Geometric Progression MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-11. We Provide Step by Step Answer of Exe-MCQs on G.P. as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Geometric Progression MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-11
Board | ICSE |
Subject | Maths |
Class | 10th |
Chapter-11 | Geometric Progression |
Writer/Book | RS Aggarwal |
Topics | Solution of MCQs on GP |
Academic Session | 2024-2025 |
Multiple Choice Questions
( Geometric Progression MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-11 )
Que-1: The nth term of a G.P. whose first term is a and common ratio is r, is given by :
(a) Tn=ar^(n+1) (b) Tn=ar^(1-n) (c) Tn=ar^(n-1) (d) Tn=[a/r^(n-1)]
Sol: (c) Tn=ar^(n-1)
Reason: Tn=ar^(n-1)
It is a formula to find n term of the G.P.
Que-2: If the first term, common ratio and the last term of a G.P. are a,r and l respectively, then the nth term from the end of the G.P. is given by :
(a) lr^(n-1) (b) lr^(1-n) (c) [l/r^(1-n)] (d) [l/r^(n+1)]
Sol: (b) lr^(1-n)
Reason: lr^(1-n)
Que-3: The product of n terms of a G.P. with first term a and common ratio r, when r = 1 is :
(a) [{a(r^n + 1)}/(r-1)] (b) na (c) ar^n (d) a^n
Sol: (d) a^n
Reason: Tn = ar^n
= a1^n = a^n.
Que-4: The sum of n terms of a G.P. with first term a and common ratio r, when r = 1, is given by :
(a) Sn=[{a(r^n – 1)}/(1-r)] (b) Sn=[{a(r^n – 1)}/(r-1)] (c) n²a (d) na
Sol: (d) na
Reason: When the common ratio, r = 1
It means all the terms of the progression is same and each term equal to first term a
Sn = a+a+a+a+………..+a
= na.
Que-5: The sum of n term of a G.P. with first term a and common ratio r, when r>1, is :
(a) (ar^n – 1)/(r-1) (b) (1-ar^n)/(1-r) (c) [{a(r^n – 1)}/(r-1)] (d) [{a(1-r^n)}/(1-r)]
Sol: (c) [{a(r^n – 1)}/(r-1)]
Reason: [{a(r^n – 1)}/(r-1)]
This formula is derived by subtracting the terms of the series in such a way that most terms cancel out, leaving a simplified expression.
Que-6: The sum of n terms of a G.P. with first term a and common ratio r, when r<1 is :
(a) (ar^n – 1)/(r-1) (b) (1-ar^n)/(1-r) (c) [{a(r^n – 1)}/(r-1)] (d) [{a(1-r^n)}/(1-r)]
Sol: (d) [{a(1-r^n)}/(1-r)]
Reason: [{a(1-r^n)}/(1-r)]
This formula works for r<1 because the denominator 1−r is positive and keeps the expression valid.
Que-7: The general term of the G.P. (1/4),(-1/2),1,-2,4,………. is :
(a) (-1)^(n-1) × (2)^(n-3) (b) (-1)^(n-1) × (2)^(n-2) (c) (-1)^(n-1) × (-2)^(n-1) (d) (-1)^(n-1) × (-2)^(n-3)
Sol: (a) (-1)^(n-1) × (2)^(n-3)
Reason: a = 1/4
r = (-1/2)/(1/4) = -2
Tn = ar^(n-1)
= (1/4)(-2)^(n-1)
= (-1)^(n-1).(2)^(n-1).(1/4)
= (-1)^(n-1) . 2^(n-3).
Que-8: The 12th term of the G.P. 2, 4, 8, 16,……. is :
(a) 1024 (b) 2048 (c) 4096 (d) 8192
Sol: (c) 4096
Reason: a = 2
r = 4/2 = 2
Tn = ar^(n-1)
= 2×2^(12-1)
= 2×2¹¹
= 4096.
