In a geometric sequence, #a_3=-9# and #a_6=243#, find the value of #a_10#?

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Answer 1 · 2018-03-17T12:15:12

The value of # a_10=19683 #

#a_3=-9 , a_6=243 ; a_10= ?#

#n# th term in G.P series is #a_n=a_1*r^(n-1) ; a_1 and r# are

first term and common ratio of G.P. series.

#a_3=a_1*r^(3-1) or a_1*r^2= -9 ; (1)# similarly ,

#a_6=a_1*r^(6-1) or a_1*r^5= 243 ; (2)# Dividing

equation (2) by equation (1) we get, #r^5/r^2= -243/9# or

#r^3= -27 or r = -3# Putting #r= -3# in equation (1) we

get , #a_1*(-3)^2=-9 or a_1= -9/9 or a_1= -1#

#:. a_10= a_1*r^(10-1)= -1*( -3)^9=19683 #

The value of # a_10=19683 # [Ans]

Answer 2 · 2018-03-17T12:29:15

#19,683#

The general term for a GP is #a_n = a_1r^(n-1)#

where #a_1# is the first term and #r# is the common ratio.

You are given the values of two terms in a GP.

Divide the two terms: The formula and the values

#a_6/a_3 = (cancelar^5)/(cancelar^2) = 243/-9#

This gives: #r^3 = -27" "larr# find the cube root.

#color(white)(xxxxxxxx)r =-3#

Now find #a_1# from #a_3#

#a_1r^2 = a_1(-3)^2 = -9#

#a_1 = (-9)/9 = -1#

Now you can find the value of #a_10#

#a_10= -1(-3)^9#

# = 19,683#