The absolute value of the sum of first `20` terms of series, if `S_(n)=(n+1)/(2)` and `(T_(n-1))/(T_(n))=(1)/(n^(2))-1`, where `n` is odd, given `S_(n)` and `T_(n)` denotes sum of first `n` terms and `n^(th)` terms of the series
A
`340`
B
`430`
C
`230`
D
`320`
Text Solution
Verified by Experts
The correct Answer is:
B
`(b)` We have `S_(n)-S_(n-2)=T_(n)+T_(n-1)` (Taking `n` to be odd) `:.T_(n-1)(1+(T_(n))/(T_(n-1)))=((n+1)/(2))-((n-1))/(2)=1` (As `S_(n)=(n+1)/(2)`, if `n` is odd) `:.T_(n-1)(1+(n^(2))/(1-n^(2)))=1` `:.T_(n-1)=-(n^(2)-1)` when `n` is odd Also, `S_(m)=S_(m-1)+T_(m)` If `m` is even then `(m-1)` is odd. `:.S_(m)=(m)/(2)-((m+1)^(2)-1)=(-m(2m+3))/(2)` `:. |S_(20)|=|(-20)/(2)(43)|=430`
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CENGAGE-PROGRESSION AND SERIES-ARCHIVES (MATRIX MATCH TYPE )
