Sequences And Series Flashcards
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.
- ( 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
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.
- ( 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
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.
- ( 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
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 )
- 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
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.
- 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
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 )
- Third term = ar^2
- r = 0.6, so r^2 = 0.36
- ar^2 / r^2 = a
- 5.4 / 0.36 = 15
- a = 15
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 )
- ( S( infinity ) = a / 1 - r )
- ( Since r < 1, this will work )
- ( a = 15 )
- ( r = 0.6 )
- S( infinity ) = 15 / 1 - 0.6 = 37.5
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.
- ( r can be found out be dividing consecutive terms together )
- 12 / 18 = 2 / 3
- r = 2 / 3
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 )
- Third term = ar^2
- P = ar^2
- P = 18 x 2 / 3^2
- P = 8
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.
- ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
- S15 = 18( 1 - 2 / 3^15 ) / 1 - 2 / 3
- S15 = 53.877
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 = 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
Hence show that p = 5.
11p^2 - 10p - 225 = 0
- 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
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 )
- 4( 5 ) , ( 3( 5 ) + 15 ), ( 5( 5 ) + 20 )
- 20, 30, 45
- r = 30 / 20
- r = 1.5
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 )
- ( Sn = a( r^n - 1 ) / r - 1 ), ( r > 1 ) )
- S10 = 20( 1.5^10 - 1 ) / 1.5 - 1
- S10 =2267
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 ).
- ( S( infinity ) = a / 1 - r )
- S( infinity ) = 20 / 1 - 7 / 8
- S( infinity ) = 160
The sum to N terms of the series is SN.
Find, to 1 decimal place, the value of S12.
( a = 20 )
( r = 7 / 8 )
- ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
- S12 = 20( 1 - 7 / 8^12 ) / 1 - 7 / 8
- S12 = 127.8
Find the smallest value of N, for which S( infinity ) - SN < 0.5.
( S( infinity ) = 160 )
( a = 20 )
( r = 7 / 8 )
- 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 )
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.
- ( 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
Given that the fourth term of this series is 62.5, find the value of a.
( r = 5 / 6 )
- Fourth term = ar^3
- 62.5 = a( 5 / 6 )^3
- a = 62.5 / ( 5 / 6 )^3
- a = 108
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 )
- ( 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
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 + 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
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 + ar = 34
- a + a( 8 / 9 ) = 34
- 17 / 9a = 34
- a = 18
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.
- ( 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
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.
- ( 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
Find the sum to infinity of the series.
a = 64
( r = 3 / 4 )
- ( S( infinity ) = a / 1 - r )
- S( infinity ) = 64 / 1 - 3 / 4
- S( infinity ) = 256
Find the difference between the 9th and 10th terms of the series.
Give your answer to 3 decimal places.
( a = 64 )
( r = 3 / 4 )
- 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
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 = 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
Given that k is not an integer,
Show that k = 9 / 11
( 11k^2 - 130k + 99 = 0 )
- 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
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 )
- 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
For this value of k, evaluate the sum of the first ten terms of the sequence.
( a = 8 / 11 )
( r = - 4 )
- ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
- S10 = 8 / 11( 1 - ( - 4 )^10 ) / 1 - ( - 4 )
- S10 = -152520
Find 10 E k = 1 ( 100( 2^k ) )
E = Sum of
- ( 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 ) ) )
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.
- 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
Show that the sum, Sn, of the first n terms of the series is given by
Sn = 25( 1 – r^n ).
( S( infinity ) = 25 )
- 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 ) )
