Arithmetic Progression MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-10. We Provide Step by Step Answer of MCQs on AP as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Arithmetic Progression MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-10
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-10 | Arithmetic Progression |
| Writer/Book | RS Aggarwal |
| Topics | Solution of MCQs |
| Academic Session | 2024-2025 |
Multiple Choice Questions on AP
Que-1: The general term of an A.P. whose first term is a and the common difference is d, is given by :
(a) Tn=2a+(n-1)d (b) Tn=(n/2)[2a+(n-1)d] (c) Tn=a+(n-1)d (d) Tn=(n/2)[a+(n-1)d]
Sol: (c) Tn=a+(n-1)d
Reason: Tn=a+(n-1)d
It is a correct formula to find general terms of the A.P.
Que-2: The nth term from the end of an A.P. whose first term, last term and common difference are a, l and d respectively, is given by :
(a) l+(n+1)d (b) l-(n-1)d (c) l+(n-1)d (d) l-(n+1)d
Sol: (b) l-(n-1)d
Reason: If d,a and l denote the common difference, first and last terms respectively, then nth term from the end is l-(n-1)d.
Que-3: Sum of an term of an A.P. whose first term is a and common difference is d, is given by :
(a) Sn=(n/2)[a+(n-1)d] (b) Sn=(n/2)[2a+(n+1)d] (c) Sn=(n/2)[2a-(n-1)d] (d) Sn=(n/2)[2a+(n-1)d]
Sol: (d) Sn=(n/2)[2a+(n-1)d]
Reason: Sum of an term of an A.P. whose first term is a and common difference is d, is given by : Sn=(n/2)[2a+(n-1)d].
Que-4: Sum of n term of an A.P. whose first term and last term are a and l respectively is given by :
(a) Sn=(n/2)(a+l) (b) Sn=(n/2)(a-l) (c) Sn=(n/2)(2a+l) (d) Sn=(n/2)[a+(n-1)l]
Sol: (a) Sn=(n/2)(a+l)
Reason: Sum of n term of an A.P. whose first term and last term are a and l respectively is given by : Sn=(n/2)(a+l).
Que-5: The sum of first n natural number is :
(a) [n(n-1)]/2 (b) [n(n+1)]/2 (c) [n(n+2)]/2 (d) [n(n-2)]/2
Sol: (b) [n(n+1)]/2
Reason: First n natural numbers 1,2,3,4,5,6,……………….. n terms
Sn = 1+2+3+4+5+6+……….(n-2) + (n-1) + n ————(1)
We can also write above given series in reverse order
Sn = n + (n-1) + (n-2) +……………4,3,2,1 —————-(2)
we find sum of first and last term = sum of second and second last term so,on…..
Equation 1 + Equation 2
2Sn = (n+1) + (n+1) + (n+1) ……… n times
2Sn = n.(n+1)
Sn = [n(n+1)]/2
Que-6: The sum of first 50 natural numbers is :
(a) 1050 (b) 1175 (c) 1225 (d) 1275
Sol: (d) 1275
Reason: The sum of first n natural numbers is given by
Sn = [n(n+1)]/2.
Here, n = 50.
⇒S50 = [50(50+1)]/2
= (50×51)/2
= 25×51
= 1275.
Que-7: The nth term of an Arithmetic Progression (A.P.) is 2n+5. The 10th term is : (a) 7 (b) 15 (c) 25 (d) 45
Sol: (c) 25
Reason: nth term of A.P. = 2n + 5
Put n = 10
∴ 10th term of A.P. = 2 × 10 + 5
= 20 + 5
= 25
Que-8: The first term of an A.P. is 7 and the common difference is 3. The general term of the A.P. is : (a) Tn=3n-4 (b) Tn=2n+5 (c) Tn=3n+4 (d) none of these
Sol: (c) Tn=3n+4
Reason: Tn = a+(n-1)d
= 7+(n-1)3
= 7+3n-3
= 4+3n.
