How do you find #S_n# for the geometric series #a_3=-36#, #a_6=-972#, #n=10#?

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Answer 1 · 2017-07-21T17:32:54

The general form for a Geometric progression is:

#a_n = ar^(n-1)#

Given: #a_3 = -36# and #a_6 = -972#

Divide the latter by the former:

#a_6/a_3= (ar^(6-1))/(ar^(3-1)) = (-972)/-36 = 27#

This means that:

#r^3 = 27#

Solve for r:

#r = 3#

Use #a_3# and the value of r, to find the value of "a":

#a_3 = -36= a(3)^(3-1)#

#a(3)^2 = -36#

#a = -4#

We can use the formula for the Finite Sum of Geometric Series :

#S_n = a(1-r^n)/(1-r)#

Substituting #a = -4, r = 3 and n = 10#

#S_n = -4(1-3^10)/(1-3)#

#S_n = 118096#