Question
Download Solution PDFयदि (b - c)2, (c - a)2, (a - b)2 A.P.में हैं,
तब \(\dfrac{1}{b-c},\dfrac{1}{c-a},\dfrac{1}{a-b}\) ________ में होगा।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
A.P में सामान्य अंतर समान है। अर्थात यदि a, b, c AP में हैं, तो
b - a = c - b
गणना:
दिया गया:
(b - c)2, (c - a)2, (a - b)2 are in A.P
⇒ (c - a)2 - (b - c)2 = (a - b)2 - (c - a)2
(c - a + b - c)(c - a - b + c) = (a - b + c - a)(a - b - c + a)
(b - a)(2c - a - b) = (c - b)(2a - b - c) ...(i)
अब, \(\dfrac{1}{b-c},\dfrac{1}{c-a},\dfrac{1}{a-b}\)
\(\dfrac{1}{c-a} - \dfrac{1}{b-c}=\dfrac{1}{a-b} - \dfrac{1}{c-a}\)
\(\dfrac{b-c-c+a}{(c-a)(b-c)} =\dfrac{c-a-a+b}{(a-b)(c-a)}\)
\(\dfrac{b+a-2c}{b-c} =\dfrac{c+b-2a}{a-b}\)
(b + a - 2c)(a - b) = (c + b - 2a)(b - c)
(b - a)(2c - a - b) = (c - b)(2a - b - c) ...(ii)
चूंकि समीकरण (i) और (ii) बराबर हैं,
इसलिए \(\dfrac{1}{b-c},\dfrac{1}{c-a},\dfrac{1}{a-b}\) AP में है
Last updated on May 6, 2025
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