Question

Find the sum of the first 8 terms of the following series, to the nearest integer.

Answer 1

Answer:

The sum of the first 8 terms will be:

• $\frac{1288991}{7776}$

Step-by-step explanation:

Given the sequence

$36, 30, 25, ...$

A geometric sequence has a constant ratio and is defined by

$a_n=a_0\cdot r^{n-1}$

computing the common ratio

$\frac{30}{36}=\frac{5}{6},\:\quad \frac{25}{30}=\frac{5}{6}$

$r=\frac{5}{6}$

As the first element is

$a_1=36$

so the nth term will be:

$a_n=a_0\cdot r^{n-1}$

$a_n=36\left(\frac{5}{6}\right)^{n-1}$

Geometric sequence sum formula

$a_1\frac{1-r^n}{1-r}$

Plugin the values to determine the sum of the first 8 terms

$n=8,\:\spacea_1=36,\:\spacer=\frac{5}{6}$

$=36\cdot \frac{1-\left(\frac{5}{6}\right)^8}{1-\frac{5}{6}}$

$=\frac{\left(1-\left(\frac{5}{6}\right)^8\right)\cdot \:36}{1-\frac{5}{6}}$    

$=\frac{36\left(-\left(\frac{5}{6}\right)^8+1\right)}{\frac{1}{6}}$

$=\frac{36\left(-\frac{390625}{1679616}+1\right)}{\frac{1}{6}}$

$=\frac{\left(1-\frac{390625}{1679616}\right)\cdot \:36\cdot \:6}{1}$

$=\frac{216\left(-\frac{390625}{1679616}+1\right)}{1}$

$=\frac{216\cdot \frac{1288991}{1679616}}{1}$

$=216\cdot \frac{1288991}{1679616}$

$=\frac{1288991\cdot \:216}{1679616}$

$=\frac{278422056}{1679616}$

$=\frac{1288991}{7776}$

 Therefore, the sum of the first 8 terms will be:

• $\frac{1288991}{7776}$