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 $|x| < 1$,what is the sum of the series $1 + 2x + 3x^2 + 4x^3 + \dots \infty$?

If $2^3+4^3+6^3+\ldots+(2n)^3 = h n^2(n+1)^2$,then $h$ is equal to

If $1+\sin \theta+\sin ^{2} \theta+\ldots \infty = 2 \sqrt{3}+4$,then $\theta = $

If $A = \sum_{n=1}^{\infty} \frac{1}{(3+(-1)^{n})^{n}}$ and $B = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(3+(-1)^{n})^{n}}$,then $\frac{A}{B}$ is equal to:

The sum to infinity of the following series $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module