Que-9: The 24th term of the A.P. -1, 3, 7, 11,……….. is : (a) 83 (b) 87 (c) 91 (d) 95
Sol: (c) 91
Reason: a = -1, d = 3-(-1) = 4
Tn = a+(n-1)d
T24 = -1+(24-1)4
= -1+(23)4
= -1+92
= 91.
Que-10: If 70,75,80,85 are the first four term of an arithmetic progression, then the 10th term is : (a) 35 (b) 25 (c) 115 (d) 105
Sol: (c) 115
Reason: a = 70, d = 5
Tn = a+(n-1)d
T10 = 70+(10-1)5
= 70+(9)5
= 70+45
= 115.
Que-11: The A.P. 6,13,20,………,216 has 31 terms. The middle term of the A.P. is : (a) 91 (b) 97 (c) 107 (d) 111
Sol: (d) 111
Reason: a = 6, d = 7
T31 = 216
Therefore, Middle term of the given AP
= (31+1)/2 th
= 32/2 = 16th
T16 = 6+(16-1)7
= 6+(15)7
= 6+105
= 111.
Que-12: Which term of the A.P. 7,13,19,25, ……….. is 241? (a) 40th (b) 36th (c) 44th (d) 45th
Sol: (a) 40th
Reason: a = 7, d = 6, l = 241
n = [(l-a)/d] + 1
= [(241-7)6] + 1
= [234/6] + 1
= 39+1
= 40.
Que-13: Which term of the A.P. 11,8,5,2,……… is -148? (a) 52th (b) 54th (c) 55th (d) 57th
Sol: (b) 54th
Reason: a = 11, d = -3, l = -148
n = [(l-a)/d] + 1
n = [(-148-11)/(-3)] + 1
n = [(-159)/(-3)] + 1
n = 53+1
n = 54.
Que-14: How many three digits number are divisible by 7? (a) 128 (b) 124 (c) 136 (d) 132
Sol: (a) 128
Reason: Hence the final sequence is as follows:
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d = 7
aₙ = 994
n = ?
We know that the nth term of an A.P. is, aₙ = a + (n – 1)d
994 = 105 + (n – 1)7
889 = (n – 1)7
n – 1 = 889/7
n – 1 = 127
n = 127 + 1
n = 128
Que-15: How many numbers lying between 20 and 200 are divisible by 4? (a) 43 (b) 44 (c) 45 (d) 46
Sol: (b) 44
Reason: Hence the final sequence is as follows:
24, 28, 32, 36, ………., 196
a = 24, d = 4, l = 196
l = a+(n-1)d
196 = 24+(n-1)4
196-24 = (n-1)4
172/4 = n-1
43+1 = n
n = 44.
Que-16: The common difference of the A.P. (1/k),[(1-k)/k],[(1-2k)/k], …….. is :
(a) -k (b) k (c) 1 (d) -1
Sol: (d) -1
Reason: T1 =1/k
T2 = [(1-k)/k] = (1/k) – 1
d = T2 – T1
d = [(1/k)-1]-(1/k)
d = (1/k)-1-(1/k)
d = -1.
Que-17: If the nth term of an A.P. is given by (3n+2), then the sum of its first three terms is : (a) 21 (b) 24 (c) 27 (d) 32
Sol: (b) 24
Reason: Given,
an = 3n + 2
a1 = 3(1) + 2 = 3 + 2 = 5,
a2 = 3(2) + 2 = 6 + 2 = 8,
a3 = 3(3) + 2 = 9 + 2 = 11.
By formula,
Sum of A.P. = (n/2)(a+l)
Sum of first three terms = (3/2)(5+11)
=(3/2)×16
= 3 × 8 = 24.
Que-18: The last term of the A.P. 5,12,19,………., having 60 terms is : (a) 406 (b) 412 (c) 416 (d) 418
Sol: (d) 418
Reason: a = 5, d = 7
Tn = a+(n-1)d
T60 = 5+(60-1)(7)
= 5+(59)(7)
= 5+413
= 418.
