Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Maths > Sequences And Series > Flashcards

Sequences And Series Flashcards

1
Q

A geometric series has first term a = 360 and common ratio r = 7 / 8.

Giving your answers to 3 significant figures where appropriate, find the 20th term of the series.

A
  • ( Un = ar^n - 1 )
  • ( a = first term )
  • ( r = common ratio )
  • ( n = Nth term )
  • ( Un = ( first term ) x ( common ratio )^ Nth term - 1 )
  • U20 = 360 x 7 / 8^20 - 1
  • U20 = 28.5
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

A geometric series has first term a = 360 and common ratio r = 7 / 8.

Giving your answers to 3 significant figures where appropriate, find the sum of the first 20 terms of the series.

A
  • ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
  • ( a = first term )
  • ( r = common ratio )
  • ( n = Nth term )
  • S20 = 360( 1 - 7 / 8^20 ) / 1 - 7 / 8
  • S20 = 2680
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

A geometric series has first term a = 360 and common ratio r = 7 / 8.

Giving your answers to 3 significant figures where appropriate, find
the sum to infinity of the series.

A
  • ( S( infinity ) = a / 1 - r )
  • ( a = first term )
  • ( r = common ratio )
  • ( S( infinity ) can only be used when | r | < 1 )
  • ( So - 1 < r < 1 )
  • S( infinity ) = 360 / 1 - 7 / 8
  • S( infinity ) = 2880
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

A geometric series is a + ar + ar^2 + …

Prove that the sum of the first n terms of this series is given by

Sn = a( 1 - r^n ) / 1 - r )

A
  • Sn = a + ar + ar^2 + ar^3 … ar^n - 3 + ar^n - 2 + ar^n - 1
  • rSn = ar + ar^2 + ar^3 + ar^4 … ar^n - 2 + ar^n - 1 + ar^n
  • Sn - rSn = a - ar^n ( other values cancel out )
  • Sn( 1 - r ) = a( 1 - r^n )
  • Sn = a( 1 - r^n ) / 1 - r
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

The third and fifth terms of a geometric series are 5.4 and 1.944 respectively and all the terms in the series are positive.

For this series find, the common ratio.

A
  • Third term = ar^2
  • Fifth term = ar^4
  • ar^4 / ar^2 = r^2
  • r^2 = 1.944 / 5.4 = 0.36
  • r = 0.6
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

The third and fifth terms of a geometric series are 5.4 and 1.944 respectively and all the terms in the series are positive.

For this series find, the first term.

( r = 0.6 )

A
  • Third term = ar^2
  • r = 0.6, so r^2 = 0.36
  • ar^2 / r^2 = a
  • 5.4 / 0.36 = 15
  • a = 15
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

The third and fifth terms of a geometric series are 5.4 and 1.944 respectively and all the terms in the series are positive.

For this series find, the sum to infinity.

( a = 15 )
( r = 0.6 )

A
  • ( S( infinity ) = a / 1 - r )
  • ( Since r < 1, this will work )
  • ( a = 15 )
  • ( r = 0.6 )
  • S( infinity ) = 15 / 1 - 0.6 = 37.5
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

The first three terms of a geometric series are

18, 12 and p

respectively, where p is a constant.

Find the value of the common ratio of the series.

A
  • ( r can be found out be dividing consecutive terms together )
  • 12 / 18 = 2 / 3
  • r = 2 / 3
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

The first three terms of a geometric series are

18, 12 and p

respectively, where p is a constant.

Find the value of p.

( r = 2 / 3 )

A
  • Third term = ar^2
  • P = ar^2
  • P = 18 x 2 / 3^2
  • P = 8
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

The first three terms of a geometric series are

18, 12 and p

respectively, where p is a constant.

Find the sum of the first 15 terms of the series, giving your answer to 3 decimal places.

A
  • ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
  • S15 = 18( 1 - 2 / 3^15 ) / 1 - 2 / 3
  • S15 = 53.877
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

The first three terms of a geometric series are 4p, ( 3p + 15 ) and ( 5p + 20 ) respectively, where p is a positive constant.

Show that 11p^2 - 10p - 225 = 0.

A
  • a = 4p
  • ar = 3p + 15
  • ar^2 = 5p + 20
  • ( r = ) 5p + 20 / 3p + 15 = 3p + 15 / 4p
  • ( So ) ( 5p + 20 ) ( 4p ) = ( 3p + 15 ) ( 3p + 15 )
  • 20p^2 + 80p = 9p^2 + 45p + 45p + 225
  • 11p^2 - 10p - 225 = 0
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Hence show that p = 5.

11p^2 - 10p - 225 = 0

A
  • a = 11
  • b = - 10
  • c = - 225
  • ( - b +- root ( b )^2 - 4ac / 2a ) ( Quadratic equation )
    • ( - 10 ) +- root ( - 10 )^2 - 4( 11 )( - 225 ) / 2( 11 )
  • P = 5 P = -4.0909…
  • P cannot be negative
  • So P = 5
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Find the common ratio of this series.

