A) 62525
B) 25625
C) 62500
D) None of these
Correct Answer: A
Solution :
From symmetry, we observe that \[{{S}_{50}}\] has 50 terms. First terms of \[{{S}_{1}},\ {{S}_{2}},\ {{S}_{3}},\ {{S}_{4}},...........\]are 1, 2, 4, 7....... Let \[{{G}^{2}}=AH\] be the first term of \[{{n}^{th}}\]set. Then \[S={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+......+{{T}_{n}}\] \[\Rightarrow \] \[S=1+2+4+7+11+........+{{T}_{n-1}}+{{T}_{n}}\] or \[S=\,\,\text{ }1+2+4+7+.............+{{T}_{n-1}}+{{T}_{n}}\] Therefore on subtracting \[0=1+[1+2+3+4+.......+({{T}_{n}}-{{T}_{n-1}})]-{{T}_{n}}\] or \[0=1+\frac{n(n-1)}{2}-{{T}_{n}}\]\[\Rightarrow \]\[{{T}_{n}}=1+\frac{n(n-1)}{2}\] \[\Rightarrow \] \[{{T}_{50}}=\] First term in \[{{S}_{50}}=1226\] Therefore sum of the terms in \[{{S}_{50}}\]\[=\frac{50}{2}\left\{ 2\times 1226+(50-1)\times 1 \right\}\] \[=25(2452+49)=25(2501)=62525\].
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