Question

The sum of the first $n$ terms in a G.P. is $s_{n}=100-100^{1-n}$. The common ratio $r$ is

• A. $\frac{3}{100}$
• B. $\frac{1}{500}$
• C. $\frac{1}{20}$
• D. $\frac{1}{50}$
• E. $\frac{1}{100}$

Hint:

Math Tip: You can also solve this by factoring $S_n$ into the standard form $S_n = \frac{a(1-r^n)}{1-r}$.
$S_n = 100 - 100 \cdot (100)^{-n} = 100\left(1 - \left(\frac{1}{100}\right)^n\right)$. By directly comparing this to $1 - r^n$, we immediately see $r = \frac{1}{100}$.

Answer 1

Answer: (E)

Concept:
Sequences and Series - Relationship between Sum of $n$ terms ($S_n$) and the $n$-th term ($t_n$).
The $n$-th term can be found using the formula: $t_n = S_n - S_{n-1}$.
The common ratio $r$ is the ratio of consecutive terms: $r = \frac{t_2}{t_1}$.
Step 1: Find the first term ($t_1$).
The sum of the first 1 term ($S_1$) is equal to the first term ($t_1$).
Substitute $n = 1$ into the given formula for $S_n$: $$ t_1 = S_1 = 100 - 100^{1-1} $$ $$ t_1 = 100 - 100^0 $$ $$ t_1 = 100 - 1 = 99 $$
Step 2: Find the sum of the first two terms ($S_2$).
Substitute $n = 2$ into the formula: $$ S_2 = 100 - 100^{1-2} $$ $$ S_2 = 100 - 100^{-1} $$ $$ S_2 = 100 - \frac{1}{100} $$ $$ S_2 = 100 - 0.01 = 99.99 $$
Step 3: Calculate the second term ($t_2$).
Using the relation $t_2 = S_2 - S_1$: $$ t_2 = 99.99 - 99 $$ $$ t_2 = 0.99 $$
Step 4: Calculate the common ratio ($r$).
The common ratio is $r = \frac{t_2}{t_1}$: $$ r = \frac{0.99}{99} $$ $$ r = \frac{99 / 100}{99} $$ $$ r = \frac{1}{100} $$