33. The sum of the first two terms of an exponential sequence is 135, and the sum of the third and the fourth terms is 60. Given that the common ratio is positive, calculate:

(i) The common ratio and the first term.

(ii) The limit of the sum of the first [tex]\(n\)[/tex] terms as [tex]\(n\)[/tex] becomes large.

(iii) The least number of terms for which the sum exceeds 240.



Answer :

Sure, let's solve this step-by-step!

(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.