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(i) Calculating the common ratio and the first term:
Given an exponential sequence, we can denote the first term by [tex]\( a \)[/tex] and the common ratio by [tex]\( r \)[/tex].
We are given two conditions:
1. The sum of the first two terms is 135.
[tex]\[ a + ar = 135 \][/tex]
2. The sum of the third and fourth terms is 60.
[tex]\[ ar^2 + ar^3 = 60 \][/tex]
From the first condition:
[tex]\[ a(1 + r) = 135 \][/tex]
From the second condition:
[tex]\[ ar^2(1 + r) = 60 \][/tex]
Dividing the second equation by the first equation to eliminate [tex]\( a \)[/tex]:
[tex]\[ \frac{ar^2(1 + r)}{a(1 + r)} = \frac{60}{135} \][/tex]
[tex]\[ r^2 = \frac{60}{135} \][/tex]
[tex]\[ r^2 = \frac{4}{9} \][/tex]
[tex]\[ r = \frac{2}{3} \][/tex] (since [tex]\( r \)[/tex] is positive)
Now, substitute [tex]\( r \)[/tex] back into the first equation to find [tex]\( a \)[/tex]:
[tex]\[ a(1 + \frac{2}{3}) = 135 \][/tex]
[tex]\[ a \cdot \frac{5}{3} = 135 \][/tex]
[tex]\[ a = 135 \times \frac{3}{5} \][/tex]
[tex]\[ a = 81 \][/tex]
So, the first term [tex]\( a \)[/tex] is 81, and the common ratio [tex]\( r \)[/tex] is [tex]\( \frac{2}{3} \)[/tex].
(ii) Calculating the limit of the sum of the first [tex]\( n \)[/tex] terms as [tex]\( n \)[/tex] becomes large:
In a geometric series, the sum of the first [tex]\( n \)[/tex] terms is given by:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]
As [tex]\( n \)[/tex] becomes very large, [tex]\( r^n \)[/tex] tends to 0 if [tex]\( |r| < 1 \)[/tex]. Given [tex]\( r = \frac{2}{3} \)[/tex], [tex]\( |r| < 1 \)[/tex], so:
[tex]\[ \lim_{n \to \infty} S_n = a \frac{1}{1 - r} \][/tex]
[tex]\[ \lim_{n \to \infty} S_n = 81 \frac{1}{1 - \frac{2}{3}} \][/tex]
[tex]\[ \lim_{n \to \infty} S_n = 81 \frac{1}{\frac{1}{3}} \][/tex]
[tex]\[ \lim_{n \to \infty} S_n = 81 \times 3 \][/tex]
[tex]\[ \lim_{n \to \infty} S_n = 243 \][/tex]
So, the limit of the sum of the first [tex]\( n \)[/tex] terms as [tex]\( n \)[/tex] becomes large is 243.
(iii) Finding the least number of terms for which the sum exceeds 240:
We want the smallest [tex]\( n \)[/tex] such that:
[tex]\[ a \frac{1 - r^n}{1 - r} > 240 \][/tex]
Using [tex]\( a = 81 \)[/tex] and [tex]\( r = \frac{2}{3} \)[/tex]:
[tex]\[ 81 \frac{1 - (\frac{2}{3})^n}{\frac{1}{3}} > 240 \][/tex]
[tex]\[ 81 \times 3 (1 - (\frac{2}{3})^n) > 240 \][/tex]
[tex]\[ 243(1 - (\frac{2}{3})^n) > 240 \][/tex]
[tex]\[ 1 - (\frac{2}{3})^n > \frac{240}{243} \][/tex]
[tex]\[ 1 - (\frac{2}{3})^n > \frac{80}{81} \][/tex]
[tex]\[ -(\frac{2}{3})^n > \frac{80}{81} - 1 \][/tex]
[tex]\[ -(\frac{2}{3})^n > \frac{-1}{81} \][/tex]
[tex]\[ (\frac{2}{3})^n < \frac{1}{81} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ \ln\left((\frac{2}{3})^n\right) < \ln\left(\frac{1}{81}\right) \][/tex]
[tex]\[ n \ln\left(\frac{2}{3}\right) < \ln\left(\frac{1}{81}\right) \][/tex]
Since [tex]\( \ln\left(\frac{2}{3}\right) \)[/tex] is negative, divide by it to flip the inequality:
[tex]\[ n > \frac{\ln\left(\frac{1}{81}\right)}{\ln\left(\frac{2}{3}\right)} \][/tex]
Calculating the value:
[tex]\[ n > \frac{\ln(1) - \ln(81)}{\ln(2) - \ln(3)} \][/tex]
[tex]\[ n > \frac{0 - \ln(81)}{\ln(2) - \ln(3)} \][/tex]
[tex]\[ n > \frac{-\ln(81)}{\ln(2) - \ln(3)} \][/tex]
[tex]\[ n > \frac{-4.394449}{-0.405465} \][/tex] (using approximate natural logarithm values)
[tex]\[ n > 10.838 \][/tex]
Since [tex]\( n \)[/tex] must be an integer, the least number of terms for which the sum exceeds 240 is 11.
