If (nPr-1/a) = (nPr/b) = (nPr+1/c), then

Question

If (nPr-1/a) = (nPr/b) = (nPr+1/c), then (A) b2 = a( b +c) (B) (1/a)+(1/b)+(1/c) = 1 (C) c2 = c( a +b) (D)  abc = 1

Tardigrade Admin · Accepted Answer

(1/a) (n!/(n - r + 1)!) = (1/b) (n!/(n-r)!) = (1/c) (n!/(n - r - 1)! ) ⇒ (b/a) = n - r + 1 ....(i) and (c/b) = n - r ...(ii) From (i) and (ii), we get (b/a) = (c/b) +1 ⇒ b2 = a(b+c).

Q. If $\frac{^{n} P _{r - 1}}{a} = \frac{^{n} P _{r}}{b} = \frac{^{n} P _{r + 1}}{c}$ , then

$b^{2} = a (b + c)$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$

$c^{2} = c (a + b)$

$ab c = 1$

Q. If anPr−1​​=bnPr​​=cnPr+1​​, then

b2=a(b+c)

a1​+b1​+c1​=1

c2=c(a+b)

abc=1

a1​(n−r+1)!n!​ =b1​(n−r)!n!​ =c1​(n−r−1)!n!​ ⇒ab​=n−r+1....(i) and bc​=n−r...(ii) From (i) and (ii), we get ab​=bc​+1 ⇒b2=a(b+c).

Q. If $\frac{^{n} P _{r - 1}}{a} = \frac{^{n} P _{r}}{b} = \frac{^{n} P _{r + 1}}{c}$ , then

$b^{2} = a (b + c)$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$

$c^{2} = c (a + b)$

$ab c = 1$