Question

The terms of a geometric progression are real and positive. If the \( p \)-th term of the progression is \( q \) and the \( q \)-th term is \( p \), then the logarithm of the first term is:

• A. \(\frac{(1-q)\log(p)-(1-p)\log(q)}{(p-1)}\)
• B. \(\frac{(1-q)\log(p)+(1-p)\log(q)}{(p-1)}\)
• C. \(\frac{(1-q)\log(q)+(1-p)\log(p)}{(p-1)}\)
• D. \(\frac{(1-q)\log(q)-(1-p)\log(p)}{(p-q)}\)

Hint:

For geometric progressions, use the properties of logarithms to simplify equations involving powers and terms. Logarithmic identities help in solving for unknown terms efficiently.

Answer 1

The Correct Option is D

1. Define the Geometric Progression:
Let the first term of the GP be \( a \) and the common ratio be \( r \). The \( n \)-th term of a GP is given by: \[ T_n = a \cdot r^{n-1} \] 2. Given Conditions:
The \( p \)-th term is \( q \):
\[ T_p = a \cdot r^{p-1} = q \] The \( q \)-th term is \( p \): \[ T_q = a \cdot r^{q-1} = p \] 3. Divide the Two Equations: Dividing the equation for \( T_q \) by the equation for \( T_p \): \[ \frac{T_q}{T_p} = \frac{a \cdot r^{q-1}}{a \cdot r^{p-1}} = \frac{p}{q} \] Simplifying: \[ r^{q-p} = \frac{p}{q} \] Taking the logarithm on both sides: \[ (q - p) \log r = \log \left( \frac{p}{q} \right) \] Solving for \( \log r \): \[ \log r = \frac{\log \left( \frac{p}{q} \right)}{q - p} \] 4. Express \( \log a \): From the equation for \( T_p \): \[ a = \frac{q}{r^{p-1}} \] Taking the logarithm: \[ \log a = \log q - (p - 1) \log r \] Substitute \( \log r \) from the previous step: \[ \log a = \log q - (p - 1) \cdot \frac{\log \left( \frac{p}{q} \right)}{q - p} \] Simplifying: \[ \log a = \log q - \frac{(p - 1)(\log p - \log q)}{q - p} \] \[ \log a = \log q + \frac{(p - 1)(\log q - \log p)}{q - p} \] \[ \log a = \log q + \frac{(p - 1)\log q - (p - 1)\log p}{q - p} \] \[ \log a = \log q \left(1 + \frac{p - 1}{q - p}\right) - \frac{(p - 1)\log p}{q - p} \] \[ \log a = \log q \left(\frac{q - p + p - 1}{q - p}\right) - \frac{(p - 1)\log p}{q - p} \] \[ \log a = \log q \left(\frac{q - 1}{q - p}\right) - \frac{(p - 1)\log p}{q - p} \] \[ \log a = \frac{(q - 1)\log q - (p - 1)\log p}{q - p} \] 5. Compare with Given Options: The expression for \( \log a \) matches option (d): \[ \log a = \frac{(1 - q)\log q - (1 - p)\log p}{p - q} \] Final Answer The logarithm of the first term is given by option (d). Answer: (d) \[ \frac{(1-q)\log(q)-(1-p)\log(p)}{(p-q)} \]