20. Sum of the first n terms of an ...
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20. Sum of the first n terms of an A.P (having positive terms) is given by Sn=(1+212) (1 -TR) where T is the term of the series. The value of T2 is 12-1 (C) 252 (D) None of these 12+1 (A) 212 (B) 252

JEE/Engineering Exams
Maths
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(2) Wote: ( 7 mathrm{f}_{1}=+1 / sqrt{2}, mathrm{T}_{2} ) will be ( sqrt{frac{sqrt{2}-1}{2 sqrt{2}}} ) which is smaller than T, i, It is an A ing A.P, SoT, ( 2 T_{1} ) [ begin{aligned} text { Put } n={&, quad s_{1}=left(1+2 T_{1}right)left(1-T_{1}right) quad text { where } S_{1}=T_{1} & therefore T_{1}=1-pi_{1}+2 pi_{1}-2 T_{1}^{2} therefore & 2 T_{1}^{2}=1 Rightarrow T_{1}=pm frac{1}{sqrt{2}} Rightarrow text { Jabe } T_{1}=-1 / sqrt{2} end{aligned} ] as sn is increasing wroucte Ale. Put ( n=2, quad S_{2}=1-T_{2}+2 T_{2}-2 T_{2}^{2} ) where [ begin{array}{c} T_{1}+T_{2}=1+T / 2-2 T_{2}^{2} 2 T_{2}^{2}=1-T_{1} Rightarrow T_{2}^{2}=frac{1-left(frac{-1}{sqrt{2}}right)}{2}=frac{sqrt{2}+1}{2 sqrt{2}}(1) end{array} ]
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