Question:
If \(\frac{{}^nP_{r-1}}{a} = \frac{{}^nP_r}{b} = \frac{{}^nP_{r+1}}{c}\), then
If \(\frac{{}^nP_{r-1}}{a} = \frac{{}^nP_r}{b} = \frac{{}^nP_{r+1}}{c}\), then
Show Hint
Use recursive relation: ${}^nP_r = {}^nP_{r-1}(n-r+1)$.
Updated On: Apr 30, 2026
- $c^2 = b(a + c)$
- $b^2 = a(a + b)$
- $b^2 = a(b + c)$
- $a^2 = b(a + c)$
- $a^2 = c(a + b)$
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The Correct Option is C
Solution and Explanation
Concept:
Use permutation relation:
\[
{}^nP_r = \frac{n!}{(n-r)!}
\]
Step 1: Take common value = $k$
\[ \frac{{}^nP_{r-1}}{a} = \frac{{}^nP_r}{b} = \frac{{}^nP_{r+1}}{c} = k \]
Step 2: Write in terms of $k$
\[ {}^nP_{r-1} = ak,\quad {}^nP_r = bk,\quad {}^nP_{r+1} = ck \]
Step 3: Use relation
\[ {}^nP_r = {}^nP_{r-1}(n-r+1) \] So: \[ bk = ak(n-r+1) \] Similarly: \[ ck = bk(n-r) \]
Step 4: Eliminate
From relations: \[ \frac{b}{a} = (n-r+1), \quad \frac{c}{b} = (n-r) \] \[ \Rightarrow b^2 = a(b+c) \] Final Conclusion:
Option (C)
Step 1: Take common value = $k$
\[ \frac{{}^nP_{r-1}}{a} = \frac{{}^nP_r}{b} = \frac{{}^nP_{r+1}}{c} = k \]
Step 2: Write in terms of $k$
\[ {}^nP_{r-1} = ak,\quad {}^nP_r = bk,\quad {}^nP_{r+1} = ck \]
Step 3: Use relation
\[ {}^nP_r = {}^nP_{r-1}(n-r+1) \] So: \[ bk = ak(n-r+1) \] Similarly: \[ ck = bk(n-r) \]
Step 4: Eliminate
From relations: \[ \frac{b}{a} = (n-r+1), \quad \frac{c}{b} = (n-r) \] \[ \Rightarrow b^2 = a(b+c) \] Final Conclusion:
Option (C)
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