Question

if 1/a,1/b,1/c,1/d is in a.p then to show b=2ac/a+c and b/d=3a-c/a+c

Answer 1

1/a , 1/b , 1/c , 1/d are in AP
∴ 1/b - 1/a = 1/c - 1/b = 1/d - 1/c = common difference
⇒1/b + 1/b = 1/c + 1/a -------(1)
⇒2/b = (c + a)/ca
⇒b = 2ca/(c + a) , hence proved--------(2)

again, LHS = b/d
RHS = (3a - c)/(a + c)
= 2ca/(a + c) × (3a - c)/2ca
= b × (3/2c - 1/2a) [ from equation (2) ]
= b/2 × ( 3 × 1/c - 1/a )
= b/2 × (3/c - 2/b + 1/c) [ from equation (2) ]
= b/2 × ( 4/c - 2/b)
= b × (1/c + 1/c - 1/b )
= b × (1/c + 1/d - 1/c)
= b × 1/d
= b/d = LHS hence proved //