Question

Find the sum of the first 25 terms in this geometric series: 8 + 6 + 4.5...

Answer 1

Step-by-step explanation:

Given the geometric sequence

8 + 6 + 4.5...

A geometric sequence has a constant ratio and is defined by

$a_n=a_1\cdot r^{n-1}$

$\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{a_{n+1}}{a_n}$

$\frac{6}{8}=\frac{3}{4},\:\quad \frac{4.5}{6}=\frac{3}{4}$

$\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$r=\frac{3}{4}$

$\mathrm{The\:first\:element\:of\:the\:sequence\:is}$

$a_1=8$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:$

$a_n=8\left(\frac{3}{4}\right)^{n-1}$

$\mathrm{Geometric\:sequence\:sum\:formula:}$

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

$\mathrm{Plug\:in\:the\:values:}$

$n=25,\:\spacea_1=8,\:\spacer=\frac{3}{4}$

$=8\cdot \frac{1-\left(\frac{3}{4}\right)^{25}}{1-\frac{3}{4}}$

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}$

$=\frac{\left(1-\left(\frac{3}{4}\right)^{25}\right)\cdot \:8}{1-\frac{3}{4}}$

$=\frac{8\left(-\left(\frac{3}{4}\right)^{25}+1\right)}{\frac{1}{4}}$

$\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}$

$=\frac{8\left(-\frac{3^{25}}{4^{25}}+1\right)}{\frac{1}{4}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{\frac{b}{c}}=\frac{a\cdot \:c}{b}$

$=\frac{\left(1-\frac{3^{25}}{4^{25}}\right)\cdot \:8\cdot \:4}{1}$

$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:4=32$

$=\frac{32\left(-\frac{3^{25}}{4^{25}}+1\right)}{1}$

$=\frac{32\cdot \frac{4^{25}-3^{25}}{4^{25}}}{1}$               ∵ $\mathrm{Join}\:1-\frac{3^{25}}{4^{25}}:\quad \frac{4^{25}-3^{25}}{4^{25}}$

$=32\cdot \frac{4^{25}-3^{25}}{4^{25}}$

$=\frac{\left(4^{25}-3^{25}\right)\cdot \:32}{4^{25}}$

$=\frac{2^5\left(4^{25}-3^{25}\right)}{2^{50}}$        ∵ $\mathrm{Factor}\:32:\ 2^5$,  $\mathrm{Factor}\:4^{25}:\ 2^{50}$

so

$=\frac{4^{25}-3^{25}}{2^{45}}$        ∵ $\mathrm{Cancel\:}\frac{\left(4^{25}-3^{25}\right)\cdot \:2^5}{2^{50}}:\quad \frac{4^{25}-3^{25}}{2^{45}}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a\pm \:b}{c}=\frac{a}{c}\pm \frac{b}{c}$

$=\frac{4^{25}}{2^{45}}-\frac{3^{25}}{2^{45}}$      

$=32-\frac{3^{25}}{2^{45}}$            ∵  $\frac{4^{25}}{2^{45}}=32$

$=32-0.024$        ∵  $\frac{3^{25}}{2^{45}}=0.024$

$=31.98$            

Therefore, the sum of the first 25 terms in this geometric series: 31.98