If x1,x2 …,x(20) are in H.P and x1,2,x(2...

If `x_1,x_2 …,x_(20)` are in H.P and `x_1,2,x_(20)` are in G.P then `Sigma_(r=1)^(19)x_rr_(x+1)`

none of these

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The correct Answer is:

Clearly, `1/x_(1),1/x_(2),…1/x_(20)` will be in A.P. Hence,
`1/x_(2)-1/x_(1)=1/x_(3)-1/x_(2)=..=1/x_(r+1)-1/x_(r)=..=lamda` (say)
`rArr(x_(r)-x_(r+1))/(x_(r)x_(r+1))=lamda`
or `x_(r)x_(r+1)=-1/lamda(x_(r+1)-x_(r ))`
`rArrsum_(r=1)^(19)x_(r )x_(r+1)=-1/lamdasum_(r=1)^(19)(x_(r+1)-x_(r ))=-1/lamda(x_(20)-x_(1))`
Now, `1/x_(20)=1/x_(1)+19lamda`
or `(x_(1)-x_(20))/(x_(1)x_(20))=19lamda`
`rArrsum_(r=1)^(19)x_(r )x_(r+1)=19x_(1)x_(20)=19xx4=76`
`(becausex_(1),2,x_(20)` are in G.P., then `x_(1)x_(20)=4)`

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