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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If the sum of the first $n$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$ is $\frac{n(n+1)^2}{2}$ when $n$ is even,what is the sum when $n$ is odd?

The sum of the first $n$ terms of the series $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ is

Let $S_{k} = \frac{1+2+\ldots+k}{k}$ and $\sum_{j=1}^n S_j^2 = \frac{n}{A}(Bn^2 + Cn + D)$,where $A, B, C, D \in \mathbb{N}$ and $A$ has the least value. Then:

What is the sum of $n$ terms of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$?

If the set of natural numbers is partitioned into subsets $S_1 = \{1\}, S_2 = \{2, 3\}, S_3 = \{4, 5, 6\}$ and so on,then the sum of the terms in $S_{50}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module