If  $$a(\dfrac{1}{b}+\dfrac{1}{c}), b(\dfrac{1}{c}+\dfrac{1}{a}), c(\dfrac{1}{a}+\dfrac{1}{b}),$$are in A.P., prove that a, b, c are in A.P.

Asked by Abhisek | 1 year ago |  74

##### Solution :-

Given $$a(\dfrac{1}{b}+\dfrac{1}{c}), b(\dfrac{1}{c}+\dfrac{1}{a}), c(\dfrac{1}{a}+\dfrac{1}{b}),$$ are in A.P.

Therefore,

$$b(\dfrac{1}{c}+\dfrac{1}{a}), -a(\dfrac{1}{b}+\dfrac{1}{c})$$

$$c(\dfrac{1}{a}+\dfrac{1}{b})-b(\dfrac{1}{c}+\dfrac{1}{a})$$

$$\dfrac{b(a+c)}{ac}-\dfrac{a(b+c)}{bc}$$

$$\dfrac{c(a+b)}{ab}-\dfrac{b(a+c)}{ac}$$

$$\dfrac{b^2a+b^2c-a^2b-a^2c}{abc}$$

$$\dfrac{c^2a+c^2b-b^2a-b^2c}{abc}$$

$$b^2a+b^2c-a^2b-a^2c$$

$$c^2a+c^2b-b^2a-b^2c$$

$$ab(b-a)+c(b^2-a^2)$$

$$=a(c^2-b^2)bc(c-b)$$

$$ab(b-a)+c(b-a)(b+a)$$

$$= a(c-b)(c+b)+bc(c-b)$$

$$(b-a)(ab+cb+ca)$$

$$=(c-b)(ac+ab+bc)$$

$$b-a=c-b$$

Therefore, a, b and c are in A.P.

Answered by Pragya Singh | 1 year ago

### Related Questions

#### Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

#### Find the two numbers whose A.M. is 25 and GM is 20.

Find the two numbers whose A.M. is 25 and GM is 20.

#### If a is the G.M. of 2 and 1/4 find a.

If a is the G.M. of 2 and $$\dfrac{1}{4}$$ find a.

#### Find the geometric means of the following pairs of numbers

Find the geometric means of the following pairs of numbers:

(i) 2 and 8

(ii) a3b and ab3

(iii) –8 and –2

Insert 5 geometric means between $$\dfrac{32}{9}$$ and $$\dfrac{81}{2}$$.