1a1a\frac{1}{a} + 1b" role="presentation">1b1b\frac{1}{b} + 1c," role="presentation">1c,1c,\frac{1}{c}{\text{,}} then average of a2, b2, c2 is?" /> 1a1a\frac{1}{a} + 1b" role="presentation">1b1b\frac{1}{b} + 1c," role="presentation">1c,1c,\frac{1}{c}{\text{,}} then average of a2, b2, c2 is?" /> 1a1a\frac{1}{a} + 1b" role="presentation">1b1b\frac{1}{b} + 1c," role="presentation">1c,1c,\frac{1}{c}{\text{,}} then average of a2, b2, c2 is?" />

India’s No.1 Educational Platform For UPSC,PSC And All Competitive Exam

Download Our App

If a + b + c = m and $\frac{1}{a}$ $\frac{1}{a}$ + $\frac{1}{b}$ $\frac{1}{b}$ + $\frac{1}{c},$ $\frac{1}{c}{\text{,}}$ then average of a², b², c² is?

1 $m^{2}$ ${m^2}$
2 $\frac{m^{2}}{3}$ $\frac{{{m^2}}}{3}$
3 $\frac{m^{2}}{9}$ $\frac{{{m^2}}}{9}$
4 $\frac{m^{2}}{27}$ $\frac{{{m^2}}}{{27}}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} a + b + c = m \\ \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \\ \Rightarrow \frac{a b + b c + c a}{a b c} = 0 \\ \Rightarrow a b + b c + c a = 0 \\ {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \\ \Rightarrow m^{2} = a^{2} + b^{2} + c^{2} \\ \Rightarrow \frac{m^{2}}{3} = \frac{a^{2} + b^{2} + c^{2}}{3} \end{aligned}

$\eqalign{ & a + b + c = m \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \cr & \Rightarrow \frac{{ab + bc + ca}}{{abc}} = 0 \cr & \Rightarrow ab + bc + ca = 0 \cr & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {m^2} = {\text{ }}{a^2} + {b^2} + {c^2} \cr & \Rightarrow \frac{{{m^2}}}{3} = {\text{ }}\frac{{{a^2} + {b^2} + {c^2}}}{3} \cr}$

Workspace Report

Post your Comments

+

\frac{1}{b}

$\frac{1}{b}$ +

\frac{1}{c},

$\frac{1}{c}{\text{,}}$ then average of a², b², c² is?", "text": "If a + b + c = m and

\frac{1}{a}

$\frac{1}{a}$ +

\frac{1}{b}

$\frac{1}{b}$ +

\frac{1}{c},

$\frac{1}{c}{\text{,}}$ then average of a², b², c² is?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a + b + c = m \\ \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \\ \Rightarrow \frac{a b + b c + c a}{a b c} = 0 \\ \Rightarrow a b + b c + c a = 0 \\ {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \\ \Rightarrow m^{2} = a^{2} + b^{2} + c^{2} \\ \Rightarrow \frac{m^{2}}{3} = \frac{a^{2} + b^{2} + c^{2}}{3} \end{aligned}

$\eqalign{ & a + b + c = m \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \cr & \Rightarrow \frac{{ab + bc + ca}}{{abc}} = 0 \cr & \Rightarrow ab + bc + ca = 0 \cr & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {m^2} = {\text{ }}{a^2} + {b^2} + {c^2} \cr & \Rightarrow \frac{{{m^2}}}{3} = {\text{ }}\frac{{{a^2} + {b^2} + {c^2}}}{3} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a + b + c = m \\ \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \\ \Rightarrow \frac{a b + b c + c a}{a b c} = 0 \\ \Rightarrow a b + b c + c a = 0 \\ {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \\ \Rightarrow m^{2} = a^{2} + b^{2} + c^{2} \\ \Rightarrow \frac{m^{2}}{3} = \frac{a^{2} + b^{2} + c^{2}}{3} \end{aligned}

$\eqalign{ & a + b + c = m \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \cr & \Rightarrow \frac{{ab + bc + ca}}{{abc}} = 0 \cr & \Rightarrow ab + bc + ca = 0 \cr & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {m^2} = {\text{ }}{a^2} + {b^2} + {c^2} \cr & \Rightarrow \frac{{{m^2}}}{3} = {\text{ }}\frac{{{a^2} + {b^2} + {c^2}}}{3} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

Test

Classes

E-Book