Question

a+ b+ c not equal to 0 a/b+c,b/c+a,c/ a+ b in AP 1/b+c,1/c+a,1/a+b.​

Answer 1

Appropriate Question :-

$\rm \: If \: a + b + c \ne \: 0 \: and \: \dfrac{a}{b + c}, \dfrac{b}{c + a}, \dfrac{c}{a + b} \: are \: in \: AP$

$\rm \: prove \: that \: \dfrac{1}{b + c}, \dfrac{1}{c + a}, \dfrac{1}{a + b} \: are \: in \: AP$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: \dfrac{a}{b + c}, \dfrac{b}{c + a}, \dfrac{c}{a + b} \: are \: in \: AP$

We know,

If each term of an AP series is increased by some real constant k, the resultant series is also an AP series.

So, using this,

Adding 1 in each term, we get

$\rm\implies \: \dfrac{a}{b + c} + 1, \dfrac{b}{c + a} + 1, \dfrac{c}{a + b} + 1 \: are \: in \: AP$

can be further rewritten as

$\rm\implies \: \dfrac{a + b + c}{b + c}, \dfrac{b + c + a}{c + a}, \dfrac{c + a + b}{a + b} \: are \: in \: AP$

We know, if each term of an AP is divided by non - zero real constant k, the resultant series is also an AP series.

So, using this,

Divide each term by a + b + c, we get

$\rm\implies \: \: \dfrac{1}{b + c}, \dfrac{1}{c + a}, \dfrac{1}{a + b} \: are \: in \: AP$

$\rule{190pt}{2pt}$

Additional Information :-

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.