The sum of the first #3# terms in a geometric series with common ratio #3# is #52#. What are the values of #t_1# and #t_6#?

Question

18426 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-15T20:07:02

#t_1 = 4#
#t_6 = 972#

You start by finding #a# in the formula #s_n = (a(1 - r^n))/(1 - r)#, since we know the sum, the common ratio (#r#) and the number of terms.

#52 = (a(1 - 3^3))/(1 - 3)#

#52(-2) = a(1 - 27)#

#-104 = -26a#

#a = (-104)/(-26)#

#a = 4#

Hence, the first term is #4#. We can find the 6th term using the formula for the nth term of a geometric sequence, #t_n = a xx r^(n - 1)#.

#t_6 = 4 xx 3^(6 - 1)#

#t_6 = 972#

Hopefully this helps!