In the sequence of sets (1,2,3), (4,5,6), (7, | vedclass

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(A) Let $T_n$ be the first term of the $n^{th}$ set. The first terms are $1, 2, 4, 7, 11, \ldots$This is a sequence where the differences are $1, 2, 3

Vedclass · Accepted Answer

$62525$

Vedclass · Answer

(A) Let $T_n$ be the first term of the $n^{th}$ set. The first terms are $1, 2, 4, 7, 11, \ldots$This is a sequence where the differences are $1, 2, 3, 4, \ldots$The $n^{th}$ term is given by $T_n = T_1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2}$.For the $50^{th}$ set,$n=50$,so $T_{50} = 1 + \frac{49 	imes 50}{2} = 1 + 1225 = 1226$.The $50^{th}$ set contains $50$ consecutive integers starting from $1226$.The sum of $n$ terms in an arithmetic progression is $S_n = \frac{n}{2}[2a + (n-1)d]$.Here $n=50$,$a=1226$,and $d=1$,so $S_{50} = \frac{50}{2}[2(1226) + 49(1)] = 25[2452 + 49] = 25[2501] = 62525$.

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

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