Que-9: Which term of the G.P. √3, 3√3, 9√3, …… is 729√3?
(a) 7th (b) 6th (c) 9th (d) 8th
Sol: (a) 7th
Reason: First term = a = √3 ,
Common ratio ( r ) = a2/a1
r = 3√3/√3
r = 3
nth term = 729√3
ar^n-1 = 729√
√3 × (3)^n -1 = 729√3
(3)^n-1 = 729√3/√3
= 729
3^(n-1) = 3^6
n-1 = 6
n = 7.
Que-10: If a1, a2, a3 , ….. an is a G.P. having common ratio r and k is a natural number such that 3<k<n, then r is equal to :
(a) ak/a1 (b) a1/a2 (c) ak/[a(n-3)] (d) [a(k-1)]/[a(k-2)]
Sol: (d) [a(k-1)]/[a(k-2)]
Reason: an+1 = an⋅r
Therefore, the general form for the k-th term is:
a(k) = a1⋅r^(k−1)
r = [a(k)]/[a(k-1)]
This indicates that the common ratio rrr can be expressed as the ratio of any two consecutive terms. Thus, for r in terms of the k-th and (k−1)-th terms, we have:
r = [a(k-1)]/[a(k-2)].
Que-11: The common ratio of the G.P. (-3/4),(1/2),(-1/3),(2/9),……… is :
(a) -4/3 (b) -2/3 (c) 2/3 (d) -3/8
Sol: (b) -2/3
Reason: Here,
First term ,a = −3/4
Common ratio, r = a2/a1
= (1/2)/(−3/4) = −2/3
Que-12: The common ratio of the G.P. [1/a³x³], ax, a^5 x^5,……….. is :
(a) 1/a²x² (b) 1/a⁴x⁴ (c) a²x² (d) a⁴x⁴
Sol: (d) a⁴x⁴
Reason: a = 1/a³x³
r = t2/t1
= ax/(1/a³x³)
= a⁴x⁴.
Que-13: The common ratio of the G.P. 0.15, 0.015, 0.00015, …….. is :
(a) 0.1 (b) 0.01 (c) 1 (d) 0.001
Sol: (a) 0.1
Reason: t1 = 0.15, t2 – 0.015, t3 = 0.00015
a = 0.15
r = t2/t1
= 0.015/0.15
= 0.1
Geometric Progression MCQs Class 10 RS Aggarwal
page-176
Que-14: The 8th term of the G.P. 1,3,9,27,……… is :
(a) 729 (b) 2187 (c) 6561 (d) none of these
Sol: (b) 2187
Reason: Given a1 = 1, a2 = 3 and n = 8.
Now, r = a2/a1 = 3/1 = 3.
We know that an = ar^(n−1).
⇒ a8 = a×r^(8−1)
= 1×3^(8−1)
= 1×3^7
= 2187
Que-15: The nth term of the G.P. x³, x^5, x^7,…………. is :
(a) x^(2n-1) (b) x^(2n+3) (c) x^(2n+1) (d) x^(3n+2)
Sol: (c) x^(2n+1)
Reason: a = x³
r = x^5/x³ = x²
Tn = a.r^n-1
= x³.(x²)^(n-1)
= x³.x^(2n-2)
= x^(2n+1).
Que-16: The 10th term of the G.P. 1,-a,a²,-a³, …………. is:
(a) a^9 (b) -a^10 (c) -a^11 (d) -a^9
Sol: (d) -a^9
Reason: a = 1
r = -a/1 = -a
Tn = a.r^n-1
T10 = a.r^10-1
= 1.(-a)^(10-1)
= -a^9
Que-17: The first term and the common ratio of the G.P. 3,(3/2),(3/4),……. are respectively.
(a) 3 and 2 (b) 3 and 1/2 (c) 3 and 3/2 (d) 3/2 and 1/2
Sol: (b) 3 and 1/2
Reason: G.P. 3,(3/2),(3/4),……
a = 3
r = (3/2)/3
= 1/2.
