Question

Maths question from arithmetic progression chapter.​

Answer 1

QUESTION:

$\sf If \ \dfrac{1}{a+b}, \ \dfrac{1}{b+c},\ \dfrac{1}{c+a}\ are \ in \ A.P \ then :$

ANSWER:

$\sf{Option\ d)\ b^2, \ a^2, \ c^2 \ are\ in\ A.P}$

GIVEN:

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

TO FIND:

• The correct option.

EXPLANATION:

$\tt {Common \ difference \ (d)= t_2-t_1=t_3-t_1}$

$\tt\dfrac{1}{b+c} - \dfrac{1}{a+b} = \dfrac{1}{c+a} - \dfrac{1}{b+c}$

$\tt{Let \ L.H.S = \dfrac{1}{b+c} - \dfrac{1}{a+b}}$

$\tt Let \ R.H.S = \dfrac{1}{c+a} - \dfrac{1}{b+c}$

L.H.S

Multiply by (a + b)(c + a) on both numerator and denominator for 1 / (b + c). Similarly, Multiply by (b + c)(c + a) on both numerator and denominator for 1 / (a + b).

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

$\tt{\dfrac{(ac + {a}^{2} + bc + ab) - (bc + ab + {c}^{2} + ac)}{(b+c)( a+b )( c+a )}}$

$\tt{\dfrac{{a}^{2 }- {c}^{2}}{(b+c)( a+b )( c+a )}}$

R.H.S

Multiply by (a + b)(c + a) on both numerator and denominator for 1 / (b + c). Similarly, Multiply by (a + b)(b + c) on both numerator and denominator for 1 / (c + a).

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

$\tt{\dfrac{(ab + ac + {b}^{2}+ bc) - (ac + {a}^{2} + bc + ab)}{c + a(a +b ) (b+c ) }}$

$\tt{\dfrac{ab + ac + {b}^{2}+ bc - ac - {a}^{2} - bc - ab}{c + a(a +b ) (b+c ) }}$

$\tt{\dfrac{{b}^{2} - {a}^{2} }{c + a(a +b ) (b+c ) }}$

EQUATE L.H.S AND R.H.S

$\tt{\dfrac{{a}^{2 }- {c}^{2}}{(b+c)( a+b )( c+a )}} = \tt{\dfrac{{b}^{2} - {a}^{2} }{c + a(a +b ) (b+c ) }}$

$\tt{{a}^{2 }- {c}^{2} = {b}^{2} - {a}^{2} }$

$\tt{2{a}^{2 }= {b}^{2} + {c}^{2}}$

$\tt{{a}^{2 }- {c}^{2} = {b}^{2} - {a}^{2} }$

$\tt{This\ is\ in \ the\ form\ t_2 - t_1=t_3-t_2}$

$\boldsymbol{t_2=a^2, \ \ t_1=c^2, \ \ t_3= b^2}$

$\underline{\boxed{\bold{Hence\ b^2, \ a^2, \ c^2 \ are\ in\ A.P}}}$

VERIFICATION:

$\sf{When \ t_1=b^2, \ \ t_2=a^2, \ \ t_3= c^2 }$

$\sf{d= t_2-t_1=t_3-t_1}$

$\sf{a^2 - b^2 =c^2 - a^2 }$

$\sf{2{a}^{2 }= {b}^{2} + {c}^{2}}$

HENCE VERIFIED.