Question

4. In a geometric sequence, the first term is 1\3 and the sixth term is 1\729, find the G.P.

Answer 1

GEOMETRIC PROGRESSION :
A sequence a1, a2, a3 ………….. an Is called a geometric sequence if a(n+1) = anr, where r is a non zero constant. Here, a1 is the first term and the constant r is called common ratio. The sequence is also called a geometric progression(G.P)

a1 = a , r = a(n+1)/ an


General term of a geometric sequence is
tn = arⁿ-1
General form of G.P is a, a r , a r ²,.........

SOLUTION :
GIVEN :
first term (a)= 1/3
6th term = 1/729

a = ⅓…………… (1)
t₆ = 1/729
tn = arⁿ-1 [ n = 6]
a r⁵ = 1/729
(1/3) r⁵ = 1/729
[ From eq .1]
r⁵ = (1/729)/(1/3)
r⁵ = (1/729)x(3/1)
r⁵ = 1/243
r⁵ = (1/3)⁵
r = 1/3

General form of G.P is a, a r , a r ²,.........
1/3 ,(1/3)(1/3),(1/3)(1/3)²,..............
1/3,1/9,1/27,............

Hence, the G.P is 1/3,1/9,1/27,............

HOPE THIS WILL HELP YOU….

Answer 2

Solution :

Let a , r are first term and common

ratio of a G.P

a = 1/3 , n = 6

**************************************

nth term in G.P = an

an = ar^n-1

***************************************

a6 = ar^5 = 1/729

=> ( 1/3 ) r^5 = 1/729

=> r^5 = 3/729

=> r^5 = 1/243

=> r^5 = ( 1/3 )^5

r = 1/3

Therefore ,

Required G.P is

1/3 , 1/9, 1/27, 1/81 , 1/243 , 1/729,..

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