Question

A geometric series has common ratio \(\frac{1}{3}\). If the sum of first four terms is 200, then the first term is

• A. 90
• B. 100
• C. 115
• D. 145
• E. 135

Hint:

When the common ratio \(r < 1\), it is easier to use \(\frac{a(1-r^n)}{1-r}\). When \(r > 1\), \(\frac{a(r^n-1)}{r-1}\) is often used to avoid negative numbers in the intermediate steps.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A geometric series is defined by its first term \(a\) and common ratio \(r\). We are given the sum \(S_4\), the ratio \(r\), and the number of terms \(n\).

Step 2: Key Formula or Approach:
The sum of the first \(n\) terms of a G.P. is \(S_n = \frac{a(1 - r^n)}{1 - r}\) for \(r \neq 1\).

Step 3: Detailed Explanation:
Given: \(r = \frac{1}{3}\), \(n = 4\), \(S_4 = 200\).
Apply the formula:
\[ 200 = \frac{a(1 - (\frac{1}{3})^4)}{1 - \frac{1}{3}} \]
Simplify the bracket:
\[ 1 - (\frac{1}{3})^4 = 1 - \frac{1}{81} = \frac{80}{81} \]
Simplify the denominator:
\[ 1 - \frac{1}{3} = \frac{2}{3} \]
The equation becomes:
\[ 200 = \frac{a \cdot (80/81)}{2/3} \]
\[ 200 = a \cdot \frac{80}{81} \cdot \frac{3}{2} \]
\[ 200 = a \cdot \frac{40}{27} \]
Solve for \(a\):
\[ a = \frac{200 \cdot 27}{40} \]
\[ a = 5 \cdot 27 = 135 \]

Step 4: Final Answer:
The first term is 135.