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, ar2, ar3, ar4,…, arn-1 Where, a = First term, r = common ratio,
nth Term of GP = arn-1
Similarly, nth term, tn = arn-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 an = 640
⇒ 640 = (5) ⋅ (2)n – 1
⇒ 2(n – 1) = 128 = 27
⇒ 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.
4th term =1/18
⇒ ar³ = 1/18
7th 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, ar2, ar3, …….
= -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)
⇒ x2 – 12x + 36 = 4x + 36
⇒ x2 – 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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