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In Which quadrant Point (2,-3) will be l...

In Which quadrant Point `(2,-3)` will be lie.

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(A) `a_(1),a_(2),"......"` are in AP.
`a_(1)+a_(4)+a_(7)+a_(14)+a_(17)+a_(20)=165` [In an AP, sum of the terms equidistant from the 1st and last is equal to sum of 1st and last terms]
`implies3(a_(1)+a_(20))=165`
`implies a_(1)+a_(1)+19d=55`
d is the common difference of AP.
`2a_(1)+19d=55 " " ".......(i)"`
Now, `alpha=a_(2)+a_(6)+a_(15)+a_(19)`
`alpha=2(a_(2)+a_(19))`
`alpha=2(a_(1)+d+a_(1)+18d)`
`alpha=2(2a_(1)+19d)" " "......(ii)"`
and `beta=2(a_(9)+a_(12))-(a_(3)+a_(18))`
`beta=2(a_(1)+8d+a_(1)+11d)-(a_(1)+2d+a_(1)+17d)`
`beta=2(2a_(1)+19d)-(2a_(1)+19d)`
`beta=2a_(1)+19d " " "......(iii)"`
From Eqs. (i) and (iii), we get
`alpha =2beta`
From Eqs. (i),(ii) and (iii), we get
`alpha+2beta=4(2a_(1)+19d)=4(55)=220" "[" from Eq.(i)"]`
`alpha+beta=3(2a_(1)+19d)`
`=3xx55=165=15xx11=15mu, " where " mu in I`
`alpha-beta=2a_(1)+19d`
`=55=5xx11=5lambda, " where " lambda in I`.
(B) `a_(1),a_(2),"......"` are in AP.
`a_(1)+a_(5)+a_(10)+a_(15)+a_(20)+a_(24)=195`
`3(a_(1)+a_(24))=195`
`implies a_(1)+a_(24)=65 " " ".....(i)"`
`implies 2a_(1)+23d=65`
Now, `alpha=a_(2)+a_(7)+a_(18)+a_(23)`
`=2(a_(2)+a_(23))=2(2a_(1)+23d)`
`alpha=130 " " " " [" fromEq. (i)" ]`
`beta=2(a_(2)+a_(22))-(a_(8)+a_(17))`
`=2(2a_(2)+23d)-(2a_(2)+23d)`
`=130-65=65`
Then, `alpha=2beta`
`alpha+2beta=130+130=260`
`alpha+beta=195=15xx13=15mu, " where " mu=13`
and `alpha -beta =130-65=65`
`=5xx13=5lambda, " where " lambda=13`.
(C ) `a_(1),a_(2),"...."` are in AP.
`a_(1)+a_(7)+a_(10)+a_(21)+a_(24)+a_(30)=225`
`3(a_(1)+a_(30))=225`
`2a_(1)+29d=75" " "........(i)"`
Now, `alpha=a_(2)+a_(7)+a_(24)+a_(29)`
`alpha=4a_(1)+58d=2(2a_(1)+29d)`
`2xx75=150`
`alpha=150" " ".......(ii)"`
and `beta=2(a_(10)+a_(21))-(a_(3)+a_(28))`
`=2(2a_(1)+29d)-(2a_(1)+29d)=150-75`
`beta=75" " "......(iii)"`
Then, `alpha=2beta`
`alpha+2beta=150+150=300 " and " alpha-beta=150-75=75`
`=5xx15=5lambda, " where " lambda=15`
and `alpha+beta=150+75=225=15xx15=15mu, " where " mu=15`.
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