Sequences And Series

Question 1

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.

Answer 1

• ( 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

Question 2

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.

Answer 2

• ( 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

Question 3

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.

Answer 3

• ( 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

Question 4

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 )

Answer 4

• 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

Question 5

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.

Answer 5

• 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

Question 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 )

Answer 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

Question 7

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 )

Answer 7

• ( S( infinity ) = a / 1 - r )
• ( Since r < 1, this will work )
• ( a = 15 )
• ( r = 0.6 )
• S( infinity ) = 15 / 1 - 0.6 = 37.5

Question 8

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.

Answer 8

• ( r can be found out be dividing consecutive terms together )
• 12 / 18 = 2 / 3
• r = 2 / 3

Question 9

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 )

Answer 9

• Third term = ar^2
• P = ar^2
• P = 18 x 2 / 3^2
• P = 8

Question 10

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.

Answer 10

• ( Sn = a( 1 - r^n ) / 1 - r ), ( r < 1 ) )
• S15 = 18( 1 - 2 / 3^15 ) / 1 - 2 / 3
• S15 = 53.877

Question 11

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.

Answer 11

• 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

Question 12

Hence show that p = 5.

11p^2 - 10p - 225 = 0

Answer 12

• 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

Question 13

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 )

Answer 13

• 4( 5 ) , ( 3( 5 ) + 15 ), ( 5( 5 ) + 20 )
• 20, 30, 45
• r = 30 / 20
• r = 1.5

Question 14

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 )

Answer 14

• ( Sn = a( r^n - 1 ) / r - 1 ), ( r > 1 ) )
• S10 = 20( 1.5^10 - 1 ) / 1.5 - 1
• S10 =2267

Question 15

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 ).

Answer 15

• ( S( infinity ) = a / 1 - r )
• S( infinity ) = 20 / 1 - 7 / 8
• S( infinity ) = 160

Question 16

The sum to N terms of the series is SN.

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

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

Answer 16

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

Question 17

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

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

Answer 17

• 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 )

Question 18

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.

Answer 18

• ( 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

Question 19

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

( r = 5 / 6 )

Answer 19

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

Question 20

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 )

Answer 20

• ( 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

Question 21

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.

Answer 21

• 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

Question 22

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 )

Answer 22

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

Question 23

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.

Answer 23

• ( 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

Question 24

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.

Answer 24

• ( 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

Question 25

Find the sum to infinity of the series. | a = 64 ( r = 3 / 4 )

Answer 25

- ( S( infinity ) = a / 1 - r ) - S( infinity ) = 64 / 1 - 3 / 4 - S( infinity ) = 256

Question 26

Find the difference between the 9th and 10th terms of the series. Give your answer to 3 decimal places. ( a = 64 ) ( r = 3 / 4 )

Answer 26

- 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

Question 27

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

Answer 27

- 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

Question 28

Given that k is not an integer, Show that k = 9 / 11 ( 11k^2 - 130k + 99 = 0 )

Answer 28

- 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

Question 29

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 )

Answer 29

- 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

Question 30

For this value of k, evaluate the sum of the first ten terms of the sequence. ( a = 8 / 11 ) ( r = - 4 )

Answer 30

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

Question 31

Find 10 E k = 1 ( 100( 2^k ) ) | E = Sum of

Answer 31

- ( 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 ) ) )

Question 32

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.

Answer 32

- 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

Question 33

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

Answer 33

- 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 ) )