Chapter 12.2, Problem 56E

To determine

Calculate the products of the numbers.

Answer to Problem 56E

The product of the numbers is ${10}^{19}$ .

Explanation of Solution

Given: It is given in the question that the term are ${10}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.},{10}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$10$}\right.},{10}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.},{10}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$10$}\right.},\mathrm{.......},{10}^{\raisebox{1ex}{$19$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}$ .

Concept Used:

In this, use the concept of sum of a arithmetic sequence and the formula used is $S_n=\frac{n}{2}[2a+(n-1)d]$ .

Calculation: Here, the term are ${10}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.},{10}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$10$}\right.},{10}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.},{10}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$10$}\right.},\mathrm{.......},{10}^{\raisebox{1ex}{$19$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}$ .

In product, ${10}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}\times {10}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}\times {10}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}\times {10}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}\times \mathrm{.......}\times {10}^{\raisebox{1ex}{$19$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}$

  $={10}^{\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}\mathrm{.........}+\frac{19}{10}}$

  ${}^{\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}\mathrm{.........}+\frac{19}{10}}$

is the partial sum of ${\displaystyle \sum _{k=1}^{10}\frac{k}{10}}$ .

So, the sum is

  $\begin{array}{l}{S}_n=\frac{n}{2}[2a+(n-1)d]\\ S_n=\frac{19}{2}[2\times \frac{1}{10}+(19-1)\frac{1}{10}]=\\ S_n=\frac{19}{2}[\frac{1}{5}+\frac{9}{5}]\\ S_n=\frac{19}{2}\times 2=19\end{array}$

Hence ,

  ${10}^{\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}\mathrm{.........}+\frac{19}{10}}={10}^{19}$

Conclusion:

The product is ${10}^{19}$ .