InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If a, b, c are in A.P., then show that : (i) a2(b + c), b2(c + a), c2(a + b) are also in A.P. (ii) b + c - a, c + a - b, a + b - c are in A.P. (iii) bc – a2, ca – b2, ab – c2 are in A.P. |
|
Answer» (i) a2(b + c), b2(c + a), c2(a + b) are also in A.P. b2(c + a) - a2(b + c) = c2(a + b) - b2(c + a) b2c + b2a – a2b – a2c = c2a + c2b – b2a – b2c c(b2 - a2) + ab(b - a) = a(c2 - b2) + bc(c - b) (b - a)(ab + bc + ca) = (c - b)(ab + bc + ca) b - a = c - b And since a, b, c are in AP, b - a = c - b Hence given terms are in AP. (ii) b + c - a, c + a - b, a + b - c are in A.P. Therefore, (c + a - b) - (b + c - a) = (a + b - c) - (c + a - b) 2a - 2b = 2b - 2c b - a = c - b And since a, b, c are in AP, b - a = c - b Hence given terms are in AP. (iii) bc – a2, ca – b2, ab – c2 are in A.P. (ca - b2) – (bc - a2) = (ab - c2) – (ca - b2) (a - b)(a + b + c) = (b - c)(a + b + c) a - b = b - c b - a = c - b And since a, b, c are in AP, b - a = c - b Hence given terms are in AP |
|