Let any term in the series be a_i
Let the sum of the series be s
Let the common ratio be r = 2/3
color(blue)("Observation")
a_i->a_1 =54 "......................................... "->" "54xx(2/3)^0=54
a_i->a_2=36/54 =2/3" "=>36=2/3xx54" "->" "54xx(2/3)^1
a_i=a_3=24/36=2/3" "=> 24=2/3xx36 " "->" "54xx(2/3)^2
a_i->a_4=16/24=2/3" "=>16=2/3xx24" "->" "54xx(2/3)^3
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color(blue)("Building the standardised equation" )
=> s = 54r^0+54r^1+54r^2+54r^3+...+54r^(n-1) .......Equation(1)
=>sr = 54r^1+54r^2+54r^3+...+54r^(n-1)+54r^n....Equation(2)
Note that r^1=r" and "r^0=1
Equation(1) - Equation(2)
s-sr=54(1-r^n)
s(1-r)=54(1-r^n)
color(blue)(s=(a_1(1-r^(n) ))/(1-r))
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color(blue)("Answering the question")
Set n=6", "r=2/3" and "a_1=54" " giving:
s=(54(1-(2/3)^6))/(1-2/3)
s= 147.77bar7 by calculator
