In the sequence, $(1,2,3),(4,5,6),(7,8,9,10)$ $\ldots$ of sets, the sum of elements in the 50th set is

In the sequence, $(1,2,3),(4,5,6),(7,8,9,10)$ $\ldots$ of sets, the sum of elements in the 50th set is
  1. $62525$
  2. $65225$
  3. $56255$
  4. $557625$

Solution

First term of each sets are $1,2,4,7, \ldots$ Let $ \begin{aligned} & S=1+2+4+7+\ldots+T_n \\ & S=1+2+4+\ldots+T_n \end{aligned} $ On subtracting, we get $ \begin{aligned} & 0=1+1+2+3+\ldots-T_n \\ & T_n=1+(1+2+3+\ldots(n-1) \text { terms }) \\ & T_n=1+\frac{(n-1) n}{2} \\ & T_{50}=1+\frac{49 \times 50}{2} \\ & \therefore \quad T_{50}=1226 \\ & \end{aligned} $ $\Rightarrow$ First term of 50th set is 1226 , therefore series is $1226,1227, \ldots 50$ terms $ S=\frac{50}{2}[2 \cdot 1226+49]=62525 $

Asked in: AP EAMCET 2002

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