Que-18: Which term of the G.P. 2,8,32,128,………….. is 131072?
(a) 9th (b) 10th (c) 8th (d) 12th
Sol: (a) 9th
Reason: Given G.P., 2,8,32,… upto n terms.
Here, first term a = 2,
common ratio r = 8/2 = 4 and nth term = 131072
We know an = a.r^(n−1)
⇒ 131072 = 2×4^(n−1)
⇒ 131072/2 = 4^(n−1)
⇒ 4^(n−1) = 65536
⇒ 4^(n−1) = (4)^8
On comparing powers, we get
n−1 = 8
n = 9.
Que-19: If the 5th term of the G.P. is 2, then the product of its first nine terms is :
(a) 256 (b) 1024 (c) 512 (d) 2048
Sol: (c) 512
Reason: Let ‘a’ be the first term and ‘r’ be the common ratio.
We know that Tn = a.r^(n-1)
Given: ar4 = 2
Now, Product of 9 terms = a × ar × ar2 × ar3 × ar4 × ar5 × ar6 × ar7 × ar8
= a9 r36 = (ar4)9 = 29 = 512.
Que-20: If the (p+q)th term of a G.P. are m and n respectively, then its pth term is :
(a) mn (b) √mn (c) (mn)² (d) m/n
Sol: (b) √mn
Reason: Here term Here ,(p+q) th term = m
⇒ ar^(p+q) − 1 = m…….(i)
And term And ,(p−q)th term = n
⇒ ar^(p−q) − 1 = n…….(ii)
Dividing by Dividing (i) by (ii):
[ar^(p+q) − 1]/[ar^(p−q) − 1] = m/n
⇒ r^2q = m/n
⇒ r^q = m/n
Now, from Now, from (i):
a(r^(p−1)×r^q) = m
⇒ ar^(p−1)×√(m/n) = m
⇒ ar^(p−1) = m×√n/√m
⇒ ar^(p−1) = (m√n)/m
Thus, the term is √mn.
Que-21: If 2nd, 3rd and 6th terms of an A.P. are the three consecutive terms of a G.P. , then the common ratio of the G.P. is :
(a) 2 (b) 3 (c) 1/2 (d) 1/3
Sol: (b) 3
Reason: Here, second term Here, second term ,a2 = a+d
Third term Third term ,a3 = a+2d
Sixth term Sixth term ,a6 = a+5d
As, a_2 , a_3 and a_6 are in G . P .
∴ First term of G . P .= a2 = A = a+d
Second term of G . P .= Ar = a+2d
Third term of G . P . = Ar² = a+5d
∴ (a+2d)² = (a+d)×(a+5d)
⇒ a²+4ad+4d² = a²+6ad+5d²
⇒ 2ad+d² = 0
⇒d(2a+d) = 0
⇒ d = 0 or 2a+d = 0
But is not possible But ,d=0 is not possible .
∴d = −2a
∴r = (a+2d)/(a+d)
⇒ r = [a+2(−2a)]/[a+(−2a)]
⇒ r = 3/1 = 3
Que-22: The 6th term from the end of the G.P. 8, 4, 2, … ….., 1/1024 is :
(a) 1/32 (b) 1/16 (c) 1/64 (d) 1/128
Sol: (a) 1/32
Reason: The nth term from the end is given by:
an = [l(1/r)^(n-1)] where, l is the last term, r is the common ratio, n is the nth term
Here, last term, l = 1/1024
First term is a = 8
Common ratio r = a2/a1 = 4/8 = 1/2
6th term from the end of the GP will be
⇒ a6 = [(1/1024)×{(2)^(6-1)}]
= (1/1024)×(2)^5 = 1/32
Hence, the 6thvalue is 132.
