NDA Math Sequences and Series

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.