Explore topic-wise InterviewSolutions in .

Correct option is (C) \(\frac{n(n+1)(2n+1)}{6}\)

(1) + (1+3) + (1+3+5) + ....... + upto n terms

= 1 + (1+3) + (1+3+5) + ....... + (1+3+5+.....+n terms)

= 1 + (1+3) + (1+3+5) + ....... + (1+3+5+.....+ 2n - 1)

= 1 + (1+3) + (1+3+5) + ....... + \(\frac n2(1+(2n-1))\)

\((\because S_n=\frac n2(a+a_n)\) in A.P.)

= 1 + (1+3) + (1+3+5) + ....... + \(n^2\)

\(=\sum n^2\)

\(=\frac{n(n+1)(2n+1)}{6}\)

Correct option is C) \(\cfrac{n(n+1)(2n+1)}{6}\)

Given, AP – 2, - 7, - 12, …

Let the nth term of an AP is - 77.

Then, first term, a = - 2

common difference, d = - 7 – (2) = - 7 + 2 = - 5.

Let nth term of an AP is - 77

a_n = a + (n - 1)d

- 77 = - 2 + (n - 1) (- 5)

- 75 = (n - 1) (- 5)

15 = n - 1

n = 16

so 16^th term of AP is - 77

S₁₆ is the sum of AP upto the term - 77 .i.e. 16^th term

As, S_n = n/2(a + a_n) [ as last term is given]

= 16/2 (- 2 - 77)

= 8 (- 79) = - 632

So 16^th term is - 77 and sum upto 16^th term is - 632

1. Rs 5000

2. Production during 8^th year is (a+7d)= 5000 + 2(2200) = 20400

3. Production during first 3 year = 5000 + 7200 + 9400 = 21600

4. N=12

5. Difference= 18200 -11600 = 6600

We have given Class 10 Maths MCQ Questions of Arithmetic Progressions with Answers to assist students with understanding the ideas well indeed. All of these questions are vital as it assists the students with prepare for CBSE Class 10 Maths Board Examination. 

Practice Class 10 Maths MCQ Question of Arithmetic Progressions

This point is additionally vital from the viewpoint of serious tests. Discover the CBSE Class 10 Maths Objective Questions underneath:

1. If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is

(a) 87
(b) 88
(c) 89
(d) 90

2. In an AP, if d = –4, n = 7, \(a_n\) = 4, then a is

(a) 6
(b) 7
(c) 20
(d) 28

3. In an AP, if a = 3.5, d = 0, n = 101, then \(a_n\) will be

(a) 0
(b) 3.5
(c) 103.5
(d) 104.5

4. The first four terms of an AP, whose first term is –2 and the common difference is –2, are

(a) – 2, 0, 2, 4
(b) – 2, 4, – 8, 16
(c) – 2, – 4, – 6, – 8
(d) – 2, – 4, – 8, –16

