Question

Find the sum of the first 10 terms of the following series, to the nearest integer. 10, 40,160

Answer 1

Answer:3495250

Step-by-step explanation:

~geometric sequence

common ratio: r=4

formula: Sn=a1(1-r^n)/1-r

Sn=10(1-(4)^10)/1-4

=-10485750/-3

=3495250

Answer 2

The sum of the first 10 terms of the following series, to the nearest integer. 10, 40,160 would be 3495250.

What is the sum of terms of a geometric sequence?

Let's suppose its initial term is a multiplication factor is r

and let it has total n terms, then, its sum is given as:

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

(sum till nth term)

The given geometric sequence is

10, 40,160

common ratio: r=4

a = 10

n=10

We know that

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

$S_{10} = \dfrac{10(4^{10}-1)}{4-1}\\S_{10} = \dfrac{10\times 1048575}{3}\\S_{10} = \dfrac{10485750}{3}\\S_{10} = 3495250$

Learn more about sum of terms of geometric sequence here:

brainly.com/question/1607203

#SPJ2