Que-19: The common difference of the A.P. (1/3),[(1-3p)/3],[(1-6p)/3],…….. is :
(a) p (b) -p (c) 1/3 (d) -1/3
Sol: (b) -p
Reason: T1 = 1/3
T2 = [(1-3p)/3]
d = T2 – T1
d = [(1-3p)/3] – (1/3)
d = [1-3p-1]/3
d = -3p/3
d = -p.
Que-20: The next term of the A.P. √3,√12,√27,√48,………….. is : (a) √72 (b) √68 (c) √75 (d) √56
Sol: (c) √75
Reason: T1 = √3, T2 = √12 = 2√3
T3 = √27 = 3√3, T4 = √48 = 4√3
We want T5:
d = T2 – T1
d = 2√3 – √3
d = √3
T5 = T4 + d
T5 = 4√3 + √3
T5 = 5√3 = √75.
Que-21: The first term of an A.P. is p and its common difference is q. The 10th term of the A.P. is : (a) p-9q (b) p+10q (c) p-10q (d) p+9q
Sol: (d) p+9q
Reason: First term a = p
Common difference, d = q
10th term = a+9d
= p+9(q)
= p+9q
Que-22: The nth term of the A.P. (1/m),[(1+m)/m],[(1+2m)/m],………. is : (a) [m(n-1)+1]/m (b) [m(n+1)-1]/m (c) [m(n-1)-1]/m (d) none of these
Sol: (a) [m(n-1)+1]/m
Reason: Given: A.P. (1/m), [(1+m)/m], [(1+2m)/m] ,….
We know that the nth term of an AP is given by
an = a+(n−1)d
In the given AP
a = 1/m, d = [(1+m)/m] − (1/m) = [1+m−1]/m = 1
Thus, the nth term of the given AP is
an = (1/m)+(n−1)1
= (1/m)+(n−1)
= [m(n-1)+1]/m
Que-23: For what value of p are 2p+1,13,5p-3 three consecutive term of an A.P. ? (a) 0 (b) 1 (c) 2 (d) 4
Sol: (d) 4
Reason: If the three terms are consecutive terms of an AP,
common difference, d = a2-a1 = a3-a2
Here, a1 = 2p+1, a2 = 13 and a3 = 5p-3
=> d= 13 – (2p+1) = (5p-3) – (13)
=> 12 – 2p = 5p – 16
=> 7p = 28
=> p = 4
Que-24: For what value of k will 2k+1,3k+2 and 5k-1 be three consecutive terms of an A.P.?
(a) 1 (b) 2 (c) 4 (d) 6
Sol: (d) 6
Reason: The given terms are 2k + 1, 3k + 3 and 5k − 1.
The differences between the consecutive terms are
3k + 3 − (2k + 1) = k + 2 = d1
and
5k − 1 − (3k + 3) = 2k − 4 = d2
If the given terms are in an AP, then
d1 = d2
⇒ k + 2 = 2k − 4
⇒ k = 6
Que-25: If the sum of first n term of an A.P. is Sn=5n²+3n, then its common difference is : (a) 8 (b) 10 (c) 18 (d) 26
Sol: (b) 10
Reason: Given:
Sn = 5n² + 3n
an = Sn – S(n-1)
= 5n² + 3n – [5(n-1)² +3(n-1)]
= 5n² +3n -[ 5(n² -2n +1)+3n -3]
= 5n² +3n – 5n² +10n -1 -3n +3
= 5n² -5n² +3n -3n +10n -1+3
an = 10n +2………..(1)
Put n= 1 in eq 1
a1= 10 ×1 +2
a1= 10 +2
a1 = 12
Put n= 2 in eq 1
a2 = 10 × 2 +2
a2 = 20 +2
a2 = 22
Common Difference (d) = a2 – a1 = 22 – 12
d = 10
Que-26: The first and the last term of an A.P. are 1 and 11 respectively. If the sum of its term is 36, then the number of terms is : (a) 6 (b) 7 (c) 8 (d) 9
Sol: (a) 6
Reason: In this case, we are given:
a = 1, l = 11, Sum = 36
Sum = (n/2)(a + l)
Plugging these values into the formula, we have:
⇒ 36 = (n/2)(1 + 11)
⇒ 36 = (n/2)(12)
⇒ 6 = n
Que-27: The sum of first 40 positive integers divisible by 6 is : (a) 2460 (b) 3640 (c) 4920 (d) 4860
Sol: (c) 4920
Reason: First term, a = 6
Common difference, d = 6