5. The number of multiples lie between n and \(x^2\) which are divisible by n is

(a) n + 1
(b) n
(c) n – 1
(d) n – 2

6. If p, q, r and s are in A.P. then r – q is

(a) s – p
(b) s – q
(c) s – r
(d) none of these

7. The famous mathematician associated with finding the sum of the first 100 natural numbers is

(a) Pythagoras
(b) Newton
(c) Gauss
(d) Euclid

8. The 21st term of the AP whose first two terms are –3 and 4 is

(a) 17

(b) 137

(c) 143

(d) –143

9. If p, q, r, s, t are the terms of an A.P. with common difference -1 the relation between p and t is:​

(a) t = p – 5
(b) t = p – 4
(c) t = p – 6
(d) t = p + 4

10. If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?

(a) 30
(b) 33
(c) 37
(d) 38

11. If the common difference of an AP is 5, then what is \(a_{18} – a_{13}\)?

(a) 5
(b) 20
(c) 25
(d) 30

12. The middle most term (s) of the AP:–11, –7, –3, ..., 49 is:

(a) 18, 20
(b) 19, 23
(c) 17, 21
(d) 23, 25

13. In an Arithmetic Progression, if a=28, d=-4, n=7, then an is:

(a)4
(b)5
(c)3
(d)7

14. If a=10 and d=10, then first four terms will be:

(a)10,30,50,60
(b)10,20,30,40
(c)10,15,20,25
(d)10,18,20,30

15. The first term and common difference for the A.P. 3,1,-1,-3 is:

(a)1 and 3
(b)-1 and 3
(c)3 and -2
(d)2 and 3

16. If 17th term of an A.P. exceeds its 10th term by 7. The common difference is:

(a)1
(b)2
(c)3
(d)4

17. If the sum of n terms of an A.P. be \(3n^2 + n\) and its common difference is 6, then its first term is

(a) 2
(b) 3
(c) 1
(d) 4

18. If the sum of n terms of an A.P. is \(2n^2 + 5n\), then its nth term is

(a) 4n – 3
(b) 3n – 4
(c) 4n + 3
(d) 3n + 4

19. If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

(a) 13
(b) 9
(c) 21
(d) 17

20. The number of multiples of 4 between 10 and 250 is:

(a)50
(b)40
(c)60
(d)30

Answers & Explanations

1. Answer: (c) 89

Explanation:

Here, a₇ = 34 

a₁₃ = 64 

a₇ = a + 6d = 34……….(1)

a₁₃ = a + 12d = 64…………(2) 

Subtracting (1) from (2) 

6d = 30 

d = 5 

Multiplying (1) by 2 

2a + 12d = 68………………….(3) 

Subtracting (2) from (3) 

a = 4 

a₁₈ = a + (n–1) d 

a₁₈ = 4 + (17) 5 

a₁₈ = 89 

2. Answer: (d) 28

Explanation: For an A.P

\(a_n\) = a + (n – 1)d

4 = a + (7 – 1)( −4)

4 = a + 6(−4)

4 + 24 = a

a = 28

3. Answer: (b) 3.5

Explanation: For an A.P

\(a_n\) = a + (n – 1)d

= 3.5 + (101 – 1) × 0

= 3.5

4. Answer: (c) – 2, – 4, – 6, – 8

Explanation: Let the first four terms of an A.P are a, a+d, a+2d and a+3d

Given that the first termis −2 and difference is also −2, then the A.P would be:

– 2, (–2–2), [–2 + 2 (–2)], [–2 + 3(–2)]

= –2, –4, –6, –8

5. Answer: (d) n – 2

6. Answer: (c) s – r

7. Answer: (c) Gauss

Explanation: Gauss is the famous mathematician associated with finding the sum of the first 100 natural Numbers.

8. Answer: (b) 137

Explanation: First two terms are –3 and 4

Therefore,

a = −3

a + d = 4

⇒ d = 4 − a

⇒ d = 4 + 3

⇒ d = 7

Thus,

\(a_{21}\) = a + (21 – 1)d

\(a_{21}\) = –3 + (20)7

\(a_{21}\) = 137

9. Answer: (b) t = p – 4

10. Answer: (b) 33

Explanation: Since

\(a_2\) = 13

\(a_5\) = 25

⇒ a + d = 13 ….(i)

⇒ a + 4d = 25 ….(ii)

Solving equations (i) and (ii), we get:

a = 9; d = 4

Therefore,

\(a_7\) = 9 + 6 × 4

\(a_7\) = 9 + 24

\(a_7\) = 33

11. Answer: (c) 25

Explanation: Since, d = 5

\(a_{18} – a_{13}\) = a + 17d – a – 12d

= 5d

= 5 × 5

= 25

12. Answer: (c) 17, 21

Explanation: Here, a = −11

d = − 7 – (−11) = 4

And \(a_n\) = 49

We have,

\(a_n\) = a + (n – 1)d

⇒ 49 = −11 + (n – 1)4

⇒ 60 = (n – 1)4

⇒ n = 16

As n is an even number, there will be two middle terms which are 16/2th and [(16/2)+1]th, i.e. the 8th term and the 9th term.

