Geometric Progression Class 10 RS Aggarwal Ex-11B Goyal Brothers ICSE Maths Solutions Ch-11. We Provide Step by Step Solutions / Answer of Exe-11B Sum of n-Term of GP Questions S Chand OP Malhotra Maths. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Geometric Progression Class 10 RS Aggarwal Ex-11B Goyal Brothers ICSE Maths Solutions Ch-11
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 10th |
| Chapter-11 | Geometric Progression |
| Writer | OP Malhotra |
| Exe-11B | Sum of n-Term of GP |
| Edition | 2024-2025 |
Sum of n-Term of GP
The sum of first n terms of a GP a, ar, ar2, ar3,….arn-1,… is given be the formula,
The sum of first n term of a GP when r > 1
Sn = a[(rn-1)/(r-1)] Where, a = First term, r = Common ratio, n = Number of terms
The sum of first n term of a GP when r < 1
or Sn = [a(1 – rn)] / (1 – r) Where , a = First term, r = Common ratio, n = Number of terms
The sum of first n term of a GP when r = 1
Sn = na
Exercise- 11B
( Geometric Progression Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-11 )
Que-1: Find the sum of first 8 terms of the G.P. 1, 3, 9, 27, 81 , .. ….
Sol: Here, a = 1 and r = 3.
∴S8 = a[(r^8 −1)/(r−1)]
= 1 [(3^8 − 1)/(3−1)]
= (6561−1)/2
= 3280
Que-2: Find the sum of first 10 terms of the G.P. 1,√ 3, 3, 3√3, …….
Sol: 1,√3,3,……………..
common ratio = a2/a1 = √3/1
common ratio = √3
sum of n terms = Sn = {a(1-r^n)}/(1-r)
Here,
a = 1 , r = √3 and n = 10
Now we have,
S10 = {1(1-√3^10)}/(1-√3)
S10 = {1-243}/(1-√3)
S10 = {-242(1+√3)}/[(1-√3)(1+√3)]
S10 = {-242(1+√3)}/(-2)
S10 = 121(1+√3).
Que-3: Find the sum of first 9 terms of the G.P. 1,(-1/2),(1/4), (-1/8),…….
Sol: First term, a = 1
Common ratio, r = (-1/2)/1 = -1/2 …(∵ r < 1)
Number of terms to be added, n = 9
∴ Sn = a(1-r^n)/(1-r)
⇒ S8 = [1(1-(-1/2)^9)]/(1-(-1/2))
= [1-(-1/2)^9]/(1+(1/2))
= [1+(1/2)^9]/(3/2)
= (2/3)[1+(1/2)^9]
= (2/3)(1+(1/512))
= (2/3)×(513/512)
= 171/256
Que-4: Find the sum of first 6 terms of the G.P. 0.1, 0.01 , 0.001,……..
Sol: a = 0.1
r = 0.01/0.1 = 0.1 (r<1)
n = 6
sn = a(1 – r^n)/(1 – r)
S6 = 0.1[1 – (0.1)^6]/1 – 0.1
= 0.1[1-0.000001]/0.9
= (0.1×0.999999)/0.9
= 0.0999999/0.9
= 0.111111
Que-5: Find the sum of first 6 terms of the G.P. 1, (-1/3),(1/3²), (-1/3³), ……
Sol: First term = a = 1
Common ratio = r = a2 / a1
r = ( – 1/3 ) / 1
r = – 1/3 < 1
Sum of 6 terms = s6
S6 = [ a ( 1 – r^6 ) / ( 1 – r ) ]
= { 1 [ 1 – ( – 1/3 )^6 ]}/ [ 1 – ( – 1/3 ) ]
= [( 3^6 – 1) / 3^6] / [ ( 3 + 1 ) / 3 ]
= ( 3^6 – 1 ) / [ 3^6 ×( 4/3 )]
= ( 3^6 – 1 ) / ( 4 × 3^5 )
= (729-1)/(972)
= 728/972
= 182/243.
Que-6: The first term of the G.P. is 27 and its 8th term is 1/81. Find the sum of first seven terms of the G.P.
Sol: First term, a = 27
8th term = ar7 = 181
n = 10
Now,
ar^7/a = (1/81)/27
⇒ r^7 = 1/2187
⇒ r^7 = (1/3)^7
⇒ r = 1/3 …(∵ r < 1)
∴ Sn = a(1-r^n)(1-r)
⇒ S10 = 27(1-(1/3)^10)/(1-1/3)
= 27(1-(1/3)^10)(2/3)
= [(27×3)/2][1-(1/3)^10]
= (81/2)[1-(1/3)10]
= (81/2) [1-(1/59049)]
= (81/2)(59048/59049)
= 182/243.
Que-7: The 4th and the 7th terms of a G.P. are 1/27 and 1/729 respectively. Find the sum of first 6 terms of the G.P.
