Geometric Progression Class 10 RS Aggarwal Exe-11A Goyal Brothers ICSE Maths Solutions Ch-11. We Provide Step by Step Solutions / Answer of Exe-11A General Term of GP Questions . Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

## Geometric Progression Class 10 RS Aggarwal Exe-11A Goyal Brothers ICSE Maths Solutions Ch-11

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-11 | Geometric Progression |

Writer Book | RS Aggarwal |

Exe-11A | General Term of GP Questions |

Edition | 2024-2025 |

### General Term of GP Questions

**The general form of Geometric Progression is a, ar, ar ^{2}, ar^{3}, ar^{4},…, ar^{n-1 }**Where, a = First term, r = common ratio,

nth Term of GP = ar^{n-1}

##### Similarly, nth term, t_{n} = ar^{n-1}

**Exercise- 11A**

( Geometric Progression Class 10 RS Aggarwal Exe-11A Goyal Brothers ICSE Maths Solutions Ch-11 )

**Que-1: Show that the progression 2,6,18,54,162,…….. is a G.P. Write its (i) first term (ii) common ratio (iii) nth term (iv) 8th term**

**Sol: **2, 6, 18, 54, …

t1 = 2, t2 = 6, t3 = 18, t4 = 54, …

Here, t2/t1 = t3/t2 = t4/t3 = 3

∵ the ratio of any two consecutive terms is a constant, hence the given sequence is a Geometric progression.

(i) a = 2

(ii) r = 3

(iii) tn = ar^n-1

∴ tn = 2×3^n-1

(iv) T8 = ar^8-1

∴ t8 = 2×3^7

= 4374.

**Que-2: Show that the progression 625,125,25,5,1,(1/5),……….. is a G.P. ****Write its (i) first term (ii) common ratio (iii) nth term (iv) 10th term**

**Sol: **625,125,25,5,1,(1/5),………..

t1 = 625, t2 = 125, t3 = 25, t4 = 5, …

Here, t2/t1 = t3/t2 = t4/t3

= 125/625 = 1/5.

∵ the ratio of any two consecutive terms is a constant, hence the given sequence is a Geometric progression.

(i) a = 625

(ii) r = 1/5

(iii) tn = ar^n-1

∴ tn = 625×1/5^n-1

tn = 625/5^(n-1)

(iv) T10 = ar^10-1

∴ t10 = 625×(1/5)^9

= 1/3125.

**Que-3: Show that the progression -27,9,-3,1,(-1/3),……… is a G.P. ****Write its (i) first term (ii) common ratio (iii) nth term (iv) 9th term**

**Sol: **-27,9,-3,1,(-1/3),………

t1 = -27, t2 = 9, t3 = -3, t4 = 1, …

Here, t2/t1 = t3/t2 = t4/t3

= 9/-27 = -1/3.

∵ the ratio of any two consecutive terms is a constant, hence the given sequence is a Geometric progression.

(i) a = -27

(ii) r = -1/3

(iii) tn = ar^n-1

∴ tn = -27×(-1/3)^n-1

tn = 27/[3^(n-1)]

(iv) T9 = ar^9-1

∴ t10 = -27×(-1/3)^8

= -27/6561 = -1/243.

**Que-4: Show that the progression 2,2√2,4,4√2,…….. is a G.P.**

**Write its (i) first term (ii) common ratio (iii) nth term (iv) 11th term**

**Sol: **2,2√2,4,4√2,……..

t1 = 2, t2 = 2√2, t3 = 4, t4 = 4√2, …

Here, t2/t1 = t3/t2 = t4/t3

= 2√2/2 = √2.

∵ the ratio of any two consecutive terms is a constant, hence the given sequence is a Geometric progression.

(i) a = 2

(ii) r = √2

(iii) tn = ar^n-1

∴ tn = 2×√2^n-1

tn = √2²×√2^n-1

tn = (√2)^(n+1)

(iv) T11 = ar^11-1

∴ t11 = 2×√2^10

= 64.