\(a_8\) = a + 7d = – 11 + 7 × 4 = 17

\(a_9\) = a + 8d = – 11 + 8 × 4 = 21

13. Answer: (a) 4

Explanation: For an AP,

\(a_n\) = a+(n-1)d

= 28+(7-1)(-4)

= 28+6(-4)

= 28-24

\(a_n=4\)

14. Answer: (b)10,20,30,40

Explanation: a = 10, d = 10

\(a_1\) = a = 10

\(a_2 = a_1+d\) = 10+10 = 20

\(a_3 = a_2+d\) = 20+10 = 30

\(a_4 = a_3+d\) = 30+10 = 40

15. Answer: (c) 3 and -2

Explanation: First term, a = 3

Common difference, d = Second term – First term

⇒ 1 – 3 = -2

⇒ d = -2

16. Answer: (a) 1

Explanation: \(N^{th}\) term in AP is:

\(a_n\) = a+(n-1)d

\(a_{17}\) = a+(17−1)d

\(a_{17}\) = a +16d

In the same way,

\(a_{10}\) = a+9d

Given,

\(a_{17} − a_{10} = 7\)

Therefore,

(a +16d)−(a+9d) = 7

7d = 7

d = 1

Therefore, the common difference is 1.

17. Answer: (d) 4

Explanation: Here, \(S_n= 3n^2 + n\)
d = 6
Putting n= 1
\(S_1\) = 3 + 1 = 4
Sum of first 1 term = first term = 4

18. Answer: (c) 4n + 3

Explanation: Here S_n = 2n² + 5n 

Sum of the A.P with 1 term = S₁ = 2 + 5 = 7 = first term 

Sum of the A.P with 2 terms = 8 + 10 = 18 

Sum of the A.P with 3 terms = 18 + 15 = 33 

a₂ = S₂ – S₁ = 18 – 7 = 11 

d = a₂ –a₁ = 11 – 7 = 4 

n^th term = a + (n–1) d 

= 7 + (n–1) 4 

n^th term = 4n + 3

19. Answer: (c) 21

Explanation: Let 3 consecutive terms A.P is a –d, a, a + d. and the sum is 51 

So, (a –d) + a + (a + d) = 51 

3a –d + d = 51 

3a = 51 

a = 17 

The product of first and third terms = 273 

So, (a –d) (a + d) = 273

 a² –d² = 273 

17² –d² = 273 

289 –d² = 273 

d² = 289 –273 

d² = 16 

d = 4 

Third term = a + d = 17 + 4 = 21

20. Answer: (c) 60

Explanation: The multiples of 4 after 10 are:

12, 16, 20, 24, …

So here, a = 12 and d = 4

Now, 250/4 gives remainder 2. Hence, 250 – 2 = 248 is divisible by 2.

12, 16, 20, 24, …, 248

So, nth term, an = 248

As we know,

an = a+(n−1)d

248 = 12+(n-1)×4

236/4 = n-1

59 = n-1

n = 60

Click here to practice more MCQ Questions from Chapter Arithmetic Progression Class 10 Maths

Correct option is (B) -4

We have \(a_{25}-a_{12}=-52\)

\(\Rightarrow(a+24d)-(a+11d)=-52\)     \((\because a_n=a+(n-1)d)\)

\(\Rightarrow24d-11d=-52\)

\(\Rightarrow13d=-52\)

\(\Rightarrow d=\frac{-52}{13}=-4\)

Correct option is B) -4

Given: 1 + 4 + 7 + 10 + … + x = 287

Here a = 1, d = 3 and S_n = 287

Sum = S_n = n/2 [2a + (n-1)d]

287 = n/2 [2 + (n-1)3]

574 = 3n² – n

Which is a quadratic equation.

Solve 3n² – n – 574 = 0

3n² – 42n + 41n – 574 = 0

3n(n – 14) + 41(n-14) = 0

(3n + 41)(n-14) = 0

Either (3n + 41) = 0 or (n-14) = 0

n = -41/3 or n = 14

Since number of terms cannot be negative, so result is n = 14.

=> Total number of terms in AP are 14.

Which shows, x = a₁₄

or x = a + 13d

or x = 1 + 39

or x = 40

The value of x is 40.

Given a₁ = 15 and a₁ + a₂ + a₃ =102 

a + a + d + a + 2d = 102

⇒ 3a + 3d = 102 

⇒ 3 (a + d) = 102 

∴ (a + d) = 102/3 = 34 

⇒ d = 34 – a = 34 – 15 = 19 

∴ a = 15, 0 = 19 

then its 10^th term = a₁₀ 

= a + (10 – 1)d 

= a + 9d = 15 + 9(19) 

= 15 + 171 = 186 

∴ 10^th term of it = 186

Let the first term of that A.P. is ‘a’. 

