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

Options

A. 1
B. 1
C. 2
D. 9

Answer

A

Explanation

We have,
`a_(k)=2a_(k-1)-a_(k-2)" for "k=3,4, . . .,11`
`rArr" "a_(k)-a_(k-1)=a_(k-1)-a_(k-2)" for "k=3,4, . . . ,11`
`rArr" "a_(1),a_(2),a_(3), . . .,a_(11)` are in A.P.
Let d be the common difference of the A.P.
`a_(k)=15+(k-1)d,k=1,2, . . .,11`
`:." "(a_(1)^(2)+a_(2)^(2)+ . . . +a_(11)^(2))/(11)=90`
`rArr" "underset(k=1)overset(11)sums_(k)^(2)=990`
`rArr" "underset(k=1)overset(11)sum{15+(k-1)d}^(2)=990`
`rArr" "underset(k=1)overset(11)sum{225+30(k-1)d+(k-1)^(2)d^(2)}=990`
`rArr" "225xx11+30d xx(10xx11)/(2)+(10xx11xx21)/(6)d^(2)=990`
`rArr" "225+150d+35d^(2)=90`
`rArr" "35d^(2)+150s+135=0`
`rArr" "7d^(2)+30d+27=0`
`rArr" "(7d+9)(d+3)=0`
`rArr" "d=-3,-(9)/(7)`
It is given that `a_(2)lt(27)/(2)`. Therefore, d =-3.
Now, `(a_(1)+a_(2)+ . . . .+a_(11))/(11)=(1)/(2)(a_(1)+a_(11))=(1)/(2)(15+15+10d)=0`