How do you find the sum of a geometric series for which a1 = 48, an = 3, and r = -1/2?

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Answer 1 · 2018-06-19T19:29:40

#S_5=33#

We know that ,

#"(1)The nth term of a geometric series is :"#

#color(blue)(a_n=a_1(r)^(n-1), where, a_1=1^(st)term and r=common # # ratio#

#"(2) The sum of first n term of the geometric series is :"#

#color(red)(S_n=(a_1(1-r^n))/(1-r) ,where, r!=1#

Here, #a_1=48 ,a_n=3 and r=(-1/2)#

Using #(1)# we get

#a_n=48(-1/2)^(n-1)=3#

#=>(-1/2)^(n-1)=3/48=1/16#

#=>(-1/2)^(n-1)=(-1/2)^4#

#=>n-1=4#

#=>n=5#

Now ,using #(2)# #"we get, "color(violet)"sum of first five terms is :"#

#S_5=(48[1-(-1/2)^5])/(1-(-1/2)#

#=>S_5=(48[1-(-1/32)])/(1+1/2)#

#=>S_5=(48[1+1/32])/(3/2)#

#=>S_5=(96(33/32))/3#

#=>S_5=33#
.....................................................................

Note:

The sum of first n term of this series is :

#S_n=(48[1-(-1/2)^n])/(1-(-1/2)#

#S_n=(48[1-(-1/2)^n])/(3/2)#

#S_n=2/3xx48[1-(-1/2)^n]#

#S_n=32[1-(-1/2)^n]#