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#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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
