# Thread: Geometric progression involving logs!

1. ## Geometric progression involving logs!

Hey, can anyone help me with this please?

5. In a geometric progression, the first term is 5 and the second term is 4.8
i) I have solved this. r = 0.96
ii) The sum of the first n terms is greater than 124. Show that

0.96^n < 0.008,

and use logarithms to calculate the smallest possible value of n.

Thank you if you can help me with this.

2. Originally Posted by LoveDeathCab
Hey, can anyone help me with this please?

5. In a geometric progression, the first term is 5 and the second term is 4.8
i) I have solved this. r = 0.96
a geometric progression is one in which the terms are given by the formula:
a_n = ar^(n - 1) for n = 1,2,3,4,5...
where a_n is the nth term, a is the first term and r is the common ration

we are told that the first term of a particular geometric sequence is 5 and the second term is 4.8

now r is given by r = (a_{n+1})/(a_n)
in particular, r = (a_2)/(a_1) = 4.8/5 = 0.96

you are correct

ii) The sum of the first n terms is greater than 124. Show that

0.96^n < 0.008,

and use logarithms to calculate the smallest possible value of n.
The sum of the first n terms of a geometric sequence/progression is given by:
S_n = a(1 - r^n)/(1 - r)
we are told that S_n > 124
=> a(1 - r^n)/(1 - r) > 124
=> 5(1 - (0.96)^n)/(1 - 0.96) > 124
=>5(1 - (0.96)^n)/(0.04) > 124
=> 5(1 - (0.96)^n) > 4.96 ..............i multiplied through by 0.04
=> 1 - (0.96)^n > 0.992 ..................i divided through by 5
=> -(0.96)^n > -0.008 ....................i subtracted 1 from both sides
=> (0.96)^n < 0.008 ...............................i multiplied through by -1, so i flipped the inequality sign.

Now we will use logs to find the smallest possible n.

0.96^n < 0.008
take log to the base 10 of both sides
=> log(0.96^n) < log(0.008)
=> nlog(0.96) < log(0.008)
=> n > log(0.008)/log(0.96) ...........note that log(0.96) is negative, so when i divided by it, i flipped the inequality sign
=> n > 118.27

so the smallest possible value for n is 119

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