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Geometric Series
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Starter: Describe the recurrence relation for each sequence
Geometric Series KUS objectives BAT you need to be able to spot patterns to work out the common ratio and rule for a Geometric Sequence BAT use Percentages in Geometric Sequences BAT apply logarithms to solve problems Starter: Describe the recurrence relation for each sequence (formula for the next term) 3, 12, 48, , β¦ π’ π+1 =4Γ π’ π 4, 2, 1, , , , β¦ π’ π+1 = 1 2 Γ π’ π 6Γ 2 5 , 6Γ 4 25 , 6Γ , β¦ π’ π+1 = 2 5 Γ π’ π
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r is the common ratio, and is the same for all consecutive pairs
WB1 Common ratio In a Geometric sequence, u1, u2, u3 β¦ un we will have ππ π π+1 =π a constant value r is the common ratio, and is the same for all consecutive pairs Find the common ratio in each of these sequences: i) 4, 2, 1, , , , β¦ π«= 2 4 = 1 2 π«= = 3 ii) 3, 12, 48, , β¦ π«= = 1 3 iii) 54, 18, 6, , β¦ π«= 1 10 iv) 21, , , , β¦ βΉ π₯ 2 =1 βΉ π₯ =Β±4 βΉ π= π₯ 20 =0.2 π«= 0.8 π₯ = π₯ 20 v) 20, π₯, , β¦ π«= 51.2 π₯ = π₯ 5 βΉ π₯ 2 =256 βΉ π₯ =Β±16 βΉ π= π₯ 5 =3.2 vi) 5, π₯, , β¦
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β¦ ,4Γ 1 2 π_1 β¦ ,3Γ 4 π_1 β¦ ,54Γ 1 3 π_1 ,β¦ ,21Γ 1 10 π_1 WB2 Nth term
In a Geometric sequence, the nth term comes from multiplying the first term by the common ration (n β 1) times a, ar, ar2, ar3, β¦, arn-1 1st Term 2nd Term 3rd Term 4th Term nth Term For example. find the nth term of these: β¦ ,4Γ π_1 i) 4, 2, 1, , , , β¦ ii) 3, 12, 48, ,β¦ β¦ ,3Γ 4 π_1 β¦ ,54Γ π_1 iii) 54, 18, 6, ,β¦ ,β¦ ,21Γ π_1 iv) 21, , , , β¦
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Find the nth and 10th terms of the following sequencesβ¦
WB 3 nth term Find the nth and 10th terms of the following sequencesβ¦ i) 40, β20, , β5, β¦ π=3, π=2, π π πβ1 =3 Γ 2 πβ π‘β=1536 ii) 3, 12, 48, , β¦ π=40, π=β 1 2 , π π πβ1 =40 Γ πβ π‘β=β 5 64 iii) 2, 8, , β¦ iv) 12, β3, , β¦ v) , , , β¦ vi) 6Γ 2 5 , 6Γ 4 25 , 6Γ , β¦
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WB 4 Finding a and r The second term of a Geometric sequence is 4, and the 4th term is 8. Find the values of the common ratio and the first term 1 2nd Term ο 2 4th Term ο 2 Γ· 1 ο Square root Sub r into 1 Divide by β2 Rationalise
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a) A geometric sequence has third term 12 and sixth term -96
WB 5ab finding a and r a) A geometric sequence has third term 12 and sixth term -96 Find the first term and the common ratio π’ 6 π’ 3 = π π 5 π π 2 = π 3 π’ 6 π’ 3 = β96 12 =β8 So π=β2 So third term π π 2 =4π=12 So first term π=3 b) The third term of a geometric sequence is 324 and the fifth term is 36 Find the first term and the two possible values of the common ratio π’ 5 π’ 3 = π π 4 π π 2 = π 2 π’ 5 π’ 3 = = 1 9 So π= 1 3 So third term π π 2 = 1 9 π=324 So first term π=2916
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WB 5c finding a and r c) The numbers 3, x, and (x + 6) form the first three terms of a positive geometric sequence. Calculate the 15th term of the sequence First term = 3 Common Ratio = 2 Nth term = 3 x 2n-1 15th Term = 3 x 214 x has to be positive 15th Term = 49152
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So first term when 3 π >10000
WB 6 using logarithms Find the first term in the geometric sequence 1, 3, 9, 27 β¦ to exceed the value of 10000 π=3 π= ππ‘β π‘πππ ππ 3 π So first term when π >10000 use logarithms n> πππ So n> first term that works is n = 9
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WB7: logarithms What is the first term in the sequence 3, 6, 12, 24β¦ to exceed 1 million? π=2 π= ππ‘β π‘πππ ππ 3Γ 2 πβ1 So first term when 2 πβ1 > use logarithms nβ1 > πππ =18.35 So n> first term that works is n = 20 b) Find the first term in the geometric sequence 2, 12, 72, 432, β¦ to exceed the value of π=6 π= ππ‘β π‘πππ ππ 2Γ6 πβ1 So first term when 2Γ 6 πβ1 > use logarithms nβ1> πππ =6.81 So n> first term that works is n = 8
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Using percentage change
If I was to increase an amount by 10%, what would I multiply the value by? If I was to increase an amount by 17%, what would I multiply by? ο 1.1 ο 1.17 If I was to increase an amount by 10%, every year for 6 years, what would I multiply the value by? If I was to increase an amount by 17%, every year for 20 years, what would I multiply by? ο Γ π.π π ο Γ π.ππ ππ
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If Β£A is to be invested in a savings fund at a rate of 4%.
WB 8 percentage change If Β£A is to be invested in a savings fund at a rate of 4%. How much should be invested so the fund is worth Β£10,000 in 5 years? Y1 Y2 Y3 Y4 Y5 A Ar Ar5 Ar4 Ar3 Ar2 a = A r = 1.04 Ar5 = 10,000 A x 1.045 = 10,000 A = Β£
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One thing to improve is β
KUS objectives BAT you need to be able to spot patterns to work out the common ratio and rule for a Geometric Sequence BAT use Percentages in Geometric Sequences BAT apply logarithms to solve problems self-assess One thing learned is β One thing to improve is β
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