Then common difference (d) = 0 given then nth term = a + (n – 1) d 

= a + (n – 1) (0) 

= a + 0 = a 

∴ a_n = a

So, when d = 0 is given the ‘n’ th term of an A.P. (a_n) is equal to its first term (a).

The formula for ‘n’ th term of an A.P. is a_n = a + (n – 1) d 

The variables in. the above formula are a, d, n . 

So we should have the above three to calculate a_n.

Given: a, b, c are in AP

Since, a, b, c are in AP, we have a + c = 2b …(i)

Now, (b + c)² – a², (c + a)² – b², (a + b)² – c² will be in A.P

If (b + c – a)(b + c + a), (c + a – b)(c + a + b), (a + b – c)(a + b + c) are in AP

i.e. if b + c – a, c + a – b, a + b – c are in AP

[dividing by (a + b + c)]

if (b + c – a) + (a + b – c) = 2(c + a – b)

if 2b = 2(c + a – b)

if b = c + a – b

if a + c = 2b which is true by (i)

Hence, (b + c)² – a², (c + a)² – b², (a + b)² – c² are in A.P.

To prove: (a + 2b – c)(2b + c – a)(c + a – b) = 4abc.

Given: a, b, c are in A.P.

Proof: Since a, b, c are in A.P.

⇒ 2b = a + c … (i)

Taking LHS = (a + 2b – c) (2b + c – a) (c + a – b)

Substituting the value of 2b from eqn. (i)

= (a + a + c – c) (a + c + c – a) (c + a – b)

= (2a) (2c) (c + a – b)

Substituting the value of (a + c) from eqn. (i)

= (2a) (2c) (2b – b)

= (2a) (2c) (b) = 4abc

= RHS

Hence Proved

S_n = 3n² + 2n

Taking n = 1, we get

S₁ = 3(1)² + 2(1)

⇒ S₁ = 3 + 2

⇒ S₁ = 5

⇒ a₁ = 5

Taking n = 2, we get

S₂ = 3(2)² + 2(2)

⇒ S₂ = 12 + 4

⇒ S₂ = 16

∴ a₂ = S₂ – S₁ = 16 – 5 = 11

Taking n = 3, we get

S₃ = 3(3)² + 2(3)

⇒ S₃ = 27 + 6

⇒ S₃ = 33

∴ a₃ = S₃ – S₂ = 33 – 16 = 17

So, a = 5,

d = a₂ – a₁ = 11 – 5 = 6

Now, we have to find the 15^th term

a_n = a + (n – 1)d

a₁₅ = 5 + (15 – 1)6

a₁₅ = 5 + 14 × 6

a₁₅ = 5 + 84

a₁₅ = 89

Hence, the 15^th term is 89 and AP is 5, 11, 17, 23,…

Given AP, – 2, - 4, - 6, …, - 100.

Here,

first term, a = - 2

common difference, d = - 4 – (2) = - 2

last term, l = - 100.

We know that, the nth term 'a_n' of an AP from the end is

a_n = l - (n - 1)d

where l is the last term and d is the common difference.

12th term from the end,

a₁₂ = - 100 - (12 - 1) - 2

= - 100 + 22

= - 78

Hence, the 12th term from the end is - 78

D. 0

Let a and d be first term and common difference respectively

According to the question,

7 times the 7th term of an AP is equal to 11 times its 11th term

⇒ 7a_{7 =} 11a₁₁

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

we have,

7(a + 6d) = 11(a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 4a + 68d = 0

⇒ a + 17d = 0

⇒ a_{18 =} 0

18th term of AP is 0

Let the first term and common difference of an AP are a and d, respectively.

