Question

help me please!! this is algebra 2 ​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

Given expression ,

$:\mapsto\tt{\sum\limits_{n=0}^{38} 80.(0.99)^n}$

Express this expression to 38 terms

$:\mapsto\tt{\:80.\left((0.99)^0+(0.99)^1+(0.99)^2+(0.99)^3+(0.99)^4+........+(0.99)^{37}+(0.99)^{38}\right)} \\ \\ \\ :\mapsto\tt{\:80.\left(1+0.99^1+0.99^2+0.99^3+.........+0.99^{37}+0.99^{38}\right)}$

This , have common difference ,

So, we can say that this series performed to G.P.

$:\mapsto\tt{\red{\:Common\:difference(r)\:=\:\dfrac{second\:term}{\:first\:term}}} \\ \\ \\ \small\tt{\green{\:\:\:\:\star\:first\:term\:=\:1\:and\:\:\star\:second\:term\:=\:0.99}} \\ \\ \\ :\mapsto\tt{\red{\:(r)\:=\:\dfrac{0.99}{1}}} \\ \\ \\ :\mapsto\tt{\red{\:(r)\:=\:0.99}}$

And,

• first terms "a" = 1
• number of terms "n" = 38.
• common deference "r" = 0.99

Sum of n terms of G.P.

$\bigstar\tt{\orange{\:S_{n}\:=\:\dfrac{a(1-r^n)}{1-r}}}$

Keep all values in above formula,

$:\mapsto\tt{\:S_{38}\:=\:80.\left(\dfrac{1.(1-0.99^{38)}}{1-0.99}\right)} \\ \\ \\ :\mapsto\tt{\:S_{38}\:=\:80.\left(\dfrac{1-0.99^{38}}{0.01}\right)} \\ \\ \\ :\mapsto\tt{\:S_{38}\:=\:80\times100\times(1-0.99^{38})} \\ \\ \\ :\mapsto\tt{\red{\:S_{38}\:=\:8000\times(1-0.99^{38})\:\:\:\:\:\:Ans}}$

Here, 0.99^(38) is long multiplication terms.

So, we can write direct 0.99^(38) instead of actual value of 0.99^(38) .