Arithmetic Progression, Mean And Sum of Arithmetic Series


MCQs of Arithmetic Progression, Mean And Sum of Arithmetic Series


1.Find the 45th term of the arithmetic sequence 2,5,8,11,⋯ .

     
     
     
     
... Answer is A)
Given sequence, 2,5,8,11,⋯ .
Here, first term, a=2,
number of terms, n=45,
Common difference, d=3
So, The 45th term is:
t45=a+(n−1)d
=2+(45−1)3
=2+44×3
=134


2.Find the 20th term of the arithmetic sequence 5,8,11,54,... .

     
     
     
     
... Answer is C)
Given Sequence, 5,8,11,54,... .
a=5,d=8−5=3
So, t20 =a+(n−1)d
=5+(20−1)(3)
=5+19×3
=62


3. Find the number of terms in the series 8, 12, 16, . . .,72.

     
     
     
     
... Answer is A)
Given Sequence, 8, 12, 16, . . .,72.
First term, a = 8
Last term, l = 72
Common difference, d = 12 – 8 = 4.
Number of terms:?
We have,last term (l)= a+(n-1)d
or, 72 =8+(n-1) 4
or, 72 =8+ 4n- 4
or, 68=4n
n= 17


4. How many terms are there in the series 2+4+6+ ... +64?

     
     
     
     
... Answer is B)
First term (a) = 2
Common difference (d) =4-2=2
Here, Last term (tₙ) =64
i.e. a+ (n-1) d =64
or, 2+ (n-1) 2 =64
or, 2+2n-2 =64
∴n= 32


5. The 6th term of an A.P. is 24 and the 8th term is 30. Are 63 and 65 terms of this A.P.?

     
     
     
     
... Answer is C)
Here, 6th term =24
i.e. a+5d = 24 ...(i)
Again,8th term = 30 ...(i)
Subtracting (i) from (ii) we get,
2d = 6
d= 3
Putting the value of d in equation (i) we get,
a+5d=24
or,a+5×3 = 24
or,a = 9
If possible, let 63 is the nth term of A.P.
Then, a+(n-1)d= 63
or,9+(n-1)3 = 63
or,9+3n-3 = 63
or,3n = 57
or,n=19
∴63 is the 19th term of A.P.
If possible, let 65 is the pth term of A.P.
Then, a+(p-1)d = 65
or,9+(p-1)3 = 65
or,3p+6 = 65
or, p=59/3
∴65 is not any term of the A.P. because p is fraction, which is not possible.


6. The nth term of the series 5+3+1+ ... is same as the nth term of the series 9+6+3+ ... Find n.

     
     
     
     
... Answer is A)
For the series 5+3+1+ ...
First term (a) =5
Common difference (d) =3-5=-2
nth term = a+(n-1)d
=5+(n-1)(-2)=7-2n
For the series 9+6+3+ ...
First term (a) =9
Common difference (d) = 6-9 =-3
nth term = a+(n-1)d
=9+(n-1) (-3) = 12-3n
∴n=4


7.Find the sum of first 40 terms of the arithmetic sequence 5,7,9,11,13,⋯ .





... Answer is C)
Given sequence, 5,7,9,11,13,⋯ .
Here, first term, a=5,
number of terms, n=40,
Common difference, d=2
Now, we know,
Sₙ=n/2 {2a + (n-1)d}
=40/2 {2×5 + (40-1)2}
=20(10+78)
=20×88
=1760


8. From the series 2 + 4 + 6 + 8 +... + 40





... Answer is A)
Given series, 2 + 4 + 6 +8+...+ 40
Common difference (d) = 2
Last term (l) = 40
So,tₙ = a + (n-1)d
or, 40 = 2+(n-1)2
or,40 = 2 + 2n - 2
or2n =40
∴n=20
∴Sₙ = n/2(a+l)
=20/2(2+40) = 10×42 =420


9. Find the first term of an arithmetic series whose sum of its first seven terms is 0 and common difference is -3.




... Answer is B)
Sum of 7 terms (S₇) = -0
Common difference(d)= -3
First term (a) = ?
We, know, Sₙ= n/2{2a+(n-1)d}
or,S₇ = 7/2{2a+(7-1)(-3)}
or,0 = 7(2a-18)
or,2a=18
∴a=9


10. Find the number of terms in an arithmetic series which has its first term 49, last term -31 and the sum 153.




... Answer is C)
First term (a) =49
last term(l)=-31
Sum of the terms(Sₙ) = 153
We have, (Sₙ) =n/2(a+l)
or,153 = n/2{49+(-31)}
or,153 = n/2(18)
or,153×2 = 18n
∴n =17


11. What is the common difference when the first term is 1, the last term is 50 and the sum 204.




... Answer is D)
First term(a)=1
last term (l)=50
Sum of the terms (Sₙ)=204
or,204 =n/2(a+l)
or,204 =n/2(1+50)
or,408 =51n
∴n =8
Now,(tₙ) =a+(n-1)d
or,50 =1+(8-1)d
or,50 =1+7d
or,49 =7d
∴d =7.


