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

2 Answers
Binayaka C.
Mar 17, 2018

The value of a_10=19683

Explanation:

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]

EZ as pi
Mar 17, 2018

19,683

Explanation:

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

Related questions
See all questions in Geometric Sequences
Impact of this question
31792 views around the world
You can reuse this answer
Creative Commons License