Answer:
a = 1875
Step-by-step explanation:
Sum of geometric series formula:
[tex]S_n=\dfrac{a(1-r^n)}{1-r}[/tex]
Geometric series formula:
[tex]a_n=ar^{n-1}[/tex]
Given:
[tex]\implies 120=a(0.4)^{n-1}[/tex]
[tex]\implies a=\dfrac{120}{(0.4)^{n-1}}[/tex]
[tex]\implies 3045=\dfrac{a(1-0.4^n)}{1-0.4}[/tex]
[tex]\implies 3045=\dfrac{a(1-0.4^n)}{0.6}[/tex]
[tex]\implies 1827=a(1-0.4^n)[/tex]
[tex]\implies a=\dfrac{1827}{1-0.4^n}[/tex]
Equate equations for a:
[tex]\implies \dfrac{120}{(0.4)^{n-1}}=\dfrac{1827}{1-0.4^n}[/tex]
Apply exponent rule [tex]a^{b-c}=\dfrac{a^b}{a^c}[/tex]
[tex]\implies \dfrac{120}{\frac{0.4^n}{0.4^1}}=\dfrac{1827}{1-0.4^n}[/tex]
[tex]\implies \dfrac{48}{0.4^n}=\dfrac{1827}{1-0.4^n}[/tex]
[tex]\implies 48(1-0.4^n)=1827(0.4^n)[/tex]
[tex]\implies 48-48(0.4^n)=1827(0.4^n)[/tex]
[tex]\implies 48=1875(0.4^n)[/tex]
[tex]\implies 0.4^n=\dfrac{48}{1875}[/tex]
[tex]\implies 0.4^n=0.0256[/tex]
Taking natural logs:
[tex]\implies \ln0.4^n=\ln0.0256[/tex]
[tex]\implies n\ln0.4=\ln0.0256[/tex]
[tex]\implies n=\dfrac{\ln0.0256}{\ln0.4}[/tex]
[tex]\implies n=4[/tex]
Substituting found value for n into one of the equations and solving for a:
[tex]\implies a=\dfrac{120}{(0.4)^{4-1}}[/tex]
[tex]\implies a=\dfrac{120}{0.4^3}[/tex]
[tex]\implies a=\dfrac{120}{0.064}[/tex]
[tex]\implies a=1875[/tex]
