AP-GP-Q1-to-Q12-Solutions
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Transcript AP-GP-Q1-to-Q12-Solutions
Example 1:
Find the 10th term and the nth term for the
sequence 7, 10, 13, … .
Solution:
a7
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.
U 18 45
Find the common difference and the sum of
the first hundred terms.
Solution
Sn
n
2
2 a n 1 d
S 18 45
18
[ 2 a (18 1) d ]
2
18 a 153 d 45
--------- (1)
4
U 18 45
a 18 1 d 45
a 17 d 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
45
1 45
11
2
Sn
n
a l
2
11385
Required sum = SI-SII
=124750-11385
=113365
7
Example 5
S 10 145
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 30 d
(ii) the least number of terms required
for the sum to exceed 2000.
S10 U U
11
12 U 13 U 14 U 15 U 16 231 145
S16 16 2 a 15 d
2
8
Solution
S16
16
2
2 a 15 d 145
231
82 a 15 d 376
2 a 15 d 47
(1)
S 10 145
10
2
2 a 9 d 145
2 a 9 d 29
(2)
9
(1)(2):
6 d 18
d 3
From (2): a
29 9 3
1
2
U 31 a 30 d 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 S n 2000
10
n
2
n
2
2 a n 1 d
2000
2 1 n 1 3
2000
n 3 n 1 4000
3n
2
n 4000 0
2
Consider 3 n n 4000 0
1 1 4 3 4000
n
36 . 7 or 36 . 3
6
11
For
3n
2
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
progression is
81
4
and the third term is 9.
Find the first term and the common ratio if all
the terms in the G.P. are positive.
ar
4
81
(1)
4
ar
2
9
(2)
13
(1)(2):
ar
ar
4
2
r
2
81
4
1
9
9
4
Since all the terms in the G.P. are positive,
r>0
3
r
2
2
From (2): a 3 9
2
a4
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.
x 1
r
3
3
3
x
3
x 1
3
x
81
3
3
x 1
4
x 1
15
3 x
33
1 3 x
81
r
x2
3
x 1
3
4
3
3
3
Given 81 is the fifth term, find the seventh
term:
U5= ar4 = 81
a 3 81
4
a
81
3
U7 =
ar6
1
4
1( 3 ) 729
6
16
Example 8
Find the sum of the first eight terms of the
series 3 2 4 8 .........
3
Soln:
2
a 1 r
1 r
8
U2
8
2
3 1
3
2
1
3
2
U3
2 U4 2
,
,
a 3, r
3
U1 3 U 2 3 U 3 3
G.P. with
S8
9
17
8
2
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
S5
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
r
5
1
( 3 1).
U 6 U 7 ..... U 15
2
U 1 U 2 U 3 U 4 U 5 U 6 U 7 ........ U 15
U 1 U 2 U 3 U 4 U 5
S15 S 5
19
Solution
S 5 2 S15 S 5
Given
3 S 5 2 S 15
a ( r 5 1)
a ( r 15 1)
3
2
r 1
r 1
Since r 1 and a = 1, we have
5
3 ( r 1) 2 ( r
2r
15
3r
5 3
5
15
1)
1 0
5
2(r ) 3r 1 0
20
Let x r 5 .
So the equation becomes
3
2x 3x 1 0
Let f(x) = 2 x 3 3 x 1
Since f(1) = 2 – 3 + 1 = 0
(x –1) is a factor of f(x).
3
2 x 3 x 1 ( x 1)( 2 x
2
kx 1)
Comparing coefficient of x: 3 1 k
k 2
3
2 x 3 x 1 ( x 1)( 2 x
2
2 x 1)
21
( x 1)( 2 x
Hence,
2
2 x 1) 0
x 1 or 2 x
x
2
2x 1 0
2
4 4 ( 2 )( 1)
2(2)
5
r 1 or r
r 1
5
2
12
22 3
4
1
4
( 3 1) or
2
5
1
( 3 1)
2
(r<0)
1
Since r 1 and r > 0, r ( 3 1)
2
22
Sum to infinity, S (or S ) =
a
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
(b) 1
1
4
16
S
1
1
64
a
1 r
.......... .
G.P. with
r
1
r
<1
1
4
4
Series converges
1
1
1
4
4
5
24
(c) 3
9
2
27
4
.......... ..........G.P.
3
with r
2
3
>1
r
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 ) .
Solution S
a
1 r
a (2
2 1
a
1
1
2
2 ) a (2
a 2
2 1
2)
2 1
2 1
(shown)
26
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.
S
a
1 r
T ar ar 3 ar 5 .........
ar
1 r2
Given S = 4T, we have
a
1 r
4
ar
1 r
2
27
1
1 r
1
4r
(1 r )( 1 r )
4r
1 r
1 r 4r
r
1
3
28