If (b+c), (c+a), (a+b) are in H.P.,then a^2, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $(b+c), (c+a), (a+b)$ are in $H.P.$Therefore,their reciprocals $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in $A.P.$This implies:

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(A) Given that $(b+c), (c+a), (a+b)$ are in $H.P.$Therefore,their reciprocals $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in $A.P.$This implies: $\frac{2}{c+a} = \frac{1}{b+c} + \frac{1}{a+b}$$\frac{2}{c+a} = \frac{a+b+b+c}{(b+c)(a+b)} = \frac{a+2b+c}{ab+b^2+ac+bc}$$2(ab+b^2+ac+bc) = (c+a)(a+2b+c)$$2ab+2b^2+2ac+2bc = ac+2bc+c^2+a^2+2ab+ac$$2ab+2b^2+2ac+2bc = a^2+c^2+2ac+2ab+2bc$Subtracting $(2ab+2ac+2bc)$ from both sides,we get:$2b^2 = a^2+c^2$Since $2b^2 = a^2+c^2$,it follows that $a^2, b^2, c^2$ are in $A.P.$

If $(b+c), (c+a), (a+b)$ are in $H.P.$,then $a^2, b^2, c^2$ are in.......

Similar Questions

If the sum of $n$ terms of an $A.P.$ is $2n^2 + 5n$,then the $n^{th}$ term will be

If the product of three consecutive terms of a $G.P.$ is $216$ and the sum of their products taken two at a time is $156$,then the numbers are:

If the sum of the roots of the equation $ax^2 + bx + c = 0$ is equal to the sum of the reciprocals of their squares,then $bc^2, ca^2, ab^2$ will be in

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots$. If $B - 2A = 100\lambda$,then $\lambda$ is equal to:

If $2x, x + 8, 3x + 1$ are in $A.P.$,then the value of $x$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module