( The first three terms of a geometric series are 4p, ( 3p + 15 ) and ( 5p + 20 ) respectively, where p is a positive constant. )
( P = 5 )

A
  • 4( 5 ) , ( 3( 5 ) + 15 ), ( 5( 5 ) + 20 )
  • 20, 30, 45
  • r = 30 / 20
  • r = 1.5
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Find the sum of the first ten terms of the series, giving your answer to the nearest integer.

( The first three terms of a geometric series are 4p, ( 3p + 15 ) and ( 5p + 20 ) respectively, where p is a positive constant. )
( P = 5 )
( 20, 30, 45 )
( r = 1.5 )

A
  • ( Sn = a( r^n - 1 ) / r - 1 ), ( r > 1 ) )
  • S10 = 20( 1.5^10 - 1 ) / 1.5 - 1
  • S10 =2267
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

The first term of a geometric series is 20 and the common ratio is 7 / 8.
The sum to infinity of the series is S( infinity ).

Find the value of S( infinity ).

A
  • ( S( infinity ) = a / 1 - r )
  • S( infinity ) = 20 / 1 - 7 / 8
  • S( infinity ) = 160
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

The sum to N terms of the series is SN.

Find, to 1 decimal place, the value of S12.

( a = 20 )
( r = 7 / 8 )

A
  • ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
  • S12 = 20( 1 - 7 / 8^12 ) / 1 - 7 / 8
  • S12 = 127.8
17
Q

Find the smallest value of N, for which S( infinity ) - SN < 0.5.

( S( infinity ) = 160 )
( a = 20 )
( r = 7 / 8 )

A
  • 160 - Sn < 0.5
  • 160 - 20( 1 - 7 / 8^n ) / 1 - 7 / 8 < 0.5
  • 7 / 8^n < 0.5 / 160
  • ( Things cancel out ) ( JUST CLASS THIS AS A RULE OKAY )
  • ( r^n < 0.5 / S( infinity ) )
  • log( 7 / 8 )( 0.5 / 160 ) > 43.19823887
  • n = 44 ( we round up due to the “ > “ sign )
18
Q

A geometric series has first term a, where a ≠ 0, and common ratio r.
The sum to infinity of this series is 6 times the first term of the series.

Show that r = 5 / 6.

A
  • ( S( infinity ) = a / 1 - r )
  • 6a = a / 1 - r
  • 6a( 1 - r ) = a
  • 6a - 6ar = a
  • 6 - 6r = 1 ( a’s cancel out )
    • 6r = - 5
  • r = 5 / 6
19
Q

Given that the fourth term of this series is 62.5, find the value of a.

( r = 5 / 6 )

A
  • Fourth term = ar^3
  • 62.5 = a( 5 / 6 )^3
  • a = 62.5 / ( 5 / 6 )^3
  • a = 108
20
Q

Find the difference between the sum to infinity and the sum of the first 30 terms, giving your answer to 3 significant figures.

( a = 108 )
( r = 5 / 6 )

A
  • ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 )
  • S30 = 108( 1 - 5 / 6^30 ) / 1 - 5 / 6
  • S30 = 645.2701573
  • S( infinity ) = a / 1 - r
  • S( infinity ) = 108 / 1 - 5 /6
  • S( infinity ) = 648
  • 648 - 645.270…
  • = 2.729842711
  • = 2.73
21
Q

All the terms of a geometric series are positive.
The sum of the first two terms is 34 and the sum to infinity is 162.

Find the common ratio.

A
  • a + ar = 34
  • ( S( infinity ) = a / 1 - r )
  • 162( 1 - r ) = a
  • 162 - 162r = a
  • 162 - 162r + ( 162 - 162r )r = 34
  • 162 - 162r^2 = 34
    • 162r^2 = - 128
  • r^2 = 0.7901234568
  • r = 8 / 9
22
Q

All the terms of a geometric series are positive.
The sum of the first two terms is 34 and the sum to infinity is 162.

Find the first term.

( r = 8 / 9 )

A
  • a + ar = 34
  • a + a( 8 / 9 ) = 34
  • 17 / 9a = 34
  • a = 18
23
Q

A different geometric series has a first term of 42 and a common ratio of 6 / 7.
Find the smallest value of n for which the sum of the first n terms of the series exceeds 290.

A
  • ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
  • 42( 1 - 6 / 7^n ) / 1 - 6 / 7 > 290
  • 42( 1 - 6 / 7^n ) > 290 / 7
  • 1 - 6 / 7^n > 145 / 147
  • 6 / 7^n > 2 / 147
  • log( 6 / 7 )( 2 / 147 ) < n
  • n > 27.87717453
  • n = 28
24
Q

A geometric series has first term a and common ratio r = 3 / 4 .

