Question

Find the sum of the 10 terms of the geometric series with a1=4 and r=2

Answer 1

The sum of the 10 terms of geometric series is 4092

Step-by-step explanation:

Given in the question, the series that will be formed is in Geometric Progression (GP) is the common ratio is mentioned here

According to GP, the sum of a series is given by the formula;

`\mathrm{S}_{\mathrm{n}}=(a\left(r^(n)-1\right))/(r-1)`

Then, according to question,

a is given as 4, n is given as 10 and r is given as 2.

`\mathrm{S}_(10)=(4\left(2^(10)-1\right))/(2-1)`

=
`(4(1024-1))/(1)`

= 4(1023)= 4092

Therefore, the sum of the 10 terms of geometric series is 4092

Answer 2

The sum of the first 10 terms is 4092

Step-by-step explanation:

We can use the formula for a partial sum of a geometric series of first
`a_1` term and common ratio r, which is given by:

`S_n=(a_1(1-r^n))/(1-r)`

Therefore, in this particular case where we add the first 10 terms of a geometric series with first term 4 and the common ratio 2, we have:

`S_n=(a_1(1-r^n))/(1-r) \\S_10=(4(1-2^(10)))/(1-2) \\S_10=(4(1-1024))/(1-2)\\S_10=(4(-1023))/(-1)\\S_10=(-4092))/(-1)\\S_10=4092`