**Que-5: Show that the progression (-3/4),(1/2),(-1/3),(2/9),……… is a G.P. ****Write its (i) first term (ii) common ratio (iii) nth term (iv) 6th term**

**Sol: **(-3/4),(1/2),(-1/3),(2/9),………

t1 = -3/4, t2 = 1/2, t3 = -1/3, t4 = 2/9, …

Here, t2/t1 = t3/t2 = t4/t3

= (1/2)/(-3/4) = -2/3.

∵ the ratio of any two consecutive terms is a constant, hence the given sequence is a Geometric progression.

(i) a = -3/4

(ii) r = -2/3

(iii) tn = ar^n-1

∴ tn = (-3/4)×(-2/3)^n-1.

(iv) T6 = ar^6-1

∴ t10 = (-3/4)×(-2/3)^5

= (-3/4)×(-32/243)

= 8/81.

**Que-6: Show that the progression 0.4,0.8,1.6,…….. is a G.P.**

**Write its (i) first term (ii) common ratio (iii) nth term (iv) 7th term**

**Sol: **0.4,0.8,1.6,……..

t1 = 0.4, t2 = 0.8, t3 = 1.6, …

Here, t2/t1 = t3/t2 = t4/t3

= 0.8/0.4 = 2.

∵ the ratio of any two consecutive terms is a constant, hence the given sequence is a Geometric progression.

(i) a = 0.4

(ii) r = 2

(iii) tn = ar^n-1

∴ tn = 0.4×2^(n-1)

(iv) An = 2(2^(n – 1))/5

A7 = 2(2^(7 – 1))/5

A7 = 2(2^(6))/5

A7 = 128/5

**Que-7: Which term of the G.P. 3,6,12,24,……. is 768?**

**Sol: **we know that, a = 3 and r = 2

Now ar^(n-1) = 768

or r^(n-1) = 768/3 = 256 = 2^(n-1)

But 256 = 2^8 which in same as 2^(n-1).

Hence n-1 = 8

n = 9

so the 9th term of the GP is 768

**Que-8: Which term of the G.P. 5,10,20,40,…….. is 640?**

**Sol: **Given: The sequence 5, 10, 20, 40, ….. is a GP.

Here, we have to find which term of the given sequence is 640.

As we know that, the the general term of a GP is given by: an = ar^(n−1)

Here, a = 5, r = 2 and let a_{n} = 640

⇒ 640 = (5) ⋅ (2)^{n – 1
}⇒ 2^{(n – 1)} = 128 = 2^{7
}⇒ n – 1 = 7

⇒ n = 8.

**Que-9: Which term of the G.P. √3,3,3√3,9,……. is 729?**

**Sol: **It is given that, first term, a = √3 and common ratio

r = a2/a1 = 3/√3 = √3

Let the nth term be 729. i.e., an = 729

As we know that, an = ar^(n−1)

⇒ ar^(n−1) = 729

⇒ (√3).(√3)^(n−1) = 729

⇒ (√3)^n = (√3)^12

Comparing on both sides then we get,

n = 12.

**Que-10: Find the G.P. whose 5th and 8th term are 80 and 640 respectively**

**Sol: **Here it is given that 5th term is 80 and 8th term is 640

Let the first term is a and the common ratio is r

a5 = ar^4 = 80…..(i)

a8 = ar^7 = 640……(ii)

Divide (ii) and (i)

ar^7/ar^4 = 640/80

r^3 = 8

r = 2

Put the value of 2 in eq (i)

= ar^4 = 80

a x 2^4 = 80

a = 80/16

a = 5

Second term,

ar = 5 x 2 = 10

Third term

ar^2 = 5 x 2^2 = 20

Fourth term

ar^3 = 5 x 2^3 = 40

Therefore, the required GP is 5,10,20,40,80…….

