10th Grade > Mathematics
ARITHMETIC PROGRESSIONS MCQs
Total Questions : 107
| Page 6 of 11 pages
:
Since, 2x, x + 10, 3x + 2 are in AP.
2 (x + 10) = 2x + (3x + 2)
2x + 20 = 5x + 2
3x = 18
x = 6.
Answer: Option D. -> 672
:
D
As an=3+2n,
so,
a1=3+2×1=5
a2=3+2×2=7
a3=3+2×3=9
List of numbers becomes 5, 7, 9, 11 . . .
Here, 7–5=9–7=11–9=2 and so on.
So, it forms an AP with common difference d=2.
To findS24, we have n=24,a=5,d=2.
Therefore, S24=242(2(5)+(24−1)2)
=12(10+46)=672
:
D
As an=3+2n,
so,
a1=3+2×1=5
a2=3+2×2=7
a3=3+2×3=9
List of numbers becomes 5, 7, 9, 11 . . .
Here, 7–5=9–7=11–9=2 and so on.
So, it forms an AP with common difference d=2.
To findS24, we have n=24,a=5,d=2.
Therefore, S24=242(2(5)+(24−1)2)
=12(10+46)=672
Answer: Option D. -> -(p + q)
:
D
let first term be aand common difference be d
sopthterm=a+(p−1)d
andsumoffirstpterms=p2(2a+(p−1)d)=q
hence(2a+(p−1)d)=2qp....(1)andqthterm=a+(q−1)d
andsumoffirstqterms=q2(2a+(q−1)d)=phence(2a+(q−1)d)=2pq....(2)subtract(1)from(2)(p−q)d=2(qp−pq)=2(q2−p2)qso,d=−2(p+q)pq....(3)(p+q)thterm=(a+(p+q−1)d)andsumofitsfirst(p+q)term=(p+q)2(2a+(p+q−1)d)=(p+q)2(2a+(p−1)d+qd)=(2qp+qd)(p+q)2...from(1)=(2qp−2−2qp)p+q2...from(3)=−(p+q)
:
D
let first term be aand common difference be d
sopthterm=a+(p−1)d
andsumoffirstpterms=p2(2a+(p−1)d)=q
hence(2a+(p−1)d)=2qp....(1)andqthterm=a+(q−1)d
andsumoffirstqterms=q2(2a+(q−1)d)=phence(2a+(q−1)d)=2pq....(2)subtract(1)from(2)(p−q)d=2(qp−pq)=2(q2−p2)qso,d=−2(p+q)pq....(3)(p+q)thterm=(a+(p+q−1)d)andsumofitsfirst(p+q)term=(p+q)2(2a+(p+q−1)d)=(p+q)2(2a+(p−1)d+qd)=(2qp+qd)(p+q)2...from(1)=(2qp−2−2qp)p+q2...from(3)=−(p+q)
Answer: Option A. -> an AP with common difference -7.
:
A
For a sequence to be an AP, the same common difference must exist between any two consecutive terms of AP.
Here,
a1=5,a2=−2anda3=−9
a2−a1=−2−5=−7
a3−a2=−9−(−2)=−7
The difference between any two consecutive terms is-7.
Hence, the list of numbers are in AP with common difference -7.
:
A
For a sequence to be an AP, the same common difference must exist between any two consecutive terms of AP.
Here,
a1=5,a2=−2anda3=−9
a2−a1=−2−5=−7
a3−a2=−9−(−2)=−7
The difference between any two consecutive terms is-7.
Hence, the list of numbers are in AP with common difference -7.
Answer: Option D. -> 2
:
D
There are five terms in this AP
an =a1 + (n - 1)d
⇒an−a1=(n−1)d
⇒an−a1n−1=d
Here, a1=1;n=5;an=9
Therefore the common difference,
d=9−15−1=2
:
D
There are five terms in this AP
an =a1 + (n - 1)d
⇒an−a1=(n−1)d
⇒an−a1n−1=d
Here, a1=1;n=5;an=9
Therefore the common difference,
d=9−15−1=2
Answer: Option B. -> 286
:
B
34+32+30+....+10
This sequence is an AP.
Here, a=34,d=32−34=−2.
Let the number of terms of the AP be n.
