RD Chapter 20 Geometric Progressions Ex 20.3 Solutions
Question - 1 : - Find the sum of the following geometric progressions:
(i) 2, 6, 18, … to 7 terms
(ii) 1, 3, 9, 27, … to 8 terms
(iii) 1, -1/2, ¼, -1/8, …
(iv) (a2 – b2), (a – b), (a-b)/(a+b), … to nterms
(v) 4, 2, 1, ½ … to 10 terms
Answer - 1 : -
(i) 2, 6, 18, … to 7 terms
We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)
Given:
a = 2, r = t2/t1 =6/2 = 3, n = 7
Now let us substitutethe values in
a(rn –1)/(r – 1) = 2 (37 – 1)/(3-1)
= 2 (37 –1)/2
= 37 –1
= 2187 – 1
= 2186
(ii) 1, 3, 9, 27, … to 8terms
We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)
Given:
a = 1, r = t2/t1 =3/1 = 3, n = 8
Now let us substitutethe values in
a(rn –1)/(r – 1) = 1 (38 – 1)/(3-1)
= (38 –1)/2
= (6561 – 1)/2
= 6560/2
= 3280
(iii) 1, -1/2, ¼, -1/8, …
We know that, sum ofGP for infinity = a/(1 – r)
Given:
a = 1, r = t2/t1 =(-1/2)/1 = -1/2
Now let us substitutethe values in
a/(1 – r) = 1/(1 –(-1/2))
= 1/(1 + 1/2)
= 1/((2+1)/2)
= 1/(3/2)
= 2/3
(iv) (a2 –b2), (a – b), (a-b)/(a+b), … to n terms
We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)
Given:
a = (a2 –b2), r = t2/t1 = (a-b)/(a2 –b2) = (a-b)/(a-b) (a+b) = 1/(a+b), n = n
Now let us substitutethe values in
a(rn –1)/(r – 1) =
(v) 4, 2, 1, ½ … to 10terms
We know that, sum ofGP for n terms = a(rn – 1)/(r – 1)
Given:
a = 4, r = t2/t1 =2/4 = 1/2, n = 10
Now let us substitutethe values in
a(rn –1)/(r – 1) = 4 ((1/2)10 – 1)/((1/2)-1)
= 4 ((1/2)10 –1)/((1-2)/2)
= 4 ((1/2)10 –1)/(-1/2)
= 4 ((1/2)10 –1) × -2/1
= -8 [1/1024 -1]
= -8 [1 – 1024]/1024
= -8 [-1023]/1024
= 1023/128
Question - 2 : - Find the sum of the following geometric series :
(i) 0.15 + 0.015 + 0.0015 + … to 8 terms;
(ii) √2 + 1/√2 + 1/2√2 + …. to 8 terms;
(iii) 2/9 – 1/3 + ½ – ¾ + … to 5 terms;
(iv) (x + y) + (x2 + xy + y2) + (x3 +x2 y + xy2 + y3) + …. to n terms ;
(v) 3/5 + 4/52 + 3/53 + 4/54 +… to 2n terms;
Answer - 2 : -
(i) 0.15 + 0.015 + 0.0015+ … to 8 terms
Given:
a = 0.15
r = t2/t1 =0.015/0.15 = 0.1 = 1/10
n = 8
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
a(1 – rn )/(1– r) = 0.15 (1 – (1/10)8) / (1 – (1/10))
= 0.15 (1 – 1/108)/ (1/10)
= 1/6 (1 – 1/108)
(ii) √2 + 1/√2 + 1/2√2 + ….to 8 terms;
Given:
a = √2
r = t2/t1 =(1/√2)/√2 = 1/2
n = 8
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
a(1 – rn )/(1– r) = √2 (1 – (1/2)8) / (1 – (1/2))
= √2 (1 – 1/256) /(1/2)
= √2 ((256 – 1)/256) ×2
= √2 (255×2)/256
= (255√2)/128
(iii) 2/9 – 1/3 + ½ – ¾ + …to 5 terms;
Given:
a = 2/9
r = t2/t1 =(-1/3) / (2/9) = -3/2
n = 5
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
a(1 – rn )/(1– r) = (2/9) (1 – (-3/2)5) / (1 – (-3/2))
= (2/9) (1 + (3/2)5)/ (1 + 3/2)
= (2/9) (1 + (3/2)5)/ (5/2)
= (2/9) (1 + 243/32) /(5/2)
= (2/9) ((32+243)/32)/ (5/2)
= (2/9) (275/32) × 2/5
= 55/72
(iv) (x + y) + (x2 +xy + y2) + (x3 + x2 y + xy2 +y3) + …. to n terms;
Let Sn =(x + y) + (x2 + xy + y2) + (x3 + x2 y+ xy2 + y3) + …. to n terms
Let us multiply anddivide by (x – y) we get,
Sn =1/(x – y) [(x + y) (x – y) + (x2 + xy + y2) (x – y)… upto n terms]
(x – y) Sn =(x2 – y2) + x3 + x2y + xy2 –x2y – xy2 – y3..upto n terms
(x – y) Sn = (x2 +x3 + x4+…n terms) – (y2 + y3 +y4 +…n terms)
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
We have two G.Ps inabove sum, so,
(x – y) Sn =x2 [(xn – 1)/ (x – 1)] – y2 [(yn –1)/ (y – 1)]
Sn =1/(x-y) {x2 [(xn – 1)/ (x – 1)] – y2 [(yn –1)/ (y – 1)]}
(v) 3/5 + 4/52 +3/53 + 4/54 + … to 2n terms;
The series can bewritten as:
3 (1/5 + 1/53 +1/55+ … to n terms) + 4 (1/52 + 1/54 +1/56 + … to n terms)
Firstly let usconsider 3 (1/5 + 1/53 + 1/55+ … to n terms)
So, a = 1/5
r = t2/t1 =1/52 = 1/25
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
Now, Let us consider 4(1/52 + 1/54 + 1/56 + … to nterms)
So, a = 1/25
r = t2/t1 =1/52 = 1/25
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
Question - 3 : - Evaluate the following:
Answer - 3 : - 
= (2 + 31)+ (2 + 32) + (2 + 33) + … + (2 + 311)
= 2×11 + 31 +32 + 33 + … + 311
= 22 + 3(311 –1)/(3 – 1) [by using the formula, a(1 – rn )/(1 – r)]
= 22 + 3(311 –1)/2
= [44 + 3(177147 –1)]/2
= [44 + 3(177146)]/2
= 265741
= (2 + 30)+ (22 + 3) + (23 + 32) + … + (2n +3n-1)
= (2 + 22 +23 + … + 2n) + (30 + 31 +32 + …. + 3n-1)
Firstly let usconsider,
(2 + 22 +23 + … + 2n)
Where, a = 2, r = 22/2= 4/2 = 2, n = n
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
= 2 (2n –1)/(2 – 1)
= 2 (2n –1)
Now, let us consider
(30 +31 + 32 + …. + 3n)
Where, a = 30 =1, r = 3/1 = 3, n = n
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
= 1 (3n –1)/ (3 – 1)
= (3n –1)/2
So,
= (2 + 22 +23 + … + 2n) + (30 + 31 +32 + …. + 3n)
= 2 (2n –1) + (3n – 1)/2
= ½ [2n+2 +3n – 4 – 1]
= ½ [2n+2 +3n – 5]
= 42 +43 + 44 + … + 410
Where, a = 42 =16, r = 43/42 = 4, n = 9
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
= 16 (49 –1)/(4 – 1)
= 16 (49 –1)/3
= 16/3 [49 –1]
Find the sum of the following series :
(i) 5 + 55 + 555 + … to n terms.
(ii) 7 + 77 + 777 + … to n terms.
(iii) 9 + 99 + 999 + … to n terms.
(iv) 0.5 + 0.55 + 0.555 + …. to n terms
(v) 0.6 + 0.66 + 0.666 + …. to n terms.
Answer - 4 : -
(i) 5 + 55 + 555 + … to nterms.
Let us take 5 as acommon term so we get,
5 [1 + 11 + 111 + … nterms]
Now multiply anddivide by 9 we get,
5/9 [9 + 99 + 999 + …n terms]
5/9 [(10 – 1) + (102 –1) + (103 – 1) + … n terms]
5/9 [(10 + 102 +103 + … n terms) – n]
So the G.P is
5/9 [(10 + 102 +103 + … n terms) – n]
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
Where, a = 10, r = 102/10= 10, n = n
a(rn –1 )/(r – 1) =
(ii) 7 + 77 + 777 + … to nterms.
Let us take 7 as acommon term so we get,
7 [1 + 11 + 111 + … ton terms]
Now multiply and divideby 9 we get,
7/9 [9 + 99 + 999 + …n terms]
7/9 [(10 – 1) + (102 –1) + (103 – 1) + … + (10n – 1)]
7/9 [(10 + 102 +103 + … +10n)] – 7/9 [(1 + 1 + 1 + … to n terms)]
So the terms are inG.P
Where, a = 10, r = 102/10= 10, n = n
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
7/9 [10 (10n –1)/(10-1)] – n
7/9 [10/9 (10n –1) – n]
7/81 [10 (10n –1) – n]
7/81 (10n+1 –9n – 10)
(iii) 9 + 99 + 999 + … to nterms.
The given terms can bewritten as
(10 – 1) + (100 – 1) +(1000 – 1) + … + n terms
(10 + 102 +103 + … n terms) – n
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
Where, a = 10, r = 10,n = n
a(rn –1 )/(r – 1) = [10 (10n – 1)/(10-1)] – n
= 10/9 (10n –1) – n
= 1/9 [10n+1 –10 – 9n]
= 1/9 [10n+1 –9n – 10]
(iv) 0.5 + 0.55 + 0.555 +…. to n terms
Let us take 5 as acommon term so we get,
5(0.1 + 0.11 + 0.111 +…n terms)
Now multiply anddivide by 9 we get,
5/9 [0.9 + 0.99 +0.999 + …+ to n terms]
5/9 [9/10 + 9/100 +9/1000 + … + n terms]
This can be written as
5/9 [(1 – 1/10) + (1 –1/100) + (1 – 1/1000) + … + n terms]
5/9 [n – {1/10 + 1/102 +1/103 + … + n terms}]
5/9 [n – 1/10{1-(1/10)n}/{1 – 1/10}]
5/9 [n – 1/9 (1 – 1/10n)]
(v) 0.6 + 0.66 + 0.666 +…. to n terms.
Let us take 6 as acommon term so we get,
6(0.1 + 0.11 + 0.111 +…n terms)
Now multiply anddivide by 9 we get,
6/9 [0.9 + 0.99 +0.999 + …+ n terms]
6/9 [9/10 + 9/100 +9/1000 + …+ n terms]
This can be written as
6/9 [(1 – 1/10) + (1 –1/100) + (1 – 1/1000) + … + n terms]
6/9 [n – {1/10 + 1/102 +1/103 + … + n terms}]
6/9 [n – 1/10{1-(1/10)n}/{1 – 1/10}]
6/9 [n – 1/9 (1 – 1/10n)]
Question - 5 : - How many terms of the G.P. 3, 3/2, ¾,… Be taken together to make 3069/512 ?
Answer - 5 : -
Given:
Sum of G.P = 3069/512
Where, a = 3, r =(3/2)/3 = 1/2, n = ?
