Math, asked by anishservice12p825x0, 1 year ago

when b+c-2a/a . c+a-2b/b , a+b-2c/c are in AP then prove that 1/a , 1/b ,1/c are in = ?
>GP
>HP
>AP
>NONE OF THIS .

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Answered by abhi178
19
(b + c - 2a)/a , (c + a - 2b)/b , (a + b- 2c)/c are in AP.

(b + c)/a - 2 , (c + a)/b -2 , (a + b)/c - 2 are in AP.

(b + c)/a , (c + a)/b , (a + b)/c are in AP.

(b + c)/a + 1 , (c + a)/b + 1 , (a + b)/c + 1 are in AP.

(b + c + a)/a, (c + a + b)/b , (a + b + c)/c are in AP.

1/a , 1/b , 1/c are in AP.
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