Question

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

Answer 1

`\\ \sf\longmapsto S_n=(a(1-r^n))/(1-r)`

`\\ \sf\longmapsto S_n=(160(1-0.5^6))/(1-0.5)`

`\\ \sf\longmapsto S_n=(160(1-0.015625))/(0.5)`

`\\ \sf\longmapsto S_n=(160(0.984378))/(0.5)`

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

`\\ \sf\longmapsto S_n=78.75`

Answer 2

`s _6 = 315`

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 = (160 (1 - 0.015625))/(0.5) \\ \\ s _6 = (157.5)/(0.5) \\ \\ s _6 = 315`

Hope this helps you.

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