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 $(-1)+\left(\frac{-1}{2}\right)+\left(\frac{-1}{6}\right)+\left(\frac{-1}{24}\right)+\left(\frac{-1}{120}\right)+\ldots$
Suppose that all the terms of an arithmetic progression ($A.P.$) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6: 11$ and the seventh term lies in between $130$ and $140$ , then the common difference of this $A.P.$ is
Let the digits $a, b, c$ be in $A.P.$ Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in $A.P.$ at least once. How many such numbers can be formed?
Show that the sum of $(m+n)^{ th }$ and $(m-n)^{ th }$ terms of an $A.P.$ is equal to twice the $m^{\text {th }}$ term.
If three numbers be in $G.P.$, then their logarithms will be in