Number of terms, n = 40
Sₙ = n/2 [2a + (n – 1) d]
S₄₀ = 40/2 [2 × 6 + (40 – 1)6]
= 20 [12 + 39 × 6]
= 20 [12 + 234]
= 20 × 246
= 4920
Que-28: The sum of first 30 odd natural numbers is : (a) 800 (b) 900 (c) 729 (d) 1249
Sol: (b) 900
Reason: First term, a = 1
Common difference, d = 2
Then the 30th odd term will be:
⇒T30 = a+(n−1)d = 1+(30−1)2 = 59
⇒Sum of n terms = (n/2)(a+l) , where l = T30
⇒Sum of n terms = (30/2)(1+59)
⇒Sum to 30 terms = 15×60 = 900
Que-29: The 7th term of an A.P. is 4 and its common difference is -4. The first term of the A.P. is : (a) 32 (b) 28 (c) 24 (d) 36
Sol: (b) 28
Reason: T7 = a + 6d
⇒ a + 6 ⨯ (-4) = 4
⇒ a = 4 + 24 = 28
Que-30: The sum of first n terms of an A.P. is (4n²+2n). The nth of the A.P. is : (a) (6n-2) (b) (8n-2) (c) (6n+2) (d) (8n-2)
Sol: (b) (8n-2)
Reason: Given that Sn = 4n² + 2n. ————— (1)
Substitute n = 1 in (1), we get
Sn = 4(1)^2 + 2(1)
= 4 + 2
= 6.
So, Sum of the first term of AP is 6 i.e a = 6.
Now,
Substitute n = 2 in (1), we get
Sn = 4(2)^2 + 2(2)
= 4 * 4 + 2 * 2
= 16 + 4
= 20.
So, Sum of the first 2 terms = 20.
Now,
First-term + second term = 20
6 + a2 = 20
a2 = 20 – 6
a2 = 4.
Hence in AP,
first term a = 6.
common difference d = a2 – a1
= 14 – 6
= 8.
We know that sum of n terms of an AP an = a + (n – 1) * d
= 6 + (n – 1) * (8)
= 6 + 8n – 8
= 8n – 2.
Que-31: The first term of an A.P. is 7 and 13th term is 35. The common difference of the A.P. is : (a) 5/3 (b) 7/2 (c) 8/3 (d) 7/3
Sol: (d) 7/3
Reason: a = 7
13th term(a13) = 35
Let common difference be ‘d’
Therefore, a+(13-1)d = 35
a+12d = 35
Putting value of ‘a’
7+12d = 35
12d = 35-7 = 28
d = 28/12
d = 7/3
Que-32: There are total 9 terms in an A.P. If the last term and the sum of all the terms are 28 and 44 respectively, then the first term is : (a) 1 (b) 7 (c) 4 (d) 5
Sol: (c) 4
Reason: Given that, l = 28, S = 144 and there are total of 9 terms.
Sn = (n/2)(a+1)
144 = (9/2)(a+28)
⇒ a + 28 = (144×2)/9
⇒ a = 16 × 2
⇒ a = 32
⇒ a = 32 – 28
a = 4
Que-33: The 5th term of an A.P. is -3 and its common difference is -4. The sum of its first 10 terms is : (a) -50 (b) -40 (c) -60 (d) -30
Sol: (a) -50
Reason: T5 = -3
D = -4
T5 = a+(n-1)d = -3
a-16 = -3
a = 13
sum = (n/2){2a+(n-1)d}
= 5(26-36)
= -50
Que-34: The 7th term of an A.P. is -1 and its 16th term is 17. The nth term of an A.P. is : (a) (3n+12) (b) (2n-5) (c) (3n+5) (d) (2n-15)
Sol: (d) (2n-15)
Reason: Let a be the first term and d be the common difference of the AP. Then,
a7 = −1
⇒ a+(7−1)d = −1 [∵nth term is given by an = a+(n−1)d]
⇒ a+6d = −1……(i)
Also, a16 = 17
⇒ a+15d = 17……(ii)
From eq. (i) and (ii), we get
−1−6d+15d = 17 [∵a = −1−6d from eq.(i)]
⇒ 9d = 17+1
⇒ d = 2
Putting d=2 in eq.(i), we get
a+6×2 = −1
⇒a = −1−12 = −13
∴an = a+(n−1)d = −13+(n−1)×2
= −13+2n−2 = 2n−15
Hence, the nth term of the AP is (2n−15).
