Question

The first three terms in a G.P. are $a, b$ and $c$ where $a \neq b$. Then the fifth term is:

• A. $\frac{c^{2}}{2a}$
• B. $\frac{c}{2a}$
• C. $\frac{c^{2}}{a}$
• D. $\frac{c^{2}}{3a}$
• E. $\frac{c}{3a}$

Hint:

For any three consecutive terms $x, y, z$ in G.P., $y^2 = xz$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given the first three terms of a Geometric Progression (G.P.) as a, b, and c. We need to find the fifth term of this G.P. in terms of a, b, and c.
Step 2: Key Formula or Approach:
In a G.P., the ratio of any two consecutive terms is constant. This is the common ratio, \( r \). So, \( r = \frac{b}{a} = \frac{c}{b} \). The n-th term of a G.P. is given by \( T_n = a \cdot r^{(n-1)} \). We need to find \( T_5 \).
Step 3: Detailed Explanation:
The first three terms are:
\( T_1 = a \)
\( T_2 = b \)
\( T_3 = c \)
From the property of G.P., the common ratio is \( r = \frac{T_2}{T_1} = \frac{b}{a} \).
We also know that \( T_3 = a \cdot r^2 \). So, \( c = a \cdot r^2 \).
From this, we can express \( r^2 \) in terms of a and c:
\[ r^2 = \frac{c}{a} \] We need to find the fifth term, \( T_5 \). Using the formula for the n-th term:
\[ T_5 = a \cdot r^{(5-1)} = a \cdot r^4 \] We can write \( r^4 \) as \( (r^2)^2 \).
\[ T_5 = a \cdot (r^2)^2 \] Now, substitute the expression for \( r^2 \) that we found:
\[ T_5 = a \cdot \left(\frac{c}{a}\right)^2 \] \[ T_5 = a \cdot \frac{c^2}{a^2} \] \[ T_5 = \frac{c^2}{a} \] Step 4: Final Answer:
The fifth term of the G.P. is \( \frac{c^2}{a} \).