Sol: Let a be the first term and r be the common ratio of the G.P.
∴a^4 = (1/27)
⇒ ar^(4−1) = 1/27
⇒ ar³ = 1/27
⇒ (ar³)² = 1/27²
⇒ a²r^6 = 1/729
⇒ ar^6 = 1/729a …(i)
Similarly Similarly ,a^7 = 1/729
⇒ ar^(7−1) = 1/729
⇒ ar^6 = 1/729
From ⇒ ar^6 = 1/729a [ From (i)]
∴a = 1
Putting this in Putting this in a^4 = 1/27
⇒ ar³ = 1/3³
⇒ r³ = 1/3³
∴ r = 1/3
Now, sum of n terms of the G . P Now, sum of 6 terms of the G . P .,Sn = a[r^n − 1]/(r−1)
⇒ S6 = [1(1−(1/3)^6]/(1−(1/3))
⇒ S6 = [1(1-(1/729))]/(2/3)
= [1(728/729)] (3/2)
= 364/243.
Que-8: How many terms of the G.P. (2/9), (-1/3), (1/2), …….. must be taken to make the sum equal to 55/72 ?
Sol: Sn = {a(1-r^)} / (1-r) , where, Sn denotes the sum of n terms, ‘a’ is the first term.of GP, & r is common ratio.
Here, a = 2/9, r = -1/3 ÷ 2/9 = -3/2
=> 55/72 = {2/9 (1- (-3/2)^n ) } / { 1 – (-3/2) }
=> 55/72 = {2/9 ( 1 + (3/2)^n) / 5/2
=>( 55 * 5 * 9)/(72 * 2 *2) = (1+(3/2)^n )
=> (275/32) – 1 = (3/2)^n
=> 243/32 = (3/2)^n
=> (3/2)^5 = (3/2)^n
=> n = 5
Que-9: In a G.P. the ratio of the sum of first 3 terms is to that of first 6 terms is 125:152. Find the common ratio.
Sol: Let a be the first term and r be the common ratio of given G.P.
Now, sum of first three terms = S3 = a(r³-1)r-1
Now, sum of first six terms = S6 = a(r^6 – 1)r-1
It is given that
[a(r³-1)r-1]/[a(r^6 – 1)r-1] = 125/152
⇒ (r³-1)/(r^6 – 1) = 125/152
⇒ (r³-1)/{(r³)²-(1)²} = 125/152
⇒ (r³-1)/[(r³-1)(r³+1)] = 125/152
⇒ 1/(r³+1) = 125/152
⇒ r³+1 = 152/125
⇒ r³ = (152/125) – 1
= (152-125)/125
= 27/125
⇒ r = 3/5
Que-10: A manufacturer reckons that the value of a machine which costs him Rs 31250 depreciates each year by 20%. Find its value after 2 years.
Sol: Machin cost = Rs.31250
Decrease = (20/100)×31250
= Rs.6250
Therefore after 1 year
Price = Rs.31250-Rs.6250
= Rs.25000
Again decrease = (20/100)×25000
= Rs.5000
Therefore after 2 year
Price = Rs.25000-Rs.5000
= Rs.20000
Que-11: Find the sum of the following to n terms : 7 + 77 + 777 + 7777+ ……..
Sol: We have,
7 + 77 + 777 + … n terms
Sn = 7 [1 + 11 + 111 + … n terms]
n terms = (7/9)(9+99+999+… n terms )
=79{(10−1)+(10²−1)+(10³−1)+…+(10^n − 1)}
n times =(7/9){(10+10²+10³+…+1^0n)} − (1+1+1+1…n times )
= (7/9) [10×{(10^n − 1)/(10−1) − n] = (7/9){(10/9)(10^n − 1) − n}
= (7/81) {10^(n+1)−9n−10}
Que-12: Find the sum of n terms of a series whose mth term is 2^m + 2m
Sol: Given
tm = 2m + 2m
∴tn = 2n +2n
sum of n terms = ∑tn=∑(2n +2n)
=∑2n + ∑2n
make it two parts
i) ∑2n = [2+2²+2³+……….+2n]series is GP
here first term =a =2
common ratio =r = t2/t1 = 2²/2 = 2 >1
sum of n terms in GP = a(rn -1)/(r-1)
= [2(2n -1)]/(2-1)
=2(2n -1)——(1)
ii) sum of n terms =∑ 2n = 2∑n=2[1+2+3+….+n]
=2n(n+1)/2 since sum of n natural numbers = n(n+1)/2
=n(n+1)——(2)
therefore
∑tn sum of n terms in given series = (1) +(2)
= 2(2n – 1) +n(n+1)
–: End of Geometric Progression Class 10 RS Aggarwal Ex-11B Goyal Brothers ICSE Maths Solutions / Answer :–
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