Download AP-GP-Q1-to-Q12-Solutions

Example 1:
Find the 10th term and the nth term for the
sequence 7, 10, 13, … .
Solution:
a7
d 3
U10 = 7  10  13
 34
Un= 7  n  13
 3n  4
1
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)=480
(a-d) +a + (a+d) = 24
Substituting a =8,
3a = 24
we get
a=8
d  2
2
When d = 2, required numbers are 6, 8, 10
When d = -2 , required numbers are 10, 8, 6
3
Example 3
The sum of the first eighteen terms of an
arithmetic series is -45 and the eighteenth
term is also -45.
U18  45
Find the common difference and the sum of
the first hundred terms.
Solution S  n 2a  n  1d 
n
2
18
S18  45  [2a  (18  1)d ]
2
 18a 153d  45 --------- (1)
4
U18  45
a  18 1d  45
a 17d  45 ------------ (2)
Solving equation 1 and 2, we get
a  40
d  5
5
Example 4
Find the sum of the positive integers which
are less than 500 and are not multiples of
11.
Solution
Sum of integers
Required sum= from 1 to 499
(SI)
S I  1  2  3  4  ............  499
499
1  499

2
 124750
Sum of
multiples
of 11(SII)
6
S II  11  22  33  44  .......  495
 111  2  3  4  ........  45
n
 45

S n  a  l 
 11 1  45
2
2

 11385
Required sum = SI-SII
=124750-11385
=113365
7
Example 5
S10  145
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 U 31  a  30d
(ii) the least number of terms required
for the sum to exceed 2000.
S10  U11  U12  U13  U14  U15  U16  231  145
S16  16 2a  15d 
2
8
Solution
16
S16  2a  15d   145  231
2
82a  15d   376
2a  15d  47
(1)
S10  145
10
2a  9d   145
2
2a  9d  29
(2)
9
6d  18
d 3
29  93
1
From (2): a 
2
(1)(2):
U 31  a  30d  1  303  91
(ii)Find the least number of terms required
for the sum to exceed 2000.
Let n be the least number of terms
required for Sn  2000
10
n
2a  n  1d   2000
2
n
21  n  13  2000
2
n3n  1  4000
3n 2  n  4000  0
2
Consider 3n  n  4000  0
1  1  43 4000
n
 36.7 or  36.3
6
11
For
2
3n  n  4000  0

+
n  36.3 or n  36.7
 36.3
+
36.7
Since n must be a
positive integer, n > 36.7
Hence, the least number of terms required
is 37.
12
Example 6
Given that the fifth term of a geometric
81
progression is
and the third term is 9.
4
Find the first term and the common ratio if all
the terms in the G.P. are positive.
4 81
(1)
ar 
4
2
ar  9
(2)
13
4
ar
81 1
(1)(2):
 
ar 2 4 9
9
r 
4
2
Since all the terms in the G.P. are positive,
r>0
3
r
2
2
From (2): a 3   9
 2
a4
14
Example 7
Three consecutive terms of a geometric
x 1
x
progression are 3 , 3
and 81. Find the
value of x. If 81 is the fifth term of the
geometric progression, find the seventh
term.
81
3 x 1
r  x  x 1
3
3
x 1
4
3
3
 x 1
x
3
3
15
3 x
33
1 3 x
x2
4
81 3
r  x 1  3  3
3
3
Given 81 is the fifth term, find the seventh
term:
U5= ar4 = 81
4
a 3  81
81
a  4 1
3
U7 =
ar6
 1(36 )  729
16
Example 8
Find the sum of the first eight terms of the
series 3  2  4  8  .........
3 9
Soln:
2 U 2 2 U3 2 U 4 2
 ,
 ,

G.P. with a  3, r  
3 U1 3 U 2 3 U 3 3

a 1 r
S8 
1 r
8

  2 8 
31    
  3  

2
1
3
17
  2 8 
31    
  3  

2
1
3
256 

 31 
3
 6561
6305

729
18
Example 9
a 1
A geometric series has first term 1 and
S
5
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
5 1
r  ( 3  1).
U 6  U 7  .....  U15
2
 U1  U 2  U 3  U 4  U 5   U 6  U 7  ........  U15
 U1  U 2  U 3  U 4  U 5   S15  S5
19
Solution
S5  2S15  S5 
Given
3 S5  2 S15
 a(r 5  1) 
 a(r15  1) 
 3
  2

 r  1 
 r  1 
Since r  1 and a = 1, we have
5
15
3(r  1)  2(r  1)
2r15  3r 5  1  0
2(r 5 )3  3r 5  1  0
20
Let x  r 5 .
So the equation becomes
2 x3  3x  1  0
Let f(x) = 2 x3  3x  1
Since f(1) = 2 – 3 + 1 = 0
(x –1) is a factor of f(x).
2
3
2 x  3x  1  ( x  1)(2 x  kx  1)
Comparing coefficient of x:  3  1  k
k 2
3
2
2 x  3x  1  ( x  1)(2 x  2 x  1)
21
2
( x  1)(2 x  2 x  1)  0
Hence,
2
 x  1 or 2 x  2 x  1  0
 2  4  4(2)(1)
x
2(2)
 2  12  2  2 3
 r  1 or r 

4
4
r  1
1
1
 ( 3  1) or ( 3  1)
2
2 (r<0)
5 1
Since r  1 and r > 0,  r  ( 3  1)
2
5
5
22
a
Sum to infinity, S (or S  ) =
1 r
The sum to infinity exists (the series converges
or series is convergent) provided r  1
Example 10
Determine whether the series given below
converge. If they do, give their sum to
infinity.
G.P. with r  2 >1
(a) 2  4  8  16  ........
Does not converge
23
1
1 1 1
(b) 1     ........... G.P. with r  
4
4 16 64
1
r  <1
4
a
S 
Series converges
1 r
1

1

1  
 4
4

5
24
9 27
3
(c) 3    ....................G.P. with r 
2 4
2
3
r  >1
2
Does not converge
25
Example 11
A geometric series has first term a and the
common ratio 1 .
2
Show that the sum to infinity of the geometric
progression is a (2  2 ) .
a
a
2
2

1
a
Solution S 



2 1
2 1
1  r 1  1 
2

a (2  2 )  a (2  2 ) (shown)

26
2 1
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 evennumbered terms (i.e.U 2  U 4  U 6  ....) of the
series. Given that S is four times the value
2
common
ratio
r
of T, find the value of r.
3
5
a
T

ar

ar  ar  .........
S
1 r
ar

1 r 2
a  4 ar
Given S = 4T, we have
27
1 r
1 r2
1
4r


1  r (1  r )(1  r )
4r
 1
1 r
 1  r  4r
1
r
3
28