# 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,⋯ .

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,... .

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.

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?

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.?

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.

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,⋯ .

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

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.

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.

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.

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.

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.

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.

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.

Sum of first 100 odd numbers
= 100² = 10000.

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

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.

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.

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.

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.

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.