As a, b, c, are in A.P. So, b – a = c – b (1) ka, kb, kc are in A.P ⇒kb – ka = kc – kb ⇒k(b – a) = k(c – b) ⇒ b – a = c – b (2) k – a, k – b, k – c are in A.P ⇒k – b – (k – a) = k – c – (k – b) ⇒ k – b – k + a = k – c – k + b ⇒ a – b = b – c –(b – a) = –(c – b) ⇒ b – a = c – b (3)
a
k
,
b
k
,
c
k
are in AP ⇒
b
k
−
a
k
=
c
k
−
b
k
⇒
1
k
(b−a)=
1
k
(c−b) ⇒b – a = c – b Hence, (1), (2) and (3) are in A.P.