How do you find #a_2# for the geometric series given #S_n=315#, r=0.5, n=6?

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Answer 1 · 2016-11-11T20:24:23

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

#315 = (a(1 - 0.5^6))/(1- 0.5)#

#315(0.5) = (a(1- 0.015625))#

#157.5 = 0.984375a#

#a = 160#

Now, use the formula #t_n = a xx r^(n - 1)# to find the 2nd term.

#t_2 = 160 xx 0.5^(2- 1)#

#t_2 = 160 xx 0.5#

#t_2 = 80#

Hopefully this helps!