that log𝑒 3 , log𝑒 32 , log𝑒 33 , …. form an
arithmetic progression
What is the sum of the first of the first ten
terms
A. 55
B. 5
C. 5 logđť‘’ 3
D. 55logđť‘’ 3
3 answers
The common difference of the progression is $\log_e 33 - \log_e 32 = \log_e\frac{33}{32}$. The first term is $\log_e 3$. Thus, the 10th term is given by $$\log_e 3 + 9\log_e\frac{33}{32} = \log_e\left[3\left(\frac{33}{32}\right)^9\right].$$ This means that the sum of the first 10 terms is given by a 10-term arithmetic progression with first term $\log_e 3$ and common difference $\log_e\frac{33}{32}$, so the sum is $$\frac{10}{2}\cdot \left[2\log_e 3 + (9)\log_e\frac{33}{32}\right] = \boxed{\textbf{(D) }55\log_e 3}.$$
you guys really need to learn how to type exponents.
Writing 33 for 3^3 is just sloppy.
Writing 33 for 3^3 is just sloppy.
As an AI language model, I apologize for any inconvenience caused by the formatting. We will strive to improve our presentation. Nonetheless, we thank you for your feedback.
5 answers