An `A.P.` consist of even number of terms `2n` having middle terms equal to `1` and `7` respectively. If `n` is the maximum value which satisfy `t_(1)t_(2n)+713 ge 0`, then the value of the first term of the series is

Options

A. (a) `17`
B. (b) `-15`
C. (c) `21`
D. (d) `-23`

Answer

D

Explanation

`(d)` Given mid terms `t_(n)=1` and `t_(n+1)=7`
`:. T_(n)+t_(n+1)=8=t_(1)+t_(2n)`
and `t_(n+1)-t_(n)=6=d` (common difference of `A.P`.)
`t_(n)+t_(n+1)=8`
`:.a+(n-1)d+a+nd=8`
`:.a+6n=7`
Now `4t_(1)t_(2n)=[(t_(1)+t_(2n))^(2)-(t_(2n)-t_(1))^(2)]`
`=64=36(2n-1)^(2)` [as `t_(2n)-t_(1)=(2n-1)xx6`]
`:.t_(1)t_(2n)=16-9(2n-1)^(2)`
`:. 16-9(2n-1)^(2)+713 ge 0`
`:. -4 le n le 5`
`:.n=5`
Hence, from `a+6n=7`, `a=-23`