Chapter 14.5, Problem 36E

To determine

Find the sum of first ten terms of the series.

Answer to Problem 36E

  $S_{10}=\frac{145}{16}$

Explanation of Solution

Given information:

Series $\frac{1}{16}+\frac{1}{4}+1+\mathrm{....}$

Calculation:

Consider the series,

  $\frac{1}{16}+\frac{1}{4}+1+\mathrm{....}$

Now find the tenth term,

  $a_{10}=a_1+\left(n-1\right)d$

Calculate,

  $\begin{array}{l}d=a_2-a_1\\ =\frac{1}{4}-\frac{1}{16}\\ =\frac{4-1}{16}=\frac{3}{16}\end{array}$

Now,

  $\begin{array}{l}{a}_{10}=\frac{1}{16}+\left(10-1\right)\frac{3}{16}\\ a_{10}=\frac{1}{16}+\frac{27}{16}=\frac{28}{16}\end{array}$

Now calculate sum of first ten terms,

  $\begin{array}{l}{S}_n=\frac{n\left(a_1+a_n\right)}{2}\\ S_{10}=\frac{10\left(\frac{1}{16}+\frac{28}{16}\right)}{2}\\ =\frac{10\left(\frac{29}{16}\right)}{2}\\ S_{10}=\frac{145}{16}\end{array}$

Hence, $S_{10}=\frac{145}{16}$ .