If a 2 , b 2 , c 2 are in A.P, show that a b + c , b c + a , c a + b are in A.P.

Given: a 2 , b 2 , c 2 are in A.P. ⇒ b 2 − a 2 = c 2 − b 2 ⇒ ( b − a ) ( b + a ) = ( c − b ) ( c + b ) ⇒ ( b − a ) ( c + b ) = ( c − b ) ( b + a ) ⇒ ( b − a ) ( c + b ) ( c + a ) = ( c − b ) ( b + a ) ( c + a ) [Using the property if each term of a given A.P is divided by non-zero term then resulting sequence is also an A.P.] ⇒ ( b + c − c − a ) ( c + b ) ( c + a ) = ( c + a − a − b ) ( b + a ) ( c + a ) ⇒ 1 c + a − 1 b + c = 1 a + b − 1 c + a ⇒ 1 b + c , 1 c + a , 1 a + b are in A.P. ⇒ a + b + c b + c , a + b + c c + a , a + b + c a + b are in A.P. [Using the property if each term of a given A.P is multiplied by non-zero term then resulting sequence is also an A.P.] ⇒ a b + c + 1 , b c + a + 1 , c a + b + 1 are in A.P. ⇒ a b + c , b c + a , c a + b are in A.P. [Using the property if constant is subtracted from each term of a given A.P then the resulting sequence is also an A.P.] Hence proved.

If b + c a , c + a b , a + b c are in A.P, prove that 1 a , 1 b , 1 c are in A.P.

If 1 a , 1 b , 1 c are in A.P ⇒ 1 b − 1 a = 1 c − 1 b L . H . S = 1 b − 1 a = a − b a b = c ( a − b ) a b c . . . i R . H . S = 1 c − 1 b = a ( b − c ) a b c . . . i i Since, b + c a , c + a b , a + b c are in A.P. ⇒ b + c a − c + a b = c + a b − a + b c ⇒ b 2 + c b − a c − a 2 a b = c 2 + a c − a b − b 2 b c ⇒ ( b − a ) ( a + b + c ) a b = ( c − b ) ( a + b + c ) b c ⇒ a ( b − c ) a b c = c ( a − b ) a b c . . . i i i From i , i i , i i i L . H . S = R . H . S Hence, 1 a , 1 b , 1 c are in A.P.

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

If b c , c a , a b are in A.P Then, c a − b c = a b − c a ⇒ c ( a − b ) = a ( b − c ) . . . 1 If 1 a , 1 b , 1 c are in A.P ⇒ 1 b − 1 a = 1 c − 1 b ⇒ c ( a − b ) = a ( b − c ) . . . 2 Thus, the necessary condition to prove b c , c a , a b in A.P is fulfilled. Thus, b c , c a , a b are in A.P.

If 1 a , 1 b , 1 c are in A.P, prove that b + c a , c + a b , a + b c are in A.P.

We know that b + c a , c + a b , a + b c will be in A.P if c + a b − b + c a = a + b c − c + a b ⇒ c a + a 2 − b 2 − c b a b = a b + b 2 − c 2 − a c b c Consider, L . H . S : c a + a 2 − b 2 − c b a b = c 2 a + a 2 c − b 2 c − c 2 b a b c = c ( a − b ) [ a + b + c ] a b c . . . 1 Consider, R . H . S : a b + b 2 − c 2 − a c b c = a 2 b + a b 2 − a c 2 − a 2 c a b c = a ( b − c ) [ a + b + c ] a b c . . . 2 Also, 1 a , 1 b , 1 c are in A.P ⇒ 1 b − 1 a = 1 c − 1 b ⇒ c ( b − a ) = a ( b − c ) . . . 3 ∴ L . H . S = R . H . S Hence, the given terms are in A.P.

If 1 a , 1 b , 1 c are in A.P, prove that a ( b + c ) , b ( c + a ) , c ( a + b ) are in A.P.

We know that a ( b + c ) , b ( c + a ) , c ( a + b ) will be in A.P if b ( c + a ) − a ( b + c ) = c ( a + b ) − b ( c + a ) Consider, L . H . S = b ( c + a ) − a ( b + c ) = b c + a b − a b − a c = c ( b − a ) . . . 1 Consider, R . H . S = c ( a + b ) − b ( c + a ) = c a + c b − b c − b a = a ( c − b ) . . . 2 and 1 a , 1 b , 1 c are in A.P ∴ 1 a − 1 b = 1 b − 1 c ⇒ c ( b − a ) = a ( c − b ) . . . 3 From 1 , 2 , 3 a ( b + c ) , b ( c + a ) , c ( a + b ) are in A.P.

E M B I B E

WBCHSE Mathematics for Class 11>Algebra>Chapter 9 - Sequence and Series>EXERCISE 7E

R S Aggarwal Solutions for Exercise 5: EXERCISE 7E

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 5: EXERCISE 7E

Attempt the practice questions from Exercise 5: EXERCISE 7E with hints and solutions to strengthen your understanding. WBCHSE Mathematics for Class 11 solutions are prepared by Experienced Embibe Experts.