(i) Calculating the common ratio and the first term:
Given an exponential sequence, we can denote the first term by [tex]\( a \)[/tex] and the common ratio by [tex]\( r \)[/tex].
We are given two conditions:
1. The sum of the first two terms is 135.
[tex]\[ a + ar = 135 \][/tex]
2. The sum of the third and fourth terms is 60.
[tex]\[ ar^2 + ar^3 = 60 \][/tex]
From the first condition:
[tex]\[ a(1 + r) = 135 \][/tex]
From the second condition:
[tex]\[ ar^2(1 + r) = 60 \][/tex]
Dividing the second equation by the first equation to eliminate [tex]\( a \)[/tex]:
[tex]\[ \frac{ar^2(1 + r)}{a(1 + r)} = \frac{60}{135} \][/tex]
[tex]\[ r^2 = \frac{60}{135} \][/tex]
[tex]\[ r^2 = \frac{4}{9} \][/tex]
[tex]\[ r = \frac{2}{3} \][/tex] (since [tex]\( r \)[/tex] is positive)
Now, substitute [tex]\( r \)[/tex] back into the first equation to find [tex]\( a \)[/tex]:
[tex]\[ a(1 + \frac{2}{3}) = 135 \][/tex]
[tex]\[ a \cdot \frac{5}{3} = 135 \][/tex]
[tex]\[ a = 135 \times \frac{3}{5} \][/tex]
[tex]\[ a = 81 \][/tex]
So, the first term [tex]\( a \)[/tex] is 81, and the common ratio [tex]\( r \)[/tex] is [tex]\( \frac{2}{3} \)[/tex].
(ii) Calculating the limit of the sum of the first [tex]\( n \)[/tex] terms as [tex]\( n \)[/tex] becomes large:
In a geometric series, the sum of the first [tex]\( n \)[/tex] terms is given by:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]
As [tex]\( n \)[/tex] becomes very large, [tex]\( r^n \)[/tex] tends to 0 if [tex]\( |r| < 1 \)[/tex]. Given [tex]\( r = \frac{2}{3} \)[/tex], [tex]\( |r| < 1 \)[/tex], so:
[tex]\[ \lim_{n \to \infty} S_n = a \frac{1}{1 - r} \][/tex]
[tex]\[ \lim_{n \to \infty} S_n = 81 \frac{1}{1 - \frac{2}{3}} \][/tex]
[tex]\[ \lim_{n \to \infty} S_n = 81 \frac{1}{\frac{1}{3}} \][/tex]
[tex]\[ \lim_{n \to \infty} S_n = 81 \times 3 \][/tex]
[tex]\[ \lim_{n \to \infty} S_n = 243 \][/tex]
So, the limit of the sum of the first [tex]\( n \)[/tex] terms as [tex]\( n \)[/tex] becomes large is 243.
(iii) Finding the least number of terms for which the sum exceeds 240:
We want the smallest [tex]\( n \)[/tex] such that:
[tex]\[ a \frac{1 - r^n}{1 - r} > 240 \][/tex]
Using [tex]\( a = 81 \)[/tex] and [tex]\( r = \frac{2}{3} \)[/tex]:
[tex]\[ 81 \frac{1 - (\frac{2}{3})^n}{\frac{1}{3}} > 240 \][/tex]
[tex]\[ 81 \times 3 (1 - (\frac{2}{3})^n) > 240 \][/tex]
[tex]\[ 243(1 - (\frac{2}{3})^n) > 240 \][/tex]
[tex]\[ 1 - (\frac{2}{3})^n > \frac{240}{243} \][/tex]
[tex]\[ 1 - (\frac{2}{3})^n > \frac{80}{81} \][/tex]
[tex]\[ -(\frac{2}{3})^n > \frac{80}{81} - 1 \][/tex]
[tex]\[ -(\frac{2}{3})^n > \frac{-1}{81} \][/tex]
[tex]\[ (\frac{2}{3})^n < \frac{1}{81} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ \ln\left((\frac{2}{3})^n\right) < \ln\left(\frac{1}{81}\right) \][/tex]
[tex]\[ n \ln\left(\frac{2}{3}\right) < \ln\left(\frac{1}{81}\right) \][/tex]
Since [tex]\( \ln\left(\frac{2}{3}\right) \)[/tex] is negative, divide by it to flip the inequality:
[tex]\[ n > \frac{\ln\left(\frac{1}{81}\right)}{\ln\left(\frac{2}{3}\right)} \][/tex]
Calculating the value:
[tex]\[ n > \frac{\ln(1) - \ln(81)}{\ln(2) - \ln(3)} \][/tex]
[tex]\[ n > \frac{0 - \ln(81)}{\ln(2) - \ln(3)} \][/tex]
[tex]\[ n > \frac{-\ln(81)}{\ln(2) - \ln(3)} \][/tex]
[tex]\[ n > \frac{-4.394449}{-0.405465} \][/tex] (using approximate natural logarithm values)
[tex]\[ n > 10.838 \][/tex]
Since [tex]\( n \)[/tex] must be an integer, the least number of terms for which the sum exceeds 240 is 11.
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