If a, b , c are in G.P, and x, y be the arithmetic means of a, b and b, c respectively, prove that (i) a/x+c/y=2 (ii)1/x + 1/y= 2/b

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

______

Hope this helps.

Cheers!!!

  • 65

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

⇒ ( 1 / x ) + ( 1 / y ) = 2 / b

______

Hope this helps.

Cheers!!!

  • -2

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 + 2√ac + c ) } ]

= 2 [ ( 2√ac + a + c ) / { √ac ( 2√ac + a + c ) } ]

= 2 ( 1 / √ac )

= 2 * ( 1 / b ) ................ ............. ........... ........ [ Using ( 1 ) ]

= 2 / b

______

Hope this helps.

Cheers!!!

  • 12

This one is correct finally,,, huh!! ;)

____

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 + 2√ac + c ) } ]

= 2 [ ( 2√ac + a + c ) / { √ac ( 2√ac + a + c ) } ]

= 2 ( 1 / √ac )

= 2 * ( 1 / b ) ................ ............. ........... ........ [ Using ( 1 ) ]

= 2 / b

______

Hope this helps.

Cheers!!!

  • -5
I could find that the first sat questions answer is wrong in the last second step of the sub question 1 you guys hav written ab +2ac+bc.instead of ab+ac+b^2+bc.in the denominator for getting the correct answer easily...so u csn easily cut off both the nr.and the dr....
  • -13
Please giv me a reply for this..
  • -10
Finally i tried it....huRray...

  • 26
Hi

  • 1
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