Given

a₃ + a₈ = 7

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

a + 2d + a + 7d = 7

2a + 9d = 7

2a = 7 - 9d [ Eqn 1]

a₇ + a₁₄ = - 3

a + 6d + a + 13d = - 3

2a + 19d = - 3

7 - 9d + 19d = - 3 [ using eqn i]

7 + 10d = - 3

10d = - 10

d = - 1

using this value in eqn i

2a = 7 - 9 (- 1)

2a = 16

a = 8

Now,

a₁₀ = a + 9d

= 8 + 9 (- 1)

= 8 - 9 = - 1

Let the a be first term and d be common difference

As we know

a_n = a + (n - 1)d

Given,

a₇ = a₁₁ - 24

a + 6d = a + 10d - 24

10d - 6d = 24

4d = 24

d = 6

given a = 12

then, a₂₀ = a + 19d

a₂₀ = 12 + 19(6)

= 12 + 114

= 126

(B) 33

Explanation:

We know that the nth term of an AP is

a_{n =} a + (n – 1)d

Where,

a = first term

a_n is nth term

d is the common difference

a_{2 =} a + d = 13 …..(1)

a_{5 =} a + 4d = 25 …… (2)

From equation (1) we have,

a = 13 – d

Using this in equation (2), we have

13 – d + 4d = 25

13 + 3d = 25

3d = 12

d = 4

a = 13 – 4 = 9

a_{7 =} a + 6d

= 9 + 6(4)

= 9 + 24 = 33

Let the first term, common difference be a and d respectively.

As we know

a_n = a + (n - 1)d

and Given

a₅ + a₇ = 52

a + 4d + a + 6d = 52

2a + 10d = 52

a + 5d = 26

a = 26 - 5d [ eqn i]

also we have given

a₁₀ = 46

a + 9d = 46

26 - 5d + 9d = 46 [ from eqn i]

4d = 46 - 26

4d = 20

d = 5

using this value in eqn I , we get

a = 26 - 5d

a = 26 - 5(5)

a = 1

as a = 6 and d = 5

So, required AP is a, a + d, a + 2d, a + 3d, …. i.e., 1, 1 + 5, 1 + 2(5), 1 + 3(5), …. i.e., 1, 6, 11, 16, …

Let a be the first term and d be the common difference of an AP.

nth term = a_n = a + (n – 1)d

Given: 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94

Now,

a₉ = a + 8d = -32 …(1)

a₁₁ = a + 10d

a₁₃ = a + 12d

Sum of 11th and 13th terms:

a₁₁ + a₁₃ = a + 10d + a + 12d

-94 = 2a + 22d

or a + 11d =-47 …(2)

Subtracting (1) from (2) , we have

3d = -47 + 32 = -15

or d = -5

Common difference is -5.

Let a be the first term and d be the common difference of an AP.

a_n = a + (n – 1)d

a₄ = a + (4 – 1)d = a + 3d

a + 3d = 11 ………(1)

Now, a₅ = a + 4d and a₇ = a + 6d

Now, a₅ + a₇ 

= a + 4d + a + 6d 

= 2a + 10d

2a + 10d = 34

a + 5d = 17 ……..(2)

Subtracting (1) from (2), we get

2d = 17 – 11 = 6

d = 3

The common difference = 3

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 

Where n > 0 

a = a₁ = 1, a₂ = 4, l = 88 

Common difference, 

d = a₂ – a₁ 

= 4 – 1 

= 3 

We know,

n^th term from end, 

b_n = l – (n – 1)d 

Where l is last term or a₁ and d is common difference and n is any natural number 

∴ b₁₂ = 88 – (12 – 1)3 

⇒ b₁₂ = 88 – 36 + 3 

⇒ b₁₂ = 91 – 36 

⇒ b₁₂ = 55 

Hence, 

12^th term from end for the given A.P is 55.

Correct option is (A) 10,000

First odd number is 1.

We know that two consecutive odd numbers is differ by 2.

\(\therefore\) Common difference of arithmetic progression of odd numbers is d = 2.

First term of odd numbers is a = 1.

\(\therefore\) Sum of first 100 odd numbers is

\(S_{100}=\frac{100}2[2a+(100-1)d]\)     \((\because S_n=\frac n2[2a+(n-1)d])\)

\(=50\,(2\times1+99\times2)\)               \((\because a=1\;\&\;d=2)\)

\(=50\times2(1+99)\)

\(=100\times100=10000\)

Hence, sum of first 100 odd numbers is 10000.

Correct option is A) 10,000

For finding total two-digit numbers which are divisible by 3, 

Firstly we will make an A.P. of those two-digit numbers which are divisible by 3. 

First two digit number which is divisible by 3 is 12 

∴ a₁ = a = 12 

Next two digit number which is divisible by 3 is 15 

∴ a₂ = 15 

Largest two digit number which is divisible by 