Que-35: If the sum of first p term of an A.P. is ap²+bp, then its common difference is :
(a) a (b) 3a+b (c) a+b (d) 2a
Sol: (d) 2a
Reason: Let Sp denotes the sum of first p terms of the AP.
∴ sp = ap2 + bp
⇒ sp-1 = a(p-1)² +b(p-1)
= a( p2 – 2p +1 ) +b (p-1)
= ap2 – ( 2a -b) p+ (a-b)
Now,
pth term of AP ap = sp-sp-1
= (ap²+bp) – [ap²-(2a-b)p+(a-b)]
= ap²+bp-ap²+(2a-b)p-(a-b)
= 2ap – (a-b)
Let d be the common difference of the AP.
∴ d = ap-ap-1
= [ 2 ap – (a-b) ] = [ 2a (p-1) – (a-b) ]
= 2ap – (a-b) – 2a (p-1 ) + (a-b)
= 2a
Que-36: The first term of an A.P. is a and the nth is b, then the common difference of the A.P. is : (a) (b-a)/n (b) (b-a)/(n-1) (c) (b+a)/(n-1) (d) (b-a)/(n+1)
Sol: (b) (b-a)/(n-1)
Reason: As we know that,
General term of an ap.
⇒ Tₙ = a + (n – 1)d.
⇒ b = a + (n – 1)d.
⇒ b = a + nd – d.
⇒ b – a = nd – d.
⇒ (b – a) = d(n – 1).
⇒ d = (b – a)/(n – 1).
Que-37: The first three year of an A.P. are 3y-1,3y+5 and 5y+1 respectively. Then, the value of y is : (a) 1 (b) 5 (c) 8 (d) 3
Sol: (b) 5
Reason: The first three terms of an AP are 3y – 1, 3y + 5 and 5y + 1 respectively
We need to find the value of y.
We know that if a, b and c are in AP, then:
b − a = c − b ⇒ 2b = a + c
∴2(3y+5) = 3y − 1 + 5y + 1
⇒ 6y + 10 = 8y
⇒ 10 = 8y −6y
⇒ 2y = 10
⇒ y = 5
Que-38: The 8th term from the end of the A.P. 7,10,13,………., 184 is : (a) 157 (b) 160 (c) 163 (d) 166
Sol: (c) 163
Reason: First term (a) = 7
Last term (an) = 184
Common difference (d) = 10 – 7 = 3
Now, as we know,
an = a+(n-1)d
So, for the last term,
184 = 7+(n-1)³
184 = 7 + 3n – 3
184 = 4 + 3n
184 – 4 = 3n
Further simplifying,
180 = 3n
n = 180
n = 60
So, the 8th term from the end means the 53rd term from the beginning.
So, for the 53rd term (n = 53)
a53=7+(53-1)³
= 7 + (52)³
= 7 + 156
= 163
Que-39: If a=3, n=8 and Sn = 192, then the common difference of the A.P. is : (a) 4 (b) 5 (c) 6 (d) 7
Sol: (c) 6
Reason: Given that, a = 3, n = 8, S = 192
Sn = (n/2)[2a+(n-1)d]
192 = (8/2)[2×3+(8-1)d]
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 42/7
d = 6
–: End of Arithmetic Progression MCQs Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions :–
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