Write the first five terms of the following sequence and obtain the corresponding series :
$a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}, n\, \geq\, 2$
Write the first five terms of the following sequence and obtain the corresponding series :
$a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}, n\, \geq\, 2$
$a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}, n\, \geq \,2$
$\Rightarrow a_{2}=\frac{a_{1}}{2}=\frac{-1}{2}$
$a_{3}=\frac{a_{2}}{3}=\frac{-1}{6}$
$a_{4}=\frac{a_{3}}{4}=\frac{-1}{24}$
$a_{5}=\frac{a_{4}}{5}=\frac{-1}{120}$
Hence, the first five terms of the sequence are $-1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24}$ and $\frac{-1}{120}$
The corresponding series is $(-1)+\left(\frac{-1}{2}\right)+\left(\frac{-1}{6}\right)+\left(\frac{-1}{24}\right)+\left(\frac{-1}{120}\right)+\ldots$
Similar Questions
What is the $20^{\text {th }}$ term of the sequence defined by
$a_{n}=(n-1)(2-n)(3+n) ?$
What is the $20^{\text {th }}$ term of the sequence defined by
$a_{n}=(n-1)(2-n)(3+n) ?$
Suppose we have an arithmetic progression $a_1, a_2, \ldots a_n, \ldots$ with $a_1=1, a_2-a_1=5$. The median of the finite sequence $a_1, a_2, \ldots, a_k$, where $a_k \leq 2021$ and $a_{k+1} > 2021$ is
Suppose we have an arithmetic progression $a_1, a_2, \ldots a_n, \ldots$ with $a_1=1, a_2-a_1=5$. The median of the finite sequence $a_1, a_2, \ldots, a_k$, where $a_k \leq 2021$ and $a_{k+1} > 2021$ is
- [KVPY 2021]
The number of terms in an $A .P.$ is even ; the sum of the odd terms in it is $24$ and that the even terms is $30$. If the last term exceeds the first term by $10\frac{1}{2}$ , then the number of terms in the $A.P.$ is
The number of terms in an $A .P.$ is even ; the sum of the odd terms in it is $24$ and that the even terms is $30$. If the last term exceeds the first term by $10\frac{1}{2}$ , then the number of terms in the $A.P.$ is
- [JEE MAIN 2014]
Find the $20^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=\frac{n(n-2)}{n+3}$
Find the $20^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=\frac{n(n-2)}{n+3}$
If the sum of the first $n$ terms of the series $\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$ is $435\sqrt 3 $ , then $n$ equals
If the sum of the first $n$ terms of the series $\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$ is $435\sqrt 3 $ , then $n$ equals
- [JEE MAIN 2017]