Question

Find the sum of the geometric series for which a = 160, r = 0.5, and n = 6.

Answer 1

$\\ \sf\longmapsto S_n=\dfrac{a(1-r^n)}{1-r}$

$\\ \sf\longmapsto S_n=\dfrac{160(1-0.5^6)}{1-0.5}$

$\\ \sf\longmapsto S_n=\dfrac{160(1-0.015625)}{0.5}$

$\\ \sf\longmapsto S_n=\dfrac{160(0.984378)}{0.5}$

$\\ \sf\longmapsto S_n=80(0.984378)$

$\\ \sf\longmapsto S_n=78.75$

Answer 2

Answer:

$s _6 = 315$

Step-by-step explanation:

Here,

a = First term

r ,= common ratio

Now let's use this formula to find the sum of the above geometric series

$s _n = \frac{a(1 - {r}^{n} )}{(1 - r)} \\ \\ s _6 = \frac{160(1 - {0.5}^{6} )}{(1 - 0.5)} \\ \\ s _6 = \frac{160 (1 - 0.015625)}{0.5} \\ \\ s _6 = \frac{157.5}{0.5} \\ \\ s _6 = 315$

Hope this helps you.

Let me know if you have any other questions:-)