**Que-11: The 4th term of a G.P. is 16 and the 7th term is 128. Find the first term and common ratio of the series.**

**Sol: **The 4th term of a G.P. = 16

⇒ ar^(4-1) = 16

7th term of G.P. = 128

⇒ ar^(7-1) = 128

so, (ar³)/(ar^6) = 16/128

⇒ 1/r³ = 1/8

=> r = 2

ar³ = 16

a × 2³ = 16

a x 8 = 16

a = 2

**Que-12: Find the G.P. whose 4th and 7th are (1/18) and (-1/486) respectively.**

**Sol: **Let the first term of the G.P. be a and its common ratio be r.

4^{th} term =1/18

⇒ ar³ = 1/18

7^{th} term =1/486

⇒ ar^6 = 1/486

Now, ar^6/ar³ = (-1/486)/(1/18)

⇒ r³ = -1/27

⇒ r =-1/3

ar³ = 1/18

⇒ a×(-1/3)³ = 1/18

⇒ a = -27/18 = -3/2

∴ G.P. = a, ar, ar^{2}, ar^{3}, …….

= -3/2, -3/2×(-1/3), -3/2×(-1/3)², 118,…….

= -3/2, 1/2, -1/6, 1/18,…….

**Que-13: For what value of x, the numbers (x+9),(x-6) and 4 are in G.P.?**

**Sol: **Given: The numbers (x + 9), (x – 6) and 4 in GP

As we know that, if a, b and c are in GP then b² = ac

Here, a = x + 9, b = x – 6 and c = 4

⇒ (x – 6)^{2} = 4 × (x + 9)

⇒ x^{2} – 12x + 36 = 4x + 36

⇒ x^{2} – 16x = 0

⇒ x = 0 or 16

**Que-14: Find the 6th term from the end of the G.P. 16,8,4,2,……, 1/512.**

**Sol: **a = 8

r = 4/8 = 1/2

Formula for nth term in G.P. :

An = a*(r)^n

1/1024 = 16 * (1/2)^n

(1/2)^n = 1/(1024*16)

(1/2)^n = 1/16384

2^(-n) = 1/16384

But, 2^(14) = 16384

So, 2^(-14) = 1/16384

2^(-n) = 2^(-14)

Since base is the same , so the powers can be equated.

-n = -14

n = 14

An = a*(r)^n

A8 = 16 * (1/2)^(8)

A8 = (2)^(4) * (2)^(-8)

A8 = (2)^(-4)

A8 = 1/16.

**Que-15: Find the 4th term from the end of the G.P. (2/81),(2/27),(2/9),……. 54.**

**Sol: **a = 2/81

r = (2/27)/(2/81) = 3

Tn = 54

Tn = a⋅r^(n−1)

54 = (2/81).3^(n-1)

54×81 = 2.3^(n-1)

4374 = 2.3^(n-1)

2187 = 3^(n-1)

3^7 = 3^(n-1)

n-1 = 7

n = 8

The 4th term from the end in a G.P. is the same as the (n−3)th term from the beginning. Since n = 8, the 4th term from the end is the 5th term from the beginning.

T5 = (2/81).3^(5-1)

= (2/81).3^4

= (2/81)×81

= 2.

**Que-16: The 4th,6th and the last term of a geometric progressions are 10,40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series.**

**Sol: **T4 = 10

ar³ = 10 ……….. (1)

T6 = 40

ar^5 = 40 ………. (2)

Do Equation (2) ÷ equation (1)

ar^5/ar³ = 40/10

r² = 4

r = 2 ………… (3)

Substitute r = 2 in the equation (1) , we get

a×2³ = 10

a×8 = 10

a = 10/8 = 5/4 ………… (4)

Tn = 640

ar^(n-1) = 640

(5/4)×2^(n-1) = 640

2^(n-1) = 640×(4/5)

2^(n-1) = 128×4

2^(n-1) = 2^9

n-1 = 9

n = 10.

First term (a) = 5/4

common ratio (r) = 2

Number of terms in G.P. = 10.

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

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