if a,b and c are in H.P ,then the value of [(ac+ab-bc)( ab+bc-ac)÷ (abc)²]
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have to check in ncert so i cant
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given,
a, b, c are in HP
1/a, 1/b, 1/c are in AP
=>1/b - 1/a = 1/c - 1/b
=> 1/b + 1/b = 1/c + 1/a
=> 2/b = a+c/ ac
=> b( a+c ) = 2 ac
=>ab +bc =ac+ ac
=> ab - ac =ac-bc .........(1)
( ac+ab-bc) ( ab +bc-ac) / (abc) ^2
(ab+ ab -ac) (2ac -ac) /(abc) ^2
(2ab-ac) (ac) / a^2 b^2c^2
a(2b-c) (ac )/a^2b^2c^2
2b-c / b^2c
2b/b^2c -c/b^2c
2/bc - 1/b^2
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