Question

Question 5 Which term of the following sequences:

(a) 2, 2 * 2^0.5, 4,... is 128?
(b) 3^(1/2), 3, 3 * 3^0.5,... is 729?
(c) 1/3, 1/9, 1/27,... is 1/19683?

Class X1 - Maths -Sequences and Series Page 192

Answer 1

(a) the series 2, 2√2, 4 , ........128 is in GP
here, first term (a) = 2
common ratio ( r) = 2√2/2 =√2

use formula ,
T_n = ar^(n-1)
where n is number of terms .

now,
128 = 2(√2)^(n-1)
64 =(2½)^(n-1)
(2)^6 = (2)^{(n-1)/2}

we know,
x^m = x^n then, m = n , use this concept here,

6 = {(n-1)/2}
12 = n -1
n = 13

_______________________________
(b) √3, 3 , 3√3 , .....729 are in GP.
here, a = √3 , r = 3/√3 = √3
use formula ,
T_n = ar^(n-1)
729 = (√3)(√3)^(n-1)
(3)^6 = (3½)(3)^{(n-1)/2}
3^6 = (3)^{(n-1+1)/2}
3^6 = 3^{n/2}
6 = n/2
n = 12

______________________________
(c) 1/3, 1/9, 1/27, .........1/19683
here, a = 1/3 and r = (1/9)/(1/3) = 1/3
use formula,
T_n = ar^(n-1)
1/19683 = (1/3)(1/3)^(n-1)
1/6561 = (1/3)^(n-1)
(1/3)^8 = (1/3)^(n-1) [ because, 6561 = (3)^8
8 = n -1
n = 9