Two consecutive numbers from 1, 2, 3, ..., n are removed, then arithmetic mean of the remaining numbers is `105/4` then `n/10` must be equal to

Options

A. `7^(th) " term is " 16`
B. `7^(th) " term " is 18`
C. Sum of first 10 terms is `(505)/(4)`
D. Sum of first 10 terms is `(405)/(4)`

Answer

A, C

Explanation

`S=1+1/((1+3))(1+2)^(2)+1/((1+3+5))(1+2+3)^(2)+1/((1+3+5+7))(1+2+3+4)^(2)+…`
The `r^(th)` term is given by
`T_(r)=1/r^(2)(1+2+..+r)^(2)=1/r^(2){(r(r+1))/2}^(2)=(r^(2)+2r+1)/4`
`thereforeT_(7)=16`
and `S_(10)=sum_(r=1)^(10)T_(r)`
=`1/4{((10)(10+1)(20+1))/6+(10)(10+1)+10}`
`=505/4`