UPSC NDA Math Model Paper 4

Let, 1/ab, 1/ca and 1/bc are in AP
⇒

1

ca

−

1

ab

=

1

bc

−

1

ca

1

a

(

1

a

−

1

a

)=

1

a

(

1

a

−

1

a

)
⇒

b−c

abc

=

a−b

abc

⇒b−c=a−b
⇒2b=a+c
⇒ a,b, c are in AP, which is true.
Now, let 1/(b+c) ,1/(c+a) and 1/(a+b) are in AP

2

c+a

=

1

b+c

+

1

a+b

⇒2(a+b)(b+c)=(a+b+b+c) (c+a)
⇒2(ab+ac+b2+bc) =(ac+a2+2bc+2ab+c2+ca)
⇒2ab+2ac+2b2+2bc =ac+a2+2bc+2ab+ac+c2
⇒2b2=a2+c2
a,b,c are not in AP, so this is not true.
Hence, option (1) is correct.