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Since a, b, c are in GP. Therefore, b2 = ac . . . ( 1 )
Also, x is the AM between a and b ⇒ x = ( a + b ) / 2.
And, y is the AM between b and c ⇒ y = ( b + c ) / 2.
Now, ( i ) ( a / x ) + ( c / y )
= { 2a / ( a + b ) } + { 2c / ( b + c ) }
= 2 [ { a / ( a + b ) } + { c / ( b + c ) } ]
= 2 [ { a ( b + c ) + c ( a + b ) } / { ( a + b ) ( b + c ) } ]
= 2 [ ( ab + ac + ac + bc ) / ( ab + ac + b2+ bc ) ]
= 2 [ ( ab + 2ac + bc ) / ( ab + 2ac + bc ) ] ............. .............. .............. ............. ......... [ Using ( 1 ) ]
= 2 * 1
= 2
⇒ ( a / x ) + ( c / y ) = 2
And, ( ii ) ( 1 / x ) + ( 1 / y )
= { 2 / ( a + b ) } + { 2 / ( b + c ) }
= 2 [ { 1 / ( a + b ) } + { 1 / ( b + c ) } ]
= 2 [ ( b + c + a + b ) / ( a + b ) ( b + c ) ]
= 2 [ ( 2b + a + c ) / ( ab + ac + b2 + bc ) ]
= 2 [ ( 2b + a + c ) / ( ab + 2ac + bc ) ] ............ ............ ......... [ Using ( 1 ) ]
= 2 [ ( 2√ac + a + c ) / ( a√ac + 2ac + c√ac ) ]
= 2 [ ( 2√ac + a + c ) / { √ac ( a + √ac + c ) } ]
= 2 [ ( 2√ac + a + c ) / { √ac ( 2√ac + a + c ) } ]
= 2 ( 1 / √ac )
= 2 * ( 1 / b ) ................ ............. ........... ........ [ Using ( 1 ) ]
= 2 / b
Regards
Arun (askIITians forum expert)
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