The sum of the first 4 terms of this series is 175.

Show that a = 64.

A
  • ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
  • 175 = a( 1 - 3 / 4^4 ) / 1 - 3 / 4
  • a( 1 - 3 / 4^4 ) = 175 / 4
  • a = 64
25
Q

Find the sum to infinity of the series.

a = 64
( r = 3 / 4 )

A
  • ( S( infinity ) = a / 1 - r )
  • S( infinity ) = 64 / 1 - 3 / 4
  • S( infinity ) = 256
26
Q

Find the difference between the 9th and 10th terms of the series.
Give your answer to 3 decimal places.

( a = 64 )
( r = 3 / 4 )

A
  • Ninth term = ar^8
  • Tenth term = ar^9
  • Ninth term = 64( 3 / 4 )^8 = 6.407226565
  • Tenth term = 64( 3 / 4 )^9 = 4.805419922
  • 6.307… - 4.805… = 1.601806643
  • = 1.602
27
Q

The first three terms of a geometric sequence are

7k - 5, 5k - 7, 2k + 10

where k is a constant.

Show that 11k^2 - 130k + 99 = 0

A
  • a = 7k - 5
  • ar = 5k - 7
  • ar^2 = 2k + 10
  • ( r = ) 2k + 10 / 5k - 7 = 5k - 7 / 7k - 5
  • ( 2k + 10 ) ( 7k - 5 ) = ( 5k - 7 ) ( 5k - 7 )
  • = 14k^2 - 10k + 70k - 50 = 25k^2 - 35k - 35k + 49
  • 11k^2 - 130k + 99 = 0
28
Q

Given that k is not an integer,

Show that k = 9 / 11

( 11k^2 - 130k + 99 = 0 )

A
  • a = 11
  • b = - 130
  • c = 99
  • ( - b +- root ( b )^2 - 4ac / 2a ) ( Quadratic equation )
    • ( - 130 ) +-root ( - 130 )^2 - 4( 11 )( 99 ) / 2( 11 )
  • k = 11 and k = 9 / 11
  • Since k isn’t an integer, k = 9 / 11
29
Q

For this value of k, evaluate the fourth term of the sequence, giving your answer as an exact fraction.

( k = 9 / 11 )
( 7k - 5 , 5k - 7, 2k + 10 )

A
  • 7( 9 / 11 ) - 5 = 8 / 11
  • 5( 9 / 11 ) - 7 = - 32 / 11
  • 2( 9 / 11 ) + 10 = 128 / 11
  • r = - 32 / 11 / 8 / 11 = - 4
  • Fourth term = ar^3
  • ( 8 / 11 )( - 4 )^3
  • = - 512/ 11
30
Q

For this value of k, evaluate the sum of the first ten terms of the sequence.

( a = 8 / 11 )
( r = - 4 )

A
  • ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
  • S10 = 8 / 11( 1 - ( - 4 )^10 ) / 1 - ( - 4 )
  • S10 = -152520
31
Q

Find 10 E k = 1 ( 100( 2^k ) )

E = Sum of

A
  • ( Sub values of k into the equation to find the sequence or series )
  • ( From where k = 1 )
  • Sequence or series = 200, 400, 800, 1600
  • ( It’s obvious that there is a multiplier, so series is ar^n )
  • a = 200 ( Format of question = ar^n )
  • r = 2 ( 400 / 200 = 2 )
  • n = 10
  • ( Number of terms between k = 1 and k = 10 including 1 )
  • ( 10 - ( 1 - 1 ) = 10 )
  • ( Sn = a( r^n - 1) / r - 1 ), ( r > 1 ) )
  • S10 = 200( 2^10 - 1 ) / 2 - 1
  • S10 = 204600

( If series is arithmetic, we use Sn = n / 2 ( 2a + d( n - 1 ) ) )

32
Q

A geometric series has first term a and common ratio r.
The second term of the series is 4 and the sum to infinity of the series is 25.

Show that 25r^2 - 25r + 4 = 0.

A
  • ar = 4
  • S( infinity ) = 25
  • 25 = a / 1 - r
  • a = 25( 1 - r )
  • 25( 1 - r )r = 4
  • ( 25 - 25r )r = 4
    • 25r^2 + 25r = 4
  • 25r^2 - 25r + 4 = 0
33
Q

Show that the sum, Sn, of the first n terms of the series is given by
Sn = 25( 1 – r^n ).

( S( infinity ) = 25 )

A
  • Sn = a( 1 - r^n ) / 1 - r
  • However we have S( infinity ) = a / 1 - r
  • 25 = a / 1 - r
  • ( So ) Sn = 25( 1 - r^n )
  • ( Numerator of S( infinity ) is missing “ ( 1 - r^n ) “ to match )
  • ( So to match Sn, S( infinity ) needs to be multiplied by ( 1 - r^n ) )

Maths (21 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions