RD Chapter 20 Geometric Progressions Ex 20.1 Solutions
Question - 1 : - Show that each one of the following progressions is a G.P. Also, find thecommon ratio in each case:
(i) 4, -2, 1, -1/2, ….
(ii) -2/3, -6, -54, ….
(iii) a, 3a2/4, 9a3/16, ….
(iv) ½, 1/3, 2/9, 4/27, …
Answer - 1 : -
(i) 4, -2, 1, -1/2, ….
Let a = 4, b = -2, c =1
In GP,
b2 =ac
(-2)2 =4(1)
4 = 4
So, the Common ratio =r = -2/4 = -1/2
(ii) -2/3, -6, -54, ….
Let a = -2/3, b = -6,c = -54
In GP,
b2 =ac
(-6)2 =-2/3 × (-54)
36 = 36
So, the Common ratio =r = -6/(-2/3) = -6 × 3/-2 = 9
(iii) a, 3a2/4,9a3/16, ….
Let a = a, b = 3a2/4,c = 9a3/16
In GP,
b2 =ac
(3a2/4)2 =9a3/16 × a
9a4/4 = 9a4/16
So, the Common ratio =r = (3a2/4)/a = 3a2/4a = 3a/4
(iv) ½, 1/3, 2/9, 4/27, …
Let a = 1/2, b = 1/3,c = 2/9
In GP,
b2 =ac
(1/3)2 =1/2 × (2/9)
1/9 = 1/9
So, the Common ratio =r = (1/3)/(1/2) = (1/3) × 2 = 2/3
Question - 2 : - Show that the sequence defined by an = 2/3n,n ∈ N is a G.P.
Answer - 2 : -
Given:
an =2/3n
Let us consider n = 1,2, 3, 4, … since n is a natural number.
So,
a1 =2/3
a2 =2/32 = 2/9
a3 =2/33 = 2/27
a4 =2/34 = 2/81
In GP,
a3/a2 =(2/27) / (2/9)
= 2/27 × 9/2
= 1/3
a2/a1 =(2/9) / (2/3)
= 2/9 × 3/2
= 1/3
∴ Common ratio ofconsecutive term is 1/3. Hence n ∈ N is a G.P.
Question - 3 : - Find:
(i) the ninth term of the G.P. 1, 4, 16, 64, ….
(ii) the 10th term of the G.P. -3/4, ½, -1/3, 2/9, ….
(iii) the 8th term of the G.P. 0.3, 0.06, 0.012, ….
(iv) the 12th term of the G.P. 1/a3x3 ,ax, a5x5, ….
(v) nth term of the G.P. √3, 1/√3, 1/3√3, …
(vi) the 10th term of the G.P. √2, 1/√2, 1/2√2, ….
Answer - 3 : -
(i) the ninth term of theG.P. 1, 4, 16, 64, ….
We know that,
t1 = a= 1, r = t2/t1 = 4/1 = 4
By using the formula,
Tn =arn-1
T9 = 1(4)9-1
= 1 (4)8
= 48
(ii) the 10th termof the G.P. -3/4, ½, -1/3, 2/9, ….
We know that,
t1 = a= -3/4, r = t2/t1 = (1/2) / (-3/4) = ½ × -4/3 = -2/3
By using the formula,
Tn =arn-1
T10 =-3/4 (-2/3)10-1
= -3/4 (-2/3)9
= ½ (2/3)8
(iii) the 8th termof the G.P., 0.3, 0.06, 0.012, ….
We know that,
t1 = a= 0.3, r = t2/t1 = 0.06/0.3 = 0.2
By using the formula,
Tn =arn-1
T8 =0.3 (0.2)8-1
= 0.3 (0.2)7
(iv) the 12th termof the G.P. 1/a3x3 , ax, a5x5,….
We know that,
t1 = a= 1/a3x3, r = t2/t1 = ax/(1/a3x3)= ax (a3x3) = a4x4
By using the formula,
Tn =arn-1
T12 =1/a3x3 (a4x4)12-1
= 1/a3x3 (a4x4)11
= (ax)41
(v) nth term of the G.P.√3, 1/√3, 1/3√3, …
We know that,
t1 = a= √3, r = t2/t1 = (1/√3)/√3 = 1/(√3×√3) = 1/3
By using the formula,
Tn =arn-1
Tn =√3 (1/3)n-1
(vi) the 10th termof the G.P. √2, 1/√2, 1/2√2, ….
We know that,
t1 = a= √2, r = t2/t1 = (1/√2)/√2 = 1/(√2×√2) = 1/2
By using the formula,
Tn =arn-1
T10 =√2 (1/2)10-1
= √2 (1/2)9
= 1/√2 (1/2)8
Question - 4 : - Find the 4th term from the end of the G.P. 2/27, 2/9,2/3, …., 162.
Answer - 4 : -
The nth term from theend is given by:
an = l(1/r)n-1 where, l is the last term, r is the common ratio, n isthe nth term
Given: last term, l =162
r = t2/t1 =(2/9) / (2/27)
= 2/9 × 27/2
= 3
n = 4
So, an =l (1/r)n-1
a4 =162 (1/3)4-1
= 162 (1/3)3
= 162 × 1/27
= 6
∴ 4th termfrom last is 6.
Question - 5 : - Which term of the progression 0.004, 0.02, 0.1, …. is 12.5?
Answer - 5 : -
By using the formula,
Tn =arn-1
Given:
a = 0.004
r = t2/t1 =(0.02/0.004)
= 5
Tn =12.5
n = ?
So, Tn =arn-1
12.5 = (0.004) (5)n-1
12.5/0.004 = 5n-1
3000 = 5n-1
55 = 5n-1
5 = n-1
n = 5 + 1
= 6
∴ 6th termof the progression 0.004, 0.02, 0.1, …. is 12.5.
Question - 6 : - Which term of the G.P.:
(i) √2, 1/√2, 1/2√2, 1/4√2, … is 1/512√2 ?
(ii) 2, 2√2, 4, … is 128 ?
(iii) √3, 3, 3√3, … is 729 ?
(iv) 1/3, 1/9, 1/27… is 1/19683 ?
Answer - 6 : -
(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
Question - 7 : - Which term of the progression 18, -12, 8, … is512/729 ?
Answer - 7 : -
By using the formula,
Tn =arn-1
a = 18
r = t2/t1 =(-12/18)
= -2/3
Tn =512/729
n = ?
Tn =arn-1
512/729 = 18 (-2/3)n-1
29/(729 ×18) = (-2/3)n-1
29/36 ×1/2×32 = (-2/3)n-1
(2/3)8 =(-1)n-1 (2/3)n-1
8 = n – 1
n = 8 + 1
= 9
∴ 9th termof the Progression is 512/729
Question - 8 : - Find the 4th term from the end of the G.P. ½, 1/6, 1/18, 1/54, … , 1/4374
Answer - 8 : -
The nth term from theend is given by:
an = l(1/r)n-1 where, l is the last term, r is the common ratio, n isthe nth term
Given: last term, l =1/4374
r = t2/t1 =(1/6) / (1/2)
= 1/6 × 2/1
= 1/3
n = 4
So, an =l (1/r)n-1
a4 =1/4374 (1/(1/3))4-1
= 1/4374 (3/1)3
= 1/4374 × 33
= 1/4374 × 27
= 1/162
∴ 4th termfrom last is 1/162.