Let `a_1, a_2, a_3, ,a_(11)` be real numbers satisfying `a_1=15 , 27-2a_2>0 and a_k=2a_(k-1)-a_(k-2)` for `k=3,4, , 11.` If `(a_1^2+a_2^2+...+a_11^2)/(11)=90 ,` then the value of `(a1+a2++a 11)/(11)` is equals to _______.

Explanation

`:.a_(k)=2a_(k-1)-a_(k-2)" or "a_(k-1)=(a_(k-2)+a_(k))/(2)`
`:.a_(1),a_(2),a_(3),"....."` are in AP.
`:.(a_(1)^(2)+a_(2)^(2)+a_(3)^(2)+"........"+a_(11)^(2))/(11)=90 implies sum_(k=1)^(11)a_(i)^(2)=11xx90`
`implies sum_(k=1)^(11)(a_(i)+(i-1)d)^(2)=11xx90`
`implies sum_(k=1)^(11){a_(i)^(2)+2a_(1)d(i-1)+d^(2)(i-1)^(2)}=11xx90`
`implies 11xxa_(i)^(2)+2a_(1)d(0+1+2+3+"......"+10)+d^(2)(0^(2)+1^(2)+2^(2)+"......."+10^(2))=11xx90`
`implies 11xx15^(2)+2xx15xxd*((10*11)/(2))+d^(2)*((10*11*21)/(6))=11xx90`
`implies 385d^(2)+1650d+1485=0" " [:.a_(1)=15]`
`implies 7d^(2)+30d+27=0`
`implies (7d+9)(d+3)=0`
`:.d=-3,dne-(9)/(7)" " [:.27-2a_(2)gt0]`
`:.(a_(1)+a_(2)+a_(3)+"........"+a_(11))/(11)=((11)/(2){2a_(1)+(11-1)d})/(11)`
`=a_(1)+5d=15-15=0`.