Q.

Given sequence of numbers X1,X2,x3,X2005which satisfy, x1x1+1=x2x2+3=x3x3+5x2005x2005+2009 where x1+x2+.x1005=2010 then 1005164x21 is

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answer is 0.5.

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Detailed Solution

Above sequence is in A.P
x1+1x1=x2+3x2=x1005+2009x1005=kk=x1+1+x2+3+x1005+2009x1+x2.x1005k=1+(1005)22010=x21+41x21x21=821005

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Given sequence of numbers X1,X2,x3,……X2005which satisfy, x1x1+1=x2x2+3=x3x3+5……x2005x2005+2009 where x1+x2+…….x1005=2010 then 1005164x21 is