Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=\frac{2 n-3}{6}$
Substituting $n=1,2,3,4,5,$ we obtain
$a_{1}=\frac{2 \times 1-3}{6}=\frac{-1}{6}$
$a_{2}=\frac{2 \times 2-3}{6}=\frac{1}{6}$
$a_{3}=\frac{2 \times 3-3}{6}=\frac{3}{6}=\frac{1}{2}$
$a_{4}=\frac{2 \times 4-3}{6}=\frac{5}{6}$
$a_{5}=\frac{2 \times 5-3}{6}=\frac{7}{6}$
Therefore, the required terms are $\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}$ and $\frac{7}{6}$
Find the $17^{\text {th }}$ and $24^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=4 n-3$
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
If $n$ is the smallest natural number such that $n+2 n+3 n+\ldots+99 n$ is a perfect square, then the number of digits of $n^2$ is
Let the sequence $a_{n}$ be defined as follows:
${a_1} = 1,{a_n} = {a_{n - 1}} + 2$ for $n\, \ge \,2$
Find first five terms and write corresponding series.
If the sum of $n$ terms of an $A.P.$ is $nA + {n^2}B$, where $A,B$ are constants, then its common difference will be