How do you find the sum of the infinite geometric series -13and1/2 + 9 – 6 + . . . ?

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Answer 1 · 2016-04-17T08:30:22

Sum of the series is #-8.10#

As the series is #{-13 1/2,+9,-6,.......}#

its first term is #-13 1/2=-27/2# and it is a geometric series as

#(+9)/(-27/2)=-(9xx2)/27=-2/3# and also #-6/9=-2/3#

Now sum of a geometric series upto #n^(th)# term, with first term as #a# and ratio as #r# is given by

#a(r^n-1)/(r-1)# if #|r|>1# and #a(1-r^n)/(1-r)# if #|r|<1#

If #n->oo# the latter i.e. when #|r|<1# is #a/(1-r)#

Hence sum of the series is #(-27/2)/(1-(-2/3))=(-27/2)/(1+2/3)#

= #(-27/2)/(5/3)=-27/2xx3/5=-81/10=-8.10#