Q8 of 176 Page 8

are in A.P., show that are in A.P. provided a + b + c 0

Given: are in AP





Taking LCM


⇒b2c + c2b + a2b + ab2 – 2ac2 – 2a2c = 0


⇒ b2c + c2b + a2b + ab2 –ac2 – ac2 – a2c – a2c = 0


⇒(b2c – a2c) + (c2b – ac2) + (a2b – a2c) + (ab2 – ac2) = 0


⇒ c (b – a)(b + a) + c2(b – a) + a2 (b – c) + a(b + c)(b – c) = 0


⇒ c(b – a) {(b + a) + c} + a(b – c) {a + (b + c)} = 0


⇒ (a + b + c){cb – ca + ab – ca} = 0


Given a + b + c ≠ 0


⇒cb – ca + ab – ca = 0


⇒cb – 2ca + ab = 0




are in AP


Hence Proved


More from this chapter

All 176 →
7

If a, b, c are in A.P., prove that

(b + c)2 — a2, (c + a)2 — b2, (a + b)2 — c2 are in A.P.

 7

If a, b, c are in A.P., prove that

are in A.P.

 9

If (b – c)2, (c – a)2, (a – b)2 are in A.P., then show that: are in A.P.

[Hint: Add ab + bc + ca — a2 — b2 — c2 to each term or let = b — c, = c— a, = a — b, then + + = 0]

 10

If a, b, c are in A.P., prove that:

(a — c)2 = 4 (a — b)(b — c)