What is the sum of the first 10 terms given 1, 1/3, 1/9,1/27..?

2 Answers

Answer:

#\color{red}{3/2(1-(1/3)^{10})\approx1.499974597#

Explanation:

Given series #1, 1/3, 1/9, 1/27, \ldots# is a G.P. with the first term #a=1# & common ratio #r=\frac{1/3}{1}=\frac{1/9}{1/3}=\frac{1/27}{1/9}=\ldots =1/3#

Now, the sum of given series up to #n=10# terms

#=\frac{a(r^{n}-1)}{r-1}#

#=\frac{1((1/3)^{10}-1)}{1/3-1}#

#=\frac{(1/3)^{10}-1}{-2/3}#

#=3/2(1-(1/3)^{10})#

#approx1.499974597#

Jul 7, 2018

Answer:

#S_10=3/2((3^10-1)/3^10)~~1.4999#

Explanation:

Here,

#1,1/3,1/9,1/27,...,#

The first term #=a_1=1#

The common ratio #=r=(1/3)/1=(1/9)/(1/3)=(1/27)/(1/9)=1/3#

So, the given sequence is Geometric sequence with

#color(blue)(a_1=1 and r=1/3#

The sum of first #n-terms # of geometric sequence is :

#color(blue)(S_n=(a_1(1-r^n) )/(1-r) )and# we have #color(blue)(n=10#

So,

#S_10=(1(1-(1/3)^10))/(1-1/3)#

#S_10=((1-1/3^10))/(2/3)#

#S_10=3/2((3^10-1)/3^10)#

#S_10=3/2(59048/59049)#

#S_10~~1.4999#