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If 2(y-a) is HM between y-x and y-z then x-a, y-a, I-a are inApHpGpNone

Anvita Mahajan , 5 Years ago
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anser 1 Answers
Samyak Jain

Last Activity: 5 Years ago

2(y – a) is HM between y – x and y – z.  \Rightarrow  y – x ,  2(y – a) ,  y – z are in HP.
i.e. 1 / (y – x) , 1 / 2(y – a) , 1 / (y – z) are in AP.
\therefore 2 . {1 / 2(y – a)}  =  1 / (y – x)  +  1 / (y – z)
1 / (y – a)  =  1 / (y – x)  +  1 / (y – z)  \Rightarrow  1 / (y – a)  –  1 / (y – x)  =  1 / (y – z)
(y – x – y + a) / (y – a)(y – x)  =  1 / (y – z)
 – (x – a) / (y – a)(y – x)  =  1 / (y – z)    \therefore  (x – a) / (y – x)  =  (y – a) / (z – y)
(x – a) / [(y – a) – (x – a)]  =  (y – a) / [(z – a) – (y – a)]
By invertendo,  [(y – a) – (x – a)] / (x – a)  =  [(z – a) – (y – a)] / (y – a)
i.e. {(y – a) / (x – a)} – 1  =  {(z – a) / (y – a)} – 1  \Rightarrow  (y – a) / (x – a)  =  (z – a) / (y – a)
\therefore (y – a)2  =  (x – a)(z – a)  i.e.  (x – a) , (y – a) , (z – a) are in GP.

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