### 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.