Question 9
A geometric series $G$ has common ratio $r$ where $r \gt 0$
The third term of $G$ is $\frac{27}{2}$ and the sum of the first three terms of $G$ is $\frac{57}{2}$
Given that the sum to $n$ terms of $G$ is $S_n$
(a) show that $$ S_n = \sum_{j=1}^{n} 4\left(\frac{3}{2}\right)^j $$
(8)
Given that $S_k \gt 50000$
(b) show that the least value of $k$ is given by
$$ k \gt \frac{\lg \left(\frac{12503}{3}\right)}{\lg \left(\frac{3}{2}\right)} $$(3)
(c) Hence find the least value of $k$
(1)
Solution
A geometric series $G$ has a common ratio $r$, where $r > 0$. The third term of $G$ is $\frac{27}{2}$, and the sum of the first three terms is $\frac{57}{2}$.
(a) Show that $S_n = \sum_{j=1}^{n} 4 \left(\frac{3}{2}\right)^j$
Step 1: General formula for the $n$-th term
The general formula for the $n$-th term of a geometric sequence is:
$$ T_n = ar^{n-1}, $$
where $a$ is the first term and $r$ is the common ratio.
Step 2: Solve for $a$ and $r$
The third term is:
$$ T_3 = ar^2 = \frac{27}{2}. $$
The sum of the first three terms is:
$$ S_3 = a(1 + r + r^2) = \frac{57}{2}. $$
From $T_3$, we have:
$$ ar^2 = \frac{27}{2}. $$
From $S_3$, substitute $a(1 + r + r^2) = \frac{57}{2}$ and divide by $ar^2$:
$$ \frac{1 + r + r^2}{r^2} = \frac{\frac{57}{2}}{\frac{27}{2}} = \frac{57}{27} = \frac{19}{9}. $$
Simplify:
$$ 9 + 9r + 9r^2 = 19r^2. $$
Rearranging terms:
$$ 10r^2 - 9r - 9 = 0 \rightarrow (2r - 3)(5r + 3) = 0. $$
Solve for $r > 0$:
$$ r = \frac{3}{2}. $$
Substitute $r = \frac{3}{2}$ into $ar^2 = \frac{27}{2}$:
$$ a \cdot \frac{9}{4} = \frac{27}{2}. $$
Simplify:
$$ a = 6. $$
Step 3: General sum formula
The sum to $n$ terms is:
$$ S_n = \sum_{j=1}^{n} ar^{j-1}. $$
Substitute $a = 6$ and $r = \frac{3}{2}$:
$$ S_n = \sum_{j=1}^{n} 6 \left(\frac{3}{2}\right)^{j-1}. $$
Thus:
$$ S_n = \sum_{j=1}^{n} 4 \left(\frac{3}{2}\right)^j. $$
(b) Show that $k > \frac{\log\left(\frac{12,503}{3}\right)}{\log\left(\frac{3}{2}\right)}$
Step 1: Formula for the sum of a geometric series
For $r > 1$, the sum to $n$ terms is:
$$ S_n = \frac{a(r^n - 1)}{r - 1}. $$
Substitute $a = 6$ and $r = \frac{3}{2}$:
$$ S_n = \frac{6\left(\left(\frac{3}{2}\right)^n - 1\right)}{\frac{3}{2} - 1}. $$
Simplify:
$$ S_n = 12 \left(\left(\frac{3}{2}\right)^n - 1\right). $$
Step 2: Condition $S_k > 50,000$
Given $S_k > 50,000$, substitute $S_k$:
$$ 12 \left(\left(\frac{3}{2}\right)^k - 1\right) > 50,000. $$
Divide through by 12:
$$ \left(\frac{3}{2}\right)^k - 1 > \frac{50,000}{12}. $$
Simplify:
$$ \left(\frac{3}{2}\right)^k > \frac{12,503}{3}. $$
Take the logarithm of both sides:
$$ k \log\left(\frac{3}{2}\right) > \log\left(\frac{12,503}{3}\right). $$
Rearrange for $k$:
$$ k > \frac{\log\left(\frac{12,503}{3}\right)}{\log\left(\frac{3}{2}\right)}. $$
(c) Find the least value of $k$
Step 1: Compute $k$
Simplify $\frac{12,503}{3}$:
$$ \frac{12,503}{3} = 4,167.67. $$
Use approximate logarithm values:
$$ \log(4,167.67) \approx 3.62, \quad \log\left(\frac{3}{2}\right) \approx 0.176. $$
Substitute:
$$ k > \frac{3.62}{0.176} \approx 20.57. $$
The least integer $k$ is:
$$ k = 21. $$
Final Answer
- (a) $S_n = \sum_{j=1}^{n} 4 \left(\frac{3}{2}\right)^j$
- (b) $k > \frac{\log\left(\frac{12,503}{3}\right)}{\log\left(\frac{3}{2}\right)}$
- (c) The least value of $k$ is $21$
Post a Comment