Chapter 11.5, Problem 31WE

To determine

To calculate: The first three terms.

Answer to Problem 31WE

The first three terms are 16, 64, and 208 and these three terms neither A.P nor G.P.

Explanation of Solution

Given information: The sum of series $S_n=8\left(3^n-1\right)\mathrm{...}\left(i\right)$

Put, $n=1,2,\text{ and 3}$ in equation (1) to get first three terms

At $n=1$

  $\begin{array}{l}\therefore S_n=t_1=8\left(3^1-1\right)\\ \Rightarrow t_1=8\left(3-2\right)\\ \Rightarrow t_1=8\times 2\\ \Rightarrow t_1=16\end{array}$

Hence, first term is 16

At $n=2$

  $\begin{array}{l}\therefore S_n=t_2=8\left(3^2-1\right)\\ \Rightarrow t_2=8\left(9-1\right)\\ \Rightarrow t_2=8\times 8\\ \Rightarrow t_2=64\end{array}$

Hence, second term is 64

At $n=3$

  $\begin{array}{l}\therefore S_n=t_3=8\left(3^3-1\right)\\ \Rightarrow t_3=8\times \left(27-1\right)\\ \Rightarrow t_3=8\times 26\\ \Rightarrow t_3=208\end{array}$

Hence, third term is 208.

Therefore, the first three terms are, 16, 64, and 208

Now,

Let, first term, $a=16$

Second term, $a_1=64$

Third term, $a_2=208$

Common difference, $d=a_1-a=a_2-a_1$

  $d=64-16\ne 208-64$

Hence, the given series is not an A.P

Common ration $r=\frac{a_1}{a}=\frac{a_2}{a_1}$

  $\therefore r=\frac{64}{16}\ne \frac{208}{64}$

Therefore, the given series is not in G.P

Hence, the first three terms neither A.P nor G.P