Question -
Answer -
(i) √2, 1/√2, 1/2√2,1/4√2, … is 1/512√2 ?
By using the formula,
Tn =arn-1
a = √2
r = t2/t1 =(1/√2) / (√2)
= 1/2
Tn =1/512√2
n = ?
Tn =arn-1
1/512√2 = (√2) (1/2)n-1
1/512√2×√2 = (1/2)n-1
1/512×2 = (1/2)n-1
1/1024 = (1/2)n-1
(1/2)10 =(1/2)n-1
10 = n – 1
n = 10 + 1
= 11
∴ 11th termof the G.P is 1/512√2
(ii) 2, 2√2, 4, … is 128 ?
By using the formula,
Tn =arn-1
a = 2
r = t2/t1 =(2√2/2)
= √2
Tn =128
n = ?
Tn =arn-1
128 = 2 (√2)n-1
128/2 = (√2)n-1
64 = (√2)n-1
26 =(√2)n-1
12 = n – 1
n = 12 + 1
= 13
∴ 13th termof the G.P is 128
(iii) √3, 3, 3√3, … is 729 ?
By using the formula,
Tn =arn-1
a = √3
r = t2/t1 =(3/√3)
= √3
Tn =729
n = ?
Tn =arn-1
729 = √3 (√3)n-1
729 = (√3)n
36 =(√3)n
(√3)12 =(√3)n
n = 12
∴ 12th termof the G.P is 729
(iv) 1/3, 1/9, 1/27… is1/19683 ?
By using the formula,
Tn =arn-1
a = 1/3
r = t2/t1 =(1/9) / (1/3)
= 1/9 × 3/1
= 1/3
Tn =1/19683
n = ?
Tn =arn-1
1/19683 = (1/3) (1/3)n-1
1/19683 = (1/3)n
(1/3)9 =(1/3)n
n = 9
∴ 9th termof the G.P is 1/19683