How do you find #S_n# for the geometric series #a_2=-36#, a_5=972#, n=7?

1 Answer
Jan 18, 2017

#S_n=3^((n+1))(-1)^n-3#

Explanation:

The #n^(th)# term of a Geometric Series and its sum #S_n# up to n^(th)# term #a_n#, whose first term is #a_1# and common ratio is #r# is given by

#a_n=a_1xxr^((n-1))# and #S_n=a_1xx(r^n-1)/(r-1)#

as #a_2=a_1xxr=-36# and #a_5=a_1xxr^4=972#

Dividing latter by former, we get #r^3=-972/36=-27#

and hence #r=root(3)(-27)=-3#

and #a_1=36/-3=-12#

Hence, #S_n=-12xx((-3)^n-1)/((-3)-1)=-12xx((-3)^n-1)/(-4)# i.e.

#S_n=3((-3)^n-1)=3(3^n(-1)^n-1)=3^((n+1))(-1)^n-3#