Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=n \frac{n^{2}+5}{4}$
Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=n \frac{n^{2}+5}{4}$
Substituting $n=1,2,3,4,5,$ we obtain
$a_{1}=1 \cdot \frac{1^{2}+5}{4}=\frac{6}{4}=\frac{3}{2}$
$a_{2}=2 \cdot \frac{2^{2}+5}{4}=2 \cdot \frac{9}{4}=\frac{9}{2}$
$a_{3}=3 \cdot \frac{3^{2}+5}{4}=3 \cdot \frac{14}{4}=\frac{21}{2}$
$a_{4}=4 \cdot \frac{4^{2}+5}{4}=21$
$a_{5}=5 \cdot \frac{5^{2}+5}{4}=5 \cdot \frac{30}{4}=\frac{75}{2}$
Therefore, the required terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21$ and $\frac{75}{2}$
Similar Questions
If sum of $n$ terms of an $A.P.$ is $3{n^2} + 5n$ and ${T_m} = 164$ then $m = $
If sum of $n$ terms of an $A.P.$ is $3{n^2} + 5n$ and ${T_m} = 164$ then $m = $
Let $a, b, c, d, e$ be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and $b+c+d$ is square of an integer. The least possible value of the number of digits of $c$ is
Let $a, b, c, d, e$ be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and $b+c+d$ is square of an integer. The least possible value of the number of digits of $c$ is
- [KVPY 2013]
The number of terms of the $A.P. 3,7,11,15...$ to be taken so that the sum is $406$ is
The number of terms of the $A.P. 3,7,11,15...$ to be taken so that the sum is $406$ is
If the sum of $n$ terms of an $A.P.$ is $\left(p n+q n^{2}\right),$ where $p$ and $q$ are constants, find the common difference.
If the sum of $n$ terms of an $A.P.$ is $\left(p n+q n^{2}\right),$ where $p$ and $q$ are constants, find the common difference.
If the sum of first $n$ terms of an $A.P.$ be equal to the sum of its first $m$ terms, $(m \ne n)$, then the sum of its first $(m + n)$ terms will be
If the sum of first $n$ terms of an $A.P.$ be equal to the sum of its first $m$ terms, $(m \ne n)$, then the sum of its first $(m + n)$ terms will be