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 $(-1)+\left(\frac{-1}{2}\right)+\left(\frac{-1}{6}\right)+\left(\frac{-1}{24}\right)+\left(\frac{-1}{120}\right)+\ldots$
The $A.M.$ of a $50$ set of numbers is $38$. If two numbers of the set, namely $55$ and $45$ are discarded, the $A.M.$ of the remaining set of numbers is
Let $a_1, a_2, a_3, \ldots$ be an arithmetic progression with $a_1=7$ and common difference $8$ . Let $T_1, T_2, T_3, \ldots$ be such that $T_1=3$ and $T_{n+1}-T_n=a_n$ for $n \geq 1$. Then, which of the following is/are $TRUE$ ?
$(A)$ $T_{20}=1604$
$(B)$ $\sum_{ k =1}^{20} T_{ k }=10510$
$(C)$ $T_{30}=3454$
$(D)$ $\sum_{ k =1}^{30} T_{ k }=35610$
If three numbers be in $G.P.$, then their logarithms will be in
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