Explanation

(a)
From symmetry, we observe that $S_{50}$ has 50 terms. First term of $S_{1},\ S_{2},\ S_{3},\ S_{4},\ldots,S_{50}$ are 1, 2, 4, 7, … , 50.
Let $T_{n}$ 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 $S = 1 + 2 + 4 + 7 + ...\ \ \ \ \ + T_{n - 1} + T_{n}$
therefore, on subtracting
$0 = 1 + \left\lbrack 1 + 2 + 3 + 4 + ... + \left( T_{n} - T_{n - 1} \right) \right\rbrack - T_{n}$
$or\ 0 = 1 + \frac{n(n - 1)}{2} - T_{n}$
$\Rightarrow \ T_{n} = 1 + \frac{n(n - 1)}{2}$
$\Rightarrow \ T_{50} =$ First term in $S_{50} = 1226$
Therefore, sum of the terms in $S_{50}$
$= \frac{50}{2}\left\lbrack 2 \times 1226 + (50 - 1) \times 1 \right\rbrack$
$= 25(2452 + 49) = 25(2501) = 62525$