Question

Find t18 and s18 for the following series: 2,8,32,…

Answer 1

2,8,32 ARE GEOMETRIC PROGRESSION
THEREFORE THEY ARE IN THE FORM OF A,AR,AR^2 ....
A=2
R=4
T18=A(R^(N-1))=2(4^(18-1))
     =2(4^17)
     =2^35
S18=A{(R^(N-1))}/N-1
      = 2{4^17)/3

Answer 2

Answer:

$t_{18}=2^35$ and $S_{18}=4.581298449\times 10^10$

Step-by-step explanation:

Given : 2,8,32,…

To find : $t_{18}$ and $s_{18}$

Solution:

2,8,32,…

It forma a GP

first term = a = 2

Common ratio = $r =\frac{8}{2}=\frac{32}{8}=4$

Formula of nth term in GP = $t_n=ar^{n-1}$

Substitute n = 18

$t_{18}=2 \times 4^{18-1}$

$t_{18}=2 \times 4^{17}$

$t_{18}=2 \times( 2^2)^{17}$

$t_{18}=2 \times 2^34$

$t_{18}=2^35$

Sum of n terms in GP = $S_n=\frac{a(1-r^n)}{1-r}$

Substitute n = 18

$S_{18}=\frac{2(1-4^{18})}{1-4}$

$S_{18}=4.581298449\times 10^10$

Hence $t_{18}=2^35$ and $S_{18}=4.581298449\times 10^10$