Question

Find A1 in a geometric series for which sn=3045,r=2/5, and an=120

Answer 1

a = 1875

Explanation:

Sum of geometric series formula:

`S_n=(a(1-r^n))/(1-r)`

Geometric series formula:

`a_n=ar^(n-1)`

Given:

• 
  `S_n=3045`
• 
  `r=\frac25=0.4`
• 
  `a_n=120`

`\implies 120=a(0.4)^(n-1)`

`\implies a=(120)/((0.4)^(n-1))`

`\implies 3045=(a(1-0.4^n))/(1-0.4)`

`\implies 3045=(a(1-0.4^n))/(0.6)`

`\implies 1827=a(1-0.4^n)`

`\implies a=(1827)/(1-0.4^n)`

Equate equations for a:

`\implies (120)/((0.4)^(n-1))=(1827)/(1-0.4^n)`

Apply exponent rule
`a^(b-c)=(a^b)/(a^c)`

`\implies (120)/((0.4^n)/(0.4^1))=(1827)/(1-0.4^n)`

`\implies (48)/(0.4^n)=(1827)/(1-0.4^n)`

`\implies 48(1-0.4^n)=1827(0.4^n)`

`\implies 48-48(0.4^n)=1827(0.4^n)`

`\implies 48=1875(0.4^n)`

`\implies 0.4^n=(48)/(1875)`

`\implies 0.4^n=0.0256`

Taking natural logs:

`\implies \ln0.4^n=\ln0.0256`

`\implies n\ln0.4=\ln0.0256`

`\implies n=(\ln0.0256)/(\ln0.4)`

`\implies n=4`

Substituting found value for n into one of the equations and solving for a:

`\implies a=(120)/((0.4)^(4-1))`

`\implies a=(120)/(0.4^3)`

`\implies a=(120)/(0.064)`

`\implies a=1875`