Question #254f6

2 Answers
Oct 7, 2017

#S_7=1094#

Explanation:

#"the "color(blue)"sum to n terms"# for a geometric series is.

#•color(white)(x)S_n=(a(r^n-1))/(r-1)#

#"where a is the first term and r the "color(blue)"common ratio"#

#r=(a_2)/(a_1)=(-6)/2=-3#

#rArrS_7=(2((-3)^7-1))/(-4)#

#color(white)(rArrS_7)=(2xx-2188)/(-4)=1094#

Oct 7, 2017

Answer is #1094#.

Explanation:

Given the geometric series is:

#color(red)(2-6+18-54.......)# up to #7^(th)# term.

Let, #t_1=a=2#
#t_2=a_1=-6#.

So, common multiplier
#color(red)(=t_2/t_1=a_2/a=-6/2=-3)#
#:.r=-3#

Now, Let a term of the sequence be #t_n#.
#:.t_n=a.r^(n-1)#
So, #t_7=2.(-3)^(7-1)#
#:.color(red)(t_7=2.3^6=1485)#.

Now, Let sum of the series up to #n# terms be #S_n#.

#S_n=(a(r^n-1))/(r-1)#.
#:.S_7=2[(-3)^7-1]/(-3-1)#.
#:.S_7=2xx2188/4=1094#.

So, sum upto #7th# term of the sequence is #1094#. (Answer).

Hope it Helps!!