Que-23: The 4th term from the end of the G.P. (2/27), (2/9), (2/3), ………, 162 is :
(a) 18 (b) 2 (c) 6 (d) 2/3
Sol: (c) 6
Reason: Here, first term, Here, first term, a = 2/27
Common ratio Common ratio, r = a2/a1 = (2/9)/(2/27) = 3
Last term, Last term, l = 162
After reversing the given G . P . , we get another G . P . whose first term is l and common ratio is After reversing the given G . P . , we get another G . P . whose first term is l and common ratio is 1r.
term from the end ∴4th term from the end = l(1/r)^(4−1)
= (162)(1/3)^3
= 6.
Que-24: The product of the first three terms of a G.P. is -1 and the common ratio is -3/4. The sum of these three terms is :
(a) 12/13 (b) 11/13 (c) 11/12 (d) 13/12
Sol: (d) 13/12
Reason: The three terms of the G.P. will be:
a, a⋅r, a⋅r²
The product of the first three terms is −1, so:
a, a⋅r, a⋅r² = -1
a³.r³ = -1
r = -3/4
a³(-3/4)³ = -1
a³(-27/64) = -1
a³ = -1/(-27/64)
a³ = 64/27
a = 4/3
The sum of the first three terms is:
S = a + a⋅r + a⋅r²
a = 4/3, r = -3/4
S = (3/4) + (3/4)⋅(−4/3) + (3/4)⋅(−4/3)²
S = (3/4) − 1 + (3/4)⋅(16/9)
S = (3/4) − 1 + (48/36) = (3/4) − 1 + (4/3)
S = (12/16) − (12/12) + (12/9)
= (16−12+9)/12
= 12/13
Que-25: For what value of x are the numbers -2/7, x, -7/2 in G.P. ?
(a) 0,1 (b) 0,-1 (c) -1,1 (d) -2,2
Sol: (c) -1,1
Reason: We use the property that the square of the middle term is equal to the product of the other two terms:
x² = (−7/2)×(−2/7)
x² = (72)×(2/7) = 1
x² = 1
x = ±1
x = -1,1.
Que-26: How many terms of the G.P. 1,4,16,64,……………. will make the sum 5461 ?
(a) 6 (b) 9 (c) 7 (d) 8
Sol: (c) 7
Reason: First term, a = 1
Common ratio, r = 4/1 = 4 …(∵ r > 1)
Let the number of terms to be added = n
Then, Sn = 5461
⇒ [a(r^(n-1))]/(r-1) = 5461
⇒ [1(4th-1)]/(4-1) = 5461
⇒ (4th-1)/3 = 5461
⇒ 4th – 1 = 16383
⇒ 4th = 16384
⇒ n = 7
Que-27: The sum of 7 terms of the G.P. 3, 6, 12, ………. is :
(a) 181 (b) 241 (c) 381 (d) 421
Sol: (c) 381
Reason: a = 3
r = 6/3 = 2
Sn = [a(r^n − 1)](r-1)
S7 = [3(2^7 − 1)]/(2-1)
S7 = 3(128−1)
= 3×127
= 381.
Que-28: The sum of the first two terms of a G.P. is -4 and the fifth term is 4 times the third term. Then, the first term of the G.P. is :
(a) -3/4 or 4 (b) 3/4 or 1/4 (c) 4/3 or 1/4 (d) -4/3 or 4
Sol: (d) -4/3 or 4
Reason: Let a be the first term and r be the common ratio of the G.P. According to the given conditions,
S2 = -4 = [a(1-r²)]/(1-r) ……..(i)
a5 = 4 × a3
⇒ ar4 = 4ar2 ⇒ r2 = 4
∴ r = ± 2
From (i) we obtain
-4 = [a(1-(2)²)]/(1-2) for r = 2
⇒ -4 = [a(1-4)]/-1
⇒ −4 = a(3)
⇒ a = -4/3
Also, −4 = [a(1-(-2)²)]/[1-(-2)] for r = −2
⇒ -4 = [a(1-4)]1+2
⇒ -4 = a(-3)3
⇒ a = 4
a = -4/3 or 4
Que-29: If x, y and z are in G.P., then the relation between x, y and z can be :
(a) y = x+z (b) y = xz (c) 2y = x+z (d) y = √xz
Sol: (d) y = √xz
Reason: If x, y, and z are in geometric progression (G.P.), the common ratio between consecutive terms is constant. So, we have:
y/x = z/y
y² = xz
y = √xz.
