Math, asked by kolakantichethan, 9 months ago

if 1/b+C,1/C+a,1/a+b are in A.P. then prove that a^2,b^2,c^2 are also in A.P​

Answers

Answered by VishnuPriya2801
12

Answer:-

Let a² , b² , c² are in AP.

Using Arithmetic mean we get,

\sf{ 2b^2 = a^2 + c^2\:-\: equation\:(1)}

Given:

\sf{ \frac{1}{b + c}  \: , \:  \frac{1}{c + a}  \: , \:  \frac{1}{a + b}  \:\: are \: in \: AP. }\\  \\  \sf{ Using \: Arithmetic \: mean \: again \: we \: get,} \\  \\ → \sf{ \frac{2}{c + a}  =  \frac{1}{b + c}  +  \frac{1}{a + b} } \\  \\  \sf{Taking \: LCM \: in \: RHS ,\: } \\  \\  →\sf{ \frac{2}{a + c}  =  \frac{a + b + b + c}{(b + c)(a + b)} } \\  \\→ \sf{ \frac{2}{a + c}  =  \frac{a + 2b + c}{ab + bc + ac +  {b}^{2} } }\\  \\  \sf{After \: cross \: multiplication \: we \: get,} \\  \\  →\sf{2(ab + bc + ac +  {b}^{2} ) = (c + a)(a + 2b + c)} \\  \\ → \sf{2ab + 2 {b}^{2}  + 2ac  + 2bc= c(a + 2b + c) + a(a + 2b + c)} \\  \\ → \sf{2ab + 2 {b}^{2}  + 2ac  + 2bc= ac + 2bc +  {c}^{2}  +  {a}^{2}  + 2ab + ac }\\  \\ → \sf{2ab - 2ab + 2 {b}^{2}  + 2bc - 2bc  + 2ac=  {c}^{2}  +  {a}^{2}  + 2ac} \\  \\  →\sf{2 {b}^{2}  + 2ac =  {c}^{2}  +  {a}^{2}  + 2ac} \\  \\  \sf{Substitute \:  "({a}^{2}  +  {c}^{2} )" \: value \: here.} \\  \\  →\sf{2 {b}^{2}  + 2ac =2  {b}^{2}  + 2ac}\\\\\sf{LHS = RHS.}

Hence,we can say that , , are in AP.


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