We know that
an=a+(n−1)d.
⇒10=34+(n−1)(−2)
⇒(n−1)(−2)=−24
⇒n−1=242=12
⇒n=13
Also, we know that the sum to n terms of an AP with first term a and last term l is given by
Sn=n2(a+l).
∴S13=132(34+10)
S13=286
Hence, the required sum is 286.
:
B
34+32+30+....+10
This sequence is an AP.
Here, a=34,d=32−34=−2.
Let the number of terms of the AP be n.
We know that
an=a+(n−1)d.
⇒10=34+(n−1)(−2)
⇒(n−1)(−2)=−24
⇒n−1=242=12
⇒n=13
Also, we know that the sum to n terms of an AP with first term a and last term l is given by
Sn=n2(a+l).
∴S13=132(34+10)
S13=286
Hence, the required sum is 286.
Answer: Option C. -> 65th
:
C
a=3
d=15−3=12
Let the required term be the nth term.
an=a+(n−1)d
a54+132=(32+53×12)+132
=32+636+132=800=an
800=32+(n−1)12
768=(n−1)12n−1=76812=64n=64+1=65
Thus, 65th term is 132 more than 54th term.
:
C
a=3
d=15−3=12
Let the required term be the nth term.
an=a+(n−1)d
a54+132=(32+53×12)+132
=32+636+132=800=an
800=32+(n−1)12
768=(n−1)12n−1=76812=64n=64+1=65
Thus, 65th term is 132 more than 54th term.
Answer: Option C. -> 2500
:
C
Using the formula
Sn=n2[2n+(n−1)d]
Sum of first 100 natural numbers = 1002(2+99)=5050
Sum of even numbers from 1 to 100, there are 50 even numbers between 1 and 100.
So, we have
Sum of even numbers from 1 to 100 = 502(2×2+49×2) = 2550
Hence,the difference is 5050 - 2550 = 2500
:
C
Using the formula
Sn=n2[2n+(n−1)d]
Sum of first 100 natural numbers = 1002(2+99)=5050
Sum of even numbers from 1 to 100, there are 50 even numbers between 1 and 100.
So, we have
Sum of even numbers from 1 to 100 = 502(2×2+49×2) = 2550
Hence,the difference is 5050 - 2550 = 2500
Answer: Option B. -> 16
:
B
Natural numbers that lie between 24 and 101,which are divisible by 5 are 25, 30,... and 100.
This is an A.P with first term a=25,common difference d=5 and the last term l=100.
But, l=a+(n−1)d.
⇒100=25+(n−1)×5
⇒75=(n−1)×5
⇒15=(n−1)
⇒16=n
Thus, there are 16 natural numbers between 24 and 101 that are divisible by 5.
:
B
Natural numbers that lie between 24 and 101,which are divisible by 5 are 25, 30,... and 100.
This is an A.P with first term a=25,common difference d=5 and the last term l=100.
But, l=a+(n−1)d.
⇒100=25+(n−1)×5
⇒75=(n−1)×5
⇒15=(n−1)
⇒16=n
Thus, there are 16 natural numbers between 24 and 101 that are divisible by 5.
Answer: Option B. -> 178
:
B
Given that
a11=38and a16=73,
wherea11is the 11th term anda16is the 16th term of an AP.
The nth term of an AP with first term a and common difference d is given by
an=a+(n−1)d.
⇒38=a+10d....(i)
73=a+15d....(ii)
Subtracting equation (i) from equation (ii), we get
35=5d.
⇒d=7
Substituting the value of d in equation (i), we get
a=38−10d=38−10(7)=−32.
Now, a31=a+(31−1)d=−32+30(7)
=178
Therefore, the 31st term of the given AP is 178.
:
B
Given that
a11=38and a16=73,
wherea11is the 11th term anda16is the 16th term of an AP.
The nth term of an AP with first term a and common difference d is given by
an=a+(n−1)d.
⇒38=a+10d....(i)
73=a+15d....(ii)
Subtracting equation (i) from equation (ii), we get
35=5d.
⇒d=7
Substituting the value of d in equation (i), we get
a=38−10d=38−10(7)=−32.
Now, a31=a+(31−1)d=−32+30(7)
=178
Therefore, the 31st term of the given AP is 178.
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