Que-30: The product of first five terms of a G.P. with first term a and common ratio r>1 is equal to :
(a) [{a(r^5 – 1)}/(r-1)] (b) ar⁴ (c) a^5 r^10 (d) [{a(r^5 – 1)}/r]
Sol: (c) a^5 r^10
Reason: The first five terms are:
a, ar, ar², ar³, ar^4
The product of these terms is:
P = a, ar, ar², ar³, ar^4
P = a^5 ⋅ r^0+1+2+3+4
The sum of the exponents of r is:
0+1+2+3+4 = 10
P = a^5⋅r^10.
Que-31: If 5th, 8th and 11th terms of a G.P. are x, y and z respectively, then, which one of the following is correct?
(a) y² = x²z² (b) y² = x²+z² (c) y² = xz (d) xyz = 1
Sol: (c) y² = xz
Reason: Let the first term of the geometric progression = a
Common and ratio = r
5th term = ar5–1 = ar4 = x
8th term = ar8–1 = ar7 = y
11th term = ar11–1= ar10 = z
Left side = y2 = (ar7)2
= a2 × r14
Right side = xz = ar4 ar10
= a2 × r14
Hence, y2 = xz
Que-32: Consider the G.P. a, ar, ar²,…………., l. The kth term of the end is :
(a) [l/(r^(k – 1))] (b) l/k (c) al/r^k (d) [l/ar^(k-1)]
Sol: (a) [l/(r^(k – 1))]
Reason: Tn = ar^(n−1)
Now, the k-th term from the end of the G.P. can be found by considering the sequence in reverse. In this reversed G.P., the last term becomes the first term, and the common ratio becomes 1/r.
So, the k-th term from the end is:
Tk = [l(1/r)^(k−1)] = l/r^(k−1)
Tk = [l/r^(k−1)].
Que-33: If a and l are respectively the first and the last terms of a G.P. having common ratio r>1, then the sum of n terms of the G.P. is given by:
(a) lar^(n-1) (b) ar^ln (c) [lr-a]/[r-1] (d) [lr^n – a]/[lr-1]
Sol: (c) [lr-a]/[r-1]
Reason: Sn = [a(r^n − 1)]/r-1, (for r>1)
Given that the last term l of the G.P. is related to the first term and the common ratio by:
l = ar^(n−1)
To express the sum in terms of l, substitute a = l/r^(n−1) into the sum formula:
Sn = [l(r−1)/(r-1)] = (lr−a)/(r-1)
Thus, the sum of n terms of the G.P. can be expressed as:
Sn = (lr−a)/(r-1).
Que-34: The products of n terms of a G.P. with first term a and common ratio r, r>1 is :
(a) [{a^n(r^n – 1)}/r-1] (b) a^n r^n (c) a^n r^[{n(n+1)}/2] (d) a^n r^[{n(n-1)}/2]
Sol: (d) a^n r^[{n(n-1)}/2]
Reason: Tn = a.r^(n−1)
The product of the first nnn terms of the G.P. would be:
P = a⋅ar⋅ar²⋅⋯⋅ar^(n−1)
P = a^n ⋅ r^0+1+2+⋯+(n−1)
The sum of the exponents of r is:
0+1+2+⋯+(n−1) = [n(n−1)/2]
Thus, the product of the first n terms of the G.P. is:
P = a^n⋅r^[{n(n−1)}/2]
–: End of Geometric Progression MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions :–
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