TS ICET 24 May 2019 Shift 1 Paper

Question : 128 of 200

Marks: +1, -0

If the sum of first 21 terms of an arithmetic progression is 609. then the 11th term of progression is

Solution:

S _{21}=\;{21}/{2}[2 a, 20 d]=609

⇒ a+10 d=\;{609}/{21}

t_{11}=29

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