12. If the fourth term of an A.P. is 1 and the sum of its eight terms is 18,Find the tenth term of the series.




... Answer is B)
Fourth term(a+3d) = 1...(i)
Sum of its first 8 terms (S₈) = 18
or,18 = 8/2{2a+(8-1)d}
or,18/4 = (2a+7d)
or2a+7d = 9/2...(ii)
(ii) - (i)
∴d=5/2
putting the value of d in equation (i),we get
or,a+3×(5/2) = 1
or,2a + 15=2
∴a=-13/2
∴t₁₀= a+(n-1)d
=-13/2 +(10-1) 5/2
=(-13+45)/2
=32/2 =16 .


13. The sum of first 4 terms of an A.P. is 26 and the sum of first 8 terms is 100. Find the sum of first 12 terms.




... Answer is B)
S₄=26
or,4/2 {2a+(4-1)d}= 26
2(2a+3d)=26
2a+3d = 13...(i)
S₈ =100
or,8/2 {2a +(8-1)d}= 100
or, 4(2a+7d) = 100
or, 2a+7d = 25...(ii)
(ii)-(i)
or,4d=12
d=3
putting the value of d in equation (i),we get
2a+7d = 25
2a +7×3 = 25
∴a=2
S₁₂= 12/2 {2×2+(12-1)3}
=6(4+33)
=222


14. Find three numbers in A.P. such that their sum is 12 and the sum of their squares is 50.




... Answer is C)
Let the three numbers in A.P. be (a-d) ,a, (a+d)
Now,Sum of the numbers = 12
i.e. (a-d),a,(a+d) = 12
or, 3a = 12
∴a=4
Again,(a-d)²+ a² +(a+d)² =50
(4-d)² + 4² +(4+d)²= 50
16-8d+d² + 16 +16+8d+d² = 50
2d²=2
∴d=(+1) or ( -1)
when a= 4 and d=1
First number = a-d = 3
second number = a= 4
third number =a+d = 4+1=5
when a=4 and d= -1
First number=a-d =5
Second number = a= 4
Third number = a+d = 3


15. Find the sum of the first 100 odd numbers.




... Answer is A)
Sum of first 100 odd numbers
= 100² = 10000.


16. Find the sum of all numbers between 200 and 400 divisible by 7.




... Answer is B)
Here,First term (a) =203
Common difference (d) =210-203 = 7
Last term (l) 399
i.e. a+(n-1)d =399
203+(n-1)7 = 399
7n -7 = 196
7n = 203
n = 29
Now,required sum = n/2 (a+l)
=29/2 (203+399)
=29×301
=8729


17.Find the A.M.between 5 and 15.
     
     
     
     
... Answer is B)
Let m be the A.M between 5 and 15
So,a=5, b=15, m=?
We have, A.M.=(a+b)/2
or,m=(5+15)/2=10


18. If 6, p, q, and 18 are in A.P. find the value of p and q.
     
     
     
     
... Answer is C)
Here,First term (a)=6
Number of terms (n)=4
Last term (tₙ)=18
i.e. a+(n-1)d =18
or,6+(4-1)d=18
or,3d =12
or,d=4
Now,p=a+d =6+4 =10
q=a+2d =6+2×4 =14


19. The product of two numbers is 45 and their arithmetic mean is 9. Find the numbers.
     
     
     
     
... Answer is A)
Let a and b be the two required numbers. Then.
a.b =45
b =45/a ...(i)
Again,(a+b)/2 =9
or,a+b =18 ...(ii)
Putting the value of b in (ii) we get,
a+ 45/a =18
a²+45 =18a
a²-3a-15a+45=0
a(a-3)-15(a-3)=0
(a-15)(a-3)=0
∴ a=15 or 3
If a=15, then b=45/15 =3
If a=3, then b=45/3 =15
Hence, required two numbers are 15 and 3 or 3 and 15


20. There are n arithmetic means between 20 and 80 such that the ratio of the first mean to the last mean is 1:3, Find n.

     
     
     
     
... Answer is B)
Here,the sequence is 20,...,80.
Since there are n A.M's, the total number of terms =n+2(As first and last term is left)
Since,they are in A.P, we can write 20,20+d, ...,80-d,80
Where,20+d is the second term or first mean and 80-d is the second last term or last mean.
By the question,
(First mean)/last mean =1/3
or (20+d)/80-d =1/3
or,60+3d =80-d
or,4d =20 ∴d=5
Now,we know,
t =a+(n-1)d
or,80 =20+(n-1)5
or,60 =5n-5
or,5n =65
∴n= 65/5 =13
There are n-2=13-2=11
A.M.'s between 20 and 80.


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