1.

If `(a_(2)a_(3))/(a_(1)a_(4))=(a_(2)+a_(3))/(a_(1)+a_(4))=3((a_(2)-a_(3))/(a_(1)-a_(4)))`, then `a_(1),a_(2),a_(3),a_(4)` are inA. APB. GPC. HPD. none of these

Answer» Correct Answer - C
We have,
`(a_(2)a_(3))/(a_(1)a_(4))=(a_(2)+a_(3))/(a_(1)+a_(4))`
`rArr" "(a_(2)+a_(3))/(a_(2)a_(3))-(a_(1)+a_(4))/(a_(1)a_(4))`
`rArr" "(1)/(a_(3))+(1)/(a_(2))=(1)/(a_(4))+(1)/(a_(1))rArr(1)/(a_(2))-(1)/(a_(1))=(1)/(a_(4))-(1)/(a_(3))`
Again,
`(a_(2)a_(3))/(a_(1)a_(4))=3((a_(2)-a_(3))/(a_(1)-a_(4)))`
`rArr" "(a_(1)-a_(4))/(a_(1)a_(4))=3((a_(2)-a_(3))/(a_(2)a_(3)))`
`rArr" "(1)/(a_(4))-(1)/(a_(1))=3((1)/(a_(3))-(1)/(a_(2)))`
`rArr" "((1)/(a_(4))-(1)/(a_(3)))+((1)/(a_(3))-(1)/(a_(2)))+((1)/(a_(2))-(1)/(a_(1)))=3((1)/(a_(3))-(1)/(a_(2)))`
`rArr" "((1)/(a_(2))-(1)/(a_(1)))+((1)/(a_(3))-(1)/(a_(2)))+((1)/(a_(2))-(1)/(a_(1)))=3((1)/(a_(3))-(1)/(a_(2)))` [Using (i)]
`rArr" "(1)/(a_(2))-(1)/(a_(1))=(1)/(a_(3))-(1)/(a^(2))` . . . (ii)
From (i) and (ii), we obtain
`(1)/(a_(2))-(1)/(a_(1))=(1)/(a_(3))-(1)/(a_(2))=(1)/(a_(4))-(1)/(a_(3))`
`rArr" "(1)/(a_(1)),(1)/(a_(2)),(1)/(a_(3)),(1)/(a_(4))` are in A.P. `rArra_(1),a_(2),a_(3),a_(4)` are in H.P.


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