#N^th# term and sum of the geometric series #a, ar, ar^2, ar^3,...., ar^(n-1)#, where #a# is the first term and #r# is the ratio defined by a term divided by its preceding term is given by
#ar^(n-1)# and #(a(r^n-1))/(r-1)# if #r>1#) and #(a(1-r^n))/(1-r)#, if #r<1#
In the given example, #a=-25# and #r=(-30)/(-25)=6/5# and #r=6/5>1#.
Hence sum of the infinite series #{-25 - 30 - 36 - 43.2 ..............}# is given by
#(-25(6^oo-1))/(6-1)=-oo#
Note that if #r<1# geometric series is converging and as #r^oo=0#, sum of the series will be given by #a/(1-r)#.
