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
A
`17`
B
`-15`
C
`21`
D
`-23`
Text Solution
Verified by Experts
The correct Answer is:
D
`(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`
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CENGAGE-PROGRESSION AND SERIES-ARCHIVES (MATRIX MATCH TYPE )
