Question

if three distinct numbers a,b,c, are in HP, and a^2,b^2,c^2 are in AP then,
a)a+c=b
b)a+b+c=0
c)a+b=c
d)b+c=a

Answer 1

$\underline{\underline{\large{\mathfrak{Solution : }}}}$



$\underline{\textsf{Given : \: a , b and c are in H.P.}}$



$\textsf{So , reciprocals of a , b and c must be in A.P.}$



$\underline{\textsf{Now : }}$



$\mathsf{ \implies \dfrac{1}{c} \: - \: \dfrac{1}{b} \: = \: \dfrac{1}{b} \: - \: \dfrac{1}{a} } \\ \\ \\ \mathsf{ \implies \dfrac{ \: b \: - \: c \: }{ \cancel{b}c} \: = \: \frac{ \: a \: - \: b \: }{a \: \cancel{b} }} \\ \\ \\ \mathsf{ \implies \dfrac{ \: b \: - \: c \: }{c} \: = \: \dfrac{ \: a \: - \: b \: }{a} \qquad...(1)}$



$\underline{\mathsf{Now \: a^{2} , b^{2} \: and \: c^{2} \: is \: in \: A.P.}}$



$\textsf{So,} \\ \\ \mathsf{\implies c^{2} \: - \: b^{2} \: = \: b^{2} \: - \: a^{2} \: } \\ \\ \\ \mathsf{\implies ( c \: + \: b )( c \: - \: b ) \: = \: ( b \: + \: a )( b \: - \: a )} \\ \\ \\ \mathsf{\implies \cancel{ -}(c \: + \: b )( b \: - \: c ) \: = \: \cancel{- }(a \: + \: b )( a \: - \: b )} \\ \\ \\ \mathsf{\implies (b \: - \: c ) \: = \: \dfrac{ ( a \: + \: b )( a \: - \: b)}{( b \: + \: c) }}$



$\textsf{Substitute the value of (b - c) in (1):}$



$\mathsf{ \implies \dfrac{ \dfrac{(a \: + b)(a \: - \: b)}{(b \: + \: c)}}{c} \: = \: \dfrac{a \: - \: b}{a} } \\ \\ \\ \mathsf{ \implies \frac{(a \: + \: b) \cancel{(a \: - \: b)}}{(b \: + \: c)c} \: = \: \dfrac{ \cancel{(a \: - \: b)}}{a} }$




$\mathsf{ \implies (a \: + \: b)a \: = \: (b \: + \: c)c} \\ \\ \\ \mathsf{ \implies {a}^{2} \: + \: ab \: = \: bc \: + \: {c}^{2} } \\ \\ \\ \mathsf{ \implies {a}^{2} \: - \: {c}^{2} \: = \: bc \: - \: ab } \\ \\ \\ \mathsf{ \implies(a\: + \: c) \cancel{(a \: - \: c)} \: = \: - b \cancel{(a \: - \: c)} } \\ \\ \\ \mathsf{ \therefore \quad a \: + \: b \: + \: c \: = \: 0}$



$\boxed{\large{\mathsf{The \: required \: answer \: is \: (b).}}}$