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
3069/512 = 3 ((1/2)n –1)/ (1/2 – 1)
3069/512×3×2 = 1 –(1/2)n
3069/3072 – 1 = –(1/2)n
(3069 – 3072)/3072 = –(1/2)n
-3/3072 = – (1/2)n
1/1024 = (1/2)n
(1/2)10 =(1/2)n
10 = n
∴ 10 terms are requiredto make 3069/512
Question - 6 : - How many terms of the series 2 + 6 + 18 + ….Must be taken to make the sum equal to 728?
Answer - 6 : -
Given:
Sum of GP = 728
Where, a = 2, r = 6/2= 3, n = ?
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
728 = 2 (3n –1)/(3-1)
728 = 2 (3n –1)/2
728 = 3n –1
729 = 3n
36 = 3n
6 = n
∴ 6 terms are requiredto make a sum equal to 728
Question - 7 : - How many terms of the sequence √3, 3,3√3,… must be taken to make the sum 39+ 13√3 ?
Answer - 7 : -
Given:
Sum of GP = 39 +13√3
Where, a =√3, r = 3/√3= √3, n = ?
By using the formula,
Sum of GP for n terms =a(rn – 1 )/(r – 1)
39 + 13√3 = √3 (√3n –1)/ (√3 – 1)
(39 + 13√3) (√3 – 1) =√3 (√3n – 1)
Let us simplify weget,
39√3 – 39 + 13(3) –13√3 = √3 (√3n – 1)
39√3 – 39 + 39 – 13√3= √3 (√3n – 1)
39√3 – 39 + 39 – 13√3= √3n+1 – √3
26√3 + √3 = √3n+1
27√3 = √3n+1
√36 √3= √3n+1
6+1 = n + 1
7 = n + 1
7 – 1 = n
6 = n
∴ 6 terms are requiredto make a sum of 39 + 13√3
Question - 8 : - The sum of n terms of the G.P. 3, 6, 12, … is 381. Find the value of n.
Answer - 8 : -
Given:
Sum of GP = 381
Where, a = 3, r = 6/3= 2, n = ?
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
381 = 3 (2n –1)/ (2-1)
381 = 3 (2n –1)
381/3 = 2n –1
127 = 2n –1
127 + 1 = 2n
128 = 2n
27 = 2n
n = 7
∴ value of n is 7
Question - 9 : - The common ratio of a G.P. is 3, and the last term is 486. If the sum of these terms be 728, find the first term.
Answer - 9 : -
Given:
Sum of GP = 728
Where, r = 3, a = ?
Firstly,
Tn =arn-1
486 = a3n-1
486 = a3n/3
486 (3) = a3n
1458 = a3n ….Equation (i)
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
728 = a (3n –1)/2
1456 = a3n –a … equation (2)
Subtracting equation(1) from (2) we get
1458 – 1456 = a.3n –a.3n + a
a = 2.
∴ The first termis 2
Question - 10 : - The ratio of the sum of the first three terms is to that of the first 6 terms of a G.P. is 125 : 152. Find the common ratio.
Answer - 10 : -
Given:
Sum of G.P of 3 termsis 125
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
125 = a (rn –1)/(r-1)
125 = a (r3 –1)/ (r-1) … equation (1)
Now,
Sum of G.P of 6 termsis 152
By using the formula,
Sum of GP for n terms= a(rn – 1 )/(r – 1)
152 = a (rn –1)/(r-1)
152 = a (r6 –1)/ (r-1) … equation (2)
Let us divide equation(i) by (ii) we get,
125/152 = [a (r3 –1)/ (r-1)] / [a (r6 – 1)/ (r-1)]
125/152 = (r3 –1)/(r6 – 1)
125/152 = (r3 –1)/[(r3 – 1) (r3 + 1)]
125/152 = 1/(r3 +1)
125(r3 +1) = 152
125r3 +125 = 152
125r3 =152 – 125
125r3 =27
r3 =27/125
r3 = 33/53
r = 3/5
∴ The common ratiois 3/5