Question

Which term of the following sequences:(a) 2,2 2,4.. is 128?5.(b) 3,3,33...is729?'is3 9 27'19683

Answer 1

(a)

The given sequence is 2,2 2 ,4……128.

Let the first term and common ratio of the given G.P. be a and r respectively.

Let there be n terms between 2 and 128 in the given sequence.

Here,

a=2 r= 2 2 2 = 2 a n =128

The formula for n th term of a G.P. is given by,

a n =a r n−1 (1)

Substitute the values of a and r in equation (1) to obtain the corresponding terms.

128=2× ( 2 ) n−1 ( 2 ) 7 =2× ( 2 ) n−1 ( 2 ) 6 = ( 2 ) n−1 ( 2 ) 12 = ( 2 ) n−1

Equate the coefficients on both the sides.

n−1=12 n=13

Thus, the total terms between the sequence 2,2 2 ,4……128 is 13.

(b)

The given sequence is 3 ,3,3 3 ,……729.

Let the first term and common ratio of the given G.P. be a and r respectively.

Let there be n terms between 3 and 729 in the given sequence.

Here,

a= 3 r= 3 3 = 3 a n =729

Substitute the values of a and r in equation (1) to obtain the corresponding terms.

729= 3 × ( 3 ) n−1 ( 3 ) 6 = ( 3 ) 1 2 × ( 3 ) n−1 2 ( 3 ) 6 = ( 3 ) n−1 2 + 1 2

Equate the coefficients on both the sides.

n−1 2 + 1 2 =6 n−1+1 2 =6 n 2 =6 n=12

Thus, the total terms between the sequence 3 ,3,3 3 ,……729is 12.

(c)

The given sequence is 1 3 , 1 9 , 1 27 …… 1 19683

Let the first term and common ratio of the given G.P. be a and r respectively.

Let there be n terms between 1 3 and 1 19683 in the given sequence.

Here,

a= 1 3 r= 1 9 1 3 =3 a n = 1 19683

Substitute the values of a and r in equation (1) to obtain the corresponding terms.

1 19683 = 1 3 × ( 1 3 ) n−1 ( 1 3 ) 9 = ( 1 3 ) n

Equate the coefficients on both the sides.

n=9

Thus, the total terms between the sequence 1 3 , 1 9 , 1 27 …… 1 19683 is 9.