22 nnnn
)1242(2 22 nnnnn14 n
1141 nUn
6
)tan(,45414
]1)1(4[)14(1
tconsann
nnUU nn
Thus, the series is arithmetic with common difference 4.
7
Example 15
The sum of the first n terms of a series is
given by . By finding an expression
for the nth term of the series, show that this is a geometric series, and state the value of the first term and the common ratio.
1
1
3
26
n
n
nS
1 nnn SSU
Solution
]3
26[]
3
26[
1)1(
1)1(
1
1
n
n
n
n
8
21
1
3
2
3
2
n
n
n
n
9
3
2
3
3
22n
n
n
n
n
n
n
n
3
29
3
26
n
nU
3
23
To show G.P., show
tconsU
U
n
n tan1
11
3
23
3
23
n
n
n
nU
U
)tan(3
2tconsa
9
Therefore, the series is geometric with first term
3
231U
2
and common ratio 3
2
10
Example 16
The sum of the first 100 terms of an Arithmetic Progression is 10, 000; the first, second and fifth terms of this progression are three consecutive terms of a Geometric Progression. Find the first term, a , and the non-zero common difference, d, of the A.P.Solution 1000011002
2
100100 daS
200992 da ----(1)
a, a+d, a+ 4d are three consecutive terms of a Geometric Progression
11
da
da
a
daratiocommon
4
daada 42
adadada 42 222
add 22 ad 2 ----(2)
Substituting (2) into (1), 200)2(992 aa
1a
From (2), 2d
12
ru rr 23 1
Example 17
The rth term of a series is . Find the
sum of the first n terms.
rr 23 1
Solution
The sum of the first n terms is:
nn uuuuS ...321
)(23...)3(23)2(23)1(23 1131211 nS nn
Given
13
nnn
12
213
131
GP: a = 1, r = 3, n terms
AP: a = 1, d = 1, n terms
)(2...)3(2)2(2)1(23...333 1131211 nS nn
nS nn ...32123...931 1
)1()13(2
1nnn
222132
1nnn
14
Example 18Each time that a ball falls vertically on to a horizontal floor it rebounds to three-quarters of the height from which it fell. It is initially dropped from a point 4 m above the floor. Find, and simplify, an expression for the total distance the ball travels until it is about to touch the floor for the (n+1)th time. Hence find the number of times the ball has bounced when it has traveled 24 m and also the total distance it travels before coming to rest. (The dimensions of the ball are to be ignored.)
15
4m 4
4
3
1st 2nd
44
3 2
3rd nth (n+1)th
44
3 n
Total distance (in metres) that the ball travels
4
4
4
32
4
4
32
2 4
4
32
n
n
4
3.......
4
3
4
3
4
384
32
16
4
31
4
31
4
3
84
n
r
raS
n
n
1
1
callRe
n
4
31244
n
4
32428
17
Hence find the number of times the ball has bounced when it has traveled 24 m and also the total distance it travels before coming to rest
When the ball has traveled 24 m,
244
32428
n
Solution n
4
3244
6
1
4
3
n
6
1lg
4
3lg
n
04
3lg:
Note
Let the number of times the ball has bounced be n.
18
4
3lg
6
1lg
n
23.6n
Since n is an integer, least n=7. Therefore, the ball has bounced 7 times when it has traveled 24 m.
04
3,
nnAs
284
32428 ,
ntherefore
The ball travels 28 m before coming to rest.
For the ball to come to rest, n