AP-GP-Ex-13-to-18-solutions
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Transcript AP-GP-Ex-13-to-18-solutions
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.
2
2
S
S
S
a 3 and r
n
100
3
a
a(1 r n )
2
a
(
)
1 r
1 r
100 1 r
1
n
3
3(1 r )
2
3
2
2
2
100
1
1
1
3
3
3
9 9
2 9
n
1 r
5 5
100 5
9 n
2 9
r
5
100 5
2
3
n
2
100
Note :
n
n
2 2
3
3
n
2
3 2
2
3
n
0.02
2
n l g
l g 0.0 2
3
Since
n 9.65
Hence the least value of n is 10. (ans)
3
Example 14
The sum of the first n terms of a series is
2
2n n.
Find the fifth term. Prove that the series is
arithmetic and state its common difference.
Solution
2
Given S n 2n n .
Therefore, U 5 S5 S4
[2(5) 2 5] [2(4) 2 4]
19
4
To prove that the series is an A.P., prove
U n U n1 cons tan t
2
S
2
n
n ,
Given n
S n1 2n 12 n 1
U n Sn Sn1
(2n 2 n) [2(n 1) 2 (n 1)]
2
2
2n n (2n 4n 2 n 1)
4n 1
U n1 4n 1 1
5
U n U n1 (4n 1) [4(n 1) 1]
4n 1 4n 5 4 , (a cons tan t )
Thus, the series is arithmetic with common
difference 4.
6
Example 15
The sum of the first n terms of a series is
n 1
given by 6 2
. By finding an expression
n 1
3
Sn
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.
Solution
2n1
2( n1)1
U n Sn Sn1 [6 n1 ] [6 (n1)1 ]
3
3
7
n 1
2
n 1
3
2
n
n2
3
n
n
22
2
3n 3n
3 9
2n 2n
6 9
3n 3n
n
2
U n 3
3
To show G.P., show
Un
cons tan t
U n1
n
2
3
Un
3
n 1
U n1
2
3
3
2
(a cons tan t )
3
8
Therefore, the series is geometric with first term
2
U1 3
3
2
2
and common ratio
3
9
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
100
2a 100 1d 10000
S100
2
2a 99d 200 ----(1)
a, a+d, a+ 4d are three consecutive terms
10
of a Geometric Progression
a d a 4d
common ratio
a
ad
a d a a 4 d
2
a 2 2ad d 2 a 2 4ad
d 2 2ad
d 2a ----(2)
Substituting (2) into (1), 2a 99(2a) 200
a 1
From (2),
d 2
11
Example 17
r 1
The rth term of a series is 3
2r . Find the
sum of the first n terms.
Solution
Given
ur 3r 1 2r
The sum of the first n terms is:
S n u1 u 2 u3 ...u n
Sn 311 2(1) 321 2(2) 331 2(3) ... 3n1 2(n)
12
11
21
31
n1
Sn 3 3 3 ... 3
2(1) 2(2) 2(3) ... 2(n)
S n 1 3 9 ...3n1 21 2 3 ... n
GP: a = 1, r = 3, n
terms
AP: a = 1, d = 1,
n terms
1 3 1
n
2 1 n
3 1
2
n
1
(3 n 1) n (1 n)
2
1
3 n 1 2n 2n 2
2
13
Example 18
Each 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.)
14
2
3
4
4
n
3
4
4
4m
1st
2nd
3
4
4
3rd
nth (n+1)th
Total distance (in metres) that the ball travels
n
2
3
3
3
2 4
4 2 4 2 4
4
4
4
n
3 3 2 3 3
3
4 8 .......
15
4
4
4
4
3 3 n
1
4 4
4 8
3
1
4
Re call
a 1 r
Sn
1 r
n
3 n
4 241
4
n
3
28 24
4
16
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 Let the number of times the
ball has bounced be n.
Solution
n
3
4 24
When the ball has
4
traveled 24 m,
n
3
1
n
3
28 24 24
4
6
4
1
3
n lg lg
3
Note : lg 0
6
4
17
4
1
lg
6 Since n is an integer, least n=7.
n
3 Therefore, the ball has bounced
lg
4 7 times when it has traveled 24
m.
n 6.23
n
For the ball to come to
3
As n , 0
rest, n
4
n
3
therefore, 28 24 28
4
The ball travels 28 m before coming to
18
rest.