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 :

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

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

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

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