Question

5. Which term of the geometric sequence,

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 : a = 5 , r = 2/5

Let 128/15625 is nth term of the sequence
tn = 128/15625

General term of a geometric sequence is
tn = arⁿ-1
arⁿ-1 = 128/15625
5 (2/5)ⁿ-1 = 128/15625
(2/5)ⁿ-1= 128/(15625 x 5)
(2/5)ⁿ-1 = 2⁷/(5⁶ x 5¹)
(⅖)ⁿ-1= 2⁷/5⁷
(2/5)ⁿ-1 = (2/5)⁷
n - 1 = 7
n = 7 + 1
n = 8

Hence, the 8th term of the given sequence is 128/15625.

(ii) 1,2,4,8,..........is 1024?
GIVEN :
a = 1 , r = 2/1 = 2

Let 1024 is nth term of the sequence

General term of a geometric sequence is
tn = arⁿ-1
tn = 1024
arⁿ-1= 1024
1 (2)ⁿ-1 = 1024
(2)ⁿ-1 = 2^10
n - 1 = 10
n = 10 + 1
n = 11

Hence, the 11th term of the given sequence is 1024.

HOPE THIS WILL HELP YOU….

Answer 2

Solution :

i ) Given G.P is 5,2 ,4/5,8/25 ,...,125/15625

first term = a = a1 = 5 ,

common ratio = r = a2/a1 = 2/5

nth term = an = ar^n-1

an = 128/15625

=> 5 × ( 2/5 )^n-1 = 128/15625

=> (2/5)^n-1 = 128/(5 × 15725 )

=> (2/5)^n-1 = 2^7/5^7

=> (2/5)^n-1 = (2/5)^7

=> n - 1 = 7

[ Since , if a^m = a^n then m = n ]

=> n = 8

Therefore ,

8th term in G.P is 128/15625

ii ) Given G.P 1 , 2 , 4 , ....,1024

first term = a = a1 = 1 ,

Common ratio ( r ) = a2/a1 = 2/1 = 2

nth term = an = ar^n-1

1 × 2^n-1 = 1024

=> 2^n-1 = 2^10

=> n - 1 = 10

=> n = 11

Therefore ,

11th term in G.P is 1024.

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