The first term of the 11^{th} group in the fo | vedclass

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

Question

(C) Observe the first term of each group:Group $1$: $1$Group $2$: $2$Group $3$: $5$Group $4$: $10$Group $5$: $17$Let $a_n$ be the first term of the $n

Vedclass · Accepted Answer

$101$

Vedclass · Answer

(C) Observe the first term of each group:Group $1$: $1$Group $2$: $2$Group $3$: $5$Group $4$: $10$Group $5$: $17$Let $a_n$ be the first term of the $n^{th}$ group.The sequence of first terms is $1, 2, 5, 10, 17, \dots$Calculate the differences between consecutive terms:$2 - 1 = 1$$5 - 2 = 3$$10 - 5 = 5$$17 - 10 = 7$The differences are $1, 3, 5, 7, \dots$, which is an arithmetic progression with common difference $2$.The $n^{th}$ term of this sequence is given by $a_n = a_1 + \sum_{k=1}^{n-1} (2k - 1) = 1 + (n-1)^2$.For the $11^{th}$ group, $n = 11$:$a_{11} = (11 - 1)^2 + 1 = 10^2 + 1 = 100 + 1 = 101$.

The first term of the $11^{th}$ group in the following sequence of groups $(1), (2, 3, 4), (5, 6, 7, 8, 9), \dots$ is:

Similar Questions

If $1, \log_y x, \log_z y, -15 \log_x z$ are in $A.P.$,then

The $n^{th}$ term of the series $3.8 + 6.11 + 9.14 + 12.17 + .....$ will be

The sum of three consecutive terms in a geometric progression is $14$. If $1$ is added to the first and the second terms and $1$ is subtracted from the third,the resulting new terms are in arithmetic progression. Then the lowest of the original terms is

Find the sum of the first hundred even natural numbers divisible by $5$.

If $a_1 = a_2 = 2$ and $a_n = a_{n-1} - 1$ for $n > 2$,then $a_5$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module