In an AP,if S_{n} = n(4n + 1),find the AP. | vedclass

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

Question

(A) We know that the $n^{th}$ term of an $AP$ is given by $a_{n} = S_{n} - S_{n-1}$.Given $S_{n} = n(4n + 1) = 4n^{2} + n$.Now,$S_{n-1} = (n-1)(4(n-1)

Vedclass · Accepted Answer

$5, 13, 21, \ldots$

Vedclass · Answer

(A) We know that the $n^{th}$ term of an $AP$ is given by $a_{n} = S_{n} - S_{n-1}$.Given $S_{n} = n(4n + 1) = 4n^{2} + n$.Now,$S_{n-1} = (n-1)(4(n-1) + 1) = (n-1)(4n - 4 + 1) = (n-1)(4n - 3) = 4n^{2} - 3n - 4n + 3 = 4n^{2} - 7n + 3$.Therefore,$a_{n} = (4n^{2} + n) - (4n^{2} - 7n + 3) = 4n^{2} + n - 4n^{2} + 7n - 3 = 8n - 3$.To find the $AP$,we substitute values for $n$:For $n = 1$,$a_{1} = 8(1) - 3 = 5$.For $n = 2$,$a_{2} = 8(2) - 3 = 16 - 3 = 13$.For $n = 3$,$a_{3} = 8(3) - 3 = 24 - 3 = 21$.Thus,the required $AP$ is $5, 13, 21, \ldots$

In an $AP$,if $S_{n} = n(4n + 1)$,find the $AP$.

Similar Questions

For a given $A.P.$,$T_{7} = 12$ and $T_{12} = 72$. Then,$d = \dots \dots \dots \dots$

How many numbers lie between $10$ and $300,$ which when divided by $4$ leave a remainder $3?$

Show that the sum of an $AP$ whose first term is $a,$ the second term is $b,$ and the last term is $c,$ is equal to $\frac{(a+c)(b+c-2a)}{2(b-a)}.$

The $n^{th}$ term of an $A.P.$ is given by $T_{n} = 4n + 7$. Find the sum of $n$ terms of this $A.P.$

Is $0$ a term of the $AP : 31, 28, 25, \ldots ?$ Justify your answer.

Vedclass Test Series

Exam Paper Generator

Online Exam Module