Question #1df71

1 Answer
Mar 20, 2017

Given that first term of geometric series #a=-11#,
the common ratio #r=-4#,
the last or 8 th term #t_8=ar^7=180224#

Now

#S_8=a+ar+ar^2+ar^3+.......ar^7----(1)#

#rS_8=ar+ar^2+ar^3+......+ar^7+ar^8----(2)#

subtracting (2) from (1) we get

#(1-r)S_8=a-ar^8=a-rxxar^7#

#=>S_8=(a-rxxt_8)/(1-r)=(-11-(-4)xx180224)/(1-(-4))#

#=(-11+720896)/5=144177#