Question

50 points question plz solve it
of answer is write I will mark as brainlist​

Answer 1

Note : x, y, z are in AP iff : x + z = 2y .......... (1)

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∵ 1/a, 1/b, 1/c are in AP

∴ (1/a) + (1/c) = 2(1/b)

∴ (c+a) / (ca) = 2/b

∴ b(c+a) = 2(ca)

∴ (bc+ab) = 2(ca) ................... (2)

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Then, we have :

... a(b+c) + c(a+b)

= ab + ca + ca + bc

= (bc+ab) + 2(ca)

= (bc+ab) + (bc+ab) ............ from (2)

= 2(bc+ab)

= 2 · b(c+a)

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∴ a(b+c) + c(a+b) = 2 · b(c+a)

∴ from (1),

... a(b+c), b(c+a), c(a+b) are in AP. .................... Q.E.D.

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Happy To Help !

Answer 2

$\huge\mathfrak\pink{Answer}$

$note \: : \: x , y \: , z \: are \: in \: ap \: iff \: :x + z \: = \: 2y \: ------(1)$

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$⛬ \: \frac{1}{a} , \frac{1}{b} , \frac{1}{c \:\: } are \: in \: ap \:$

$⛬ (\frac{1}{a} ) \: + ( \frac{1}{c}) \: = \: 2( \frac{l}{b} )$

$⛬ \: ( \frac{c + a}{ca} ) \: = \frac{2}{b}$

$⛬ \: b(c + a) \: = 2(ca)$

$⛬ \: (bc \: + ab) \: = 2(ca) \: ---------(2)$

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$then \: we \: have \: :$

$a(b + c) \: + c(a + b)$

$= \: ab \: + ca \: + ca \: + bc \:$

$= \: (bc + ab) \: + (bc \: + ab) \: ----------------from(2)$

$2(bc + ab) \:$

$= 2 \times b \: (c + a) \:$

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$a(b + c) \: + c(a + b) \: = \ \: 2 \times b(c + a) \:$

$⛬ \: from \: (1) \:$

$⛬ \: a(b + c) \: , b(c + a) \: , c(a + b) \: are \: in \: ap \: - - - - - Q. E. D$

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