Question

1. Find out which of the following sequences are geometric sequences. For those
geometric sequences, find the common ratio.
(i) 0.12, 0.24, 0.48,g. (ii) 0.004, 0.02, 0.1,g

Answer 1

QUESTION :
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
•If the ratio of any term other than the first term to its preceding term of a sequence is a non-zero constant, then it is a geometric sequence.

SOLUTION :
i) Here a1 = 0.12, a2 = 0.24, a3 = 0.48

Common ratio ( r ) = a(ⁿ+1)/ aⁿ
r = 0.24/0.12 = 0.48/0.24 = 2

Since the common ratio (r) is 2, the given sequence is a geometric sequence.

(ii) 0.004,0.02,0.1,..........
Here a1 = 0.004, a2 = 0.02, a3 = 0.1

Common ratio ( r ) = a(ⁿ+1)/ aⁿ
r = 0.02/0.004 = 0.1/0.02 = 5

Since the common ratio (r) is 5, the given sequence is a geometric sequence.

(iii) 1/2,1/3,2/9,4/47,...........
Here a1 = 1/2, a2 = 1/3, a3 = 2/9

Common ratio ( r ) = a(ⁿ+1)/ aⁿ
r = (1/3)/(1/2) = (2/9)/(1/3)= 2/3

Since the common ratio (r) is 2/3 , the given sequence is a geometric sequence.

(iv) 12,1,1/12,............
Here a1 = 12 , a2 = 1, a3 = 1/12

Common ratio ( r ) = a(ⁿ+1)/ aⁿ
r = 1/12 = (1/12)/(1)= 1/12

Since the common ratio (r) is 1/12 , the given sequence is a geometric sequence.


(v) √2,1/√2,1/(2√2),...........
Here a1 = √2 , a2 = 1/√2, a3 = 1/(2√2)

Common ratio ( r ) = a(ⁿ+1)/ aⁿ
r = (1/√2)/√2 = (1/2√2)/(1/√2)= 1/2

Since the common ratio (r) is 1/2 , the given sequence is a geometric sequence.

(vi) 4,-2,-1,-1/2,...............
Here a1 = -4 , a2 = -2, a3 = -1

Common ratio ( r ) = a(ⁿ+1)/ aⁿ
r = -2/4 = - 1/2 , (-1)/(-2) = 1/2

Since the common ratio (r) is different(-½,½) ,the given sequence is not a geometric sequence.

HOPE THIS WILL HELP YOU...

Answer 2

Solution :

i )Given 0.12,0.24,0.48 ,...

a2/a1 = 0.24/0.12 = 2 ,

a3/a2 = 0.48/0.24 = 2

Since a2/a1 = a3/a2 ,

So, given series in G.P .

ii) Given 0.004 , 0.002 , 0.1,....

a2/a1 = 0.002/0.004 = 1/2 ,

a3/a2 = 0.1/0.002 = 100/2

Since , a2/a1 ≠ a3/a2 ,

So, given series not in G.P.

iii ) Given 1/2 , 1/3 , 2/9 , .....

a2/a1 = ( 1/3 )/( 1/2 )= 2/3

a3/a2 = ( 2/9 )/( 1/3 ) = 2/3

Since , a2/a1 = a3/a2

So, given series in G.P.

iv ) Given 12 , 1 , 1/12 , ...

a2/a1 = 1/12

a3/a2 = ( 1/12 )/1 = 1/12

Since , a3/a2 = a2/a1

So, given series is in G.P

v ) Given √2 , 1/√2 , 2√2 , ....

a2/a1 = ( 1/√2 )/( √2 ) = 1/2 ,

a3/a2 = (2√2)/(1/√2) = 4

Since , a2/a1 ≠ a3/a2 ,

So, given series is not G.P.

vi ) 4 , -2 , -1 , -1/2 , ...

a2/a1 = ( -2 )/4 = -1/2

a3/a2 = ( -1 )/(-2 ) = 1/2

Since , a2/a1 ≠ a3/a2

So, given series in not G.P

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