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, ar ^{2}, ar^{3},….ar^{n-1},… is given be the formula,**

The sum of first n term of a GP when r > 1

**S _{n} = a[(r^{n}-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 S _{n} = [a(1 – r^{n})] / (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

**S _{n} = 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

8^{th} term = ar^{7} = 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 = S_{3} = a(r³-1)r-1

Now, sum of first six terms = S_{6} = 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 = 2^{m} + 2m

∴tn = 2^{n} +2n

sum of n terms = ∑tn=∑(2^{n} +2n)

=∑2^{n} + ∑2n

make it two parts

i) ∑2^{n} = [2+2²+2³+……….+2^{n}]series is GP

here first term =a =2

common ratio =r = t2/t1 = 2²/2 = 2 >1

sum of n terms in GP = a(r^{n} -1)/(r-1)

= [2(2^{n} -1)]/(2-1)

=2(2^{n} -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(2^{n} – 1) +n(n+1)

–: End of Geometric Progression Class 10 RS Aggarwal Ex-11B Goyal Brothers ICSE Maths Solutions / Answer :–

Return to : OP Malhotra S Chand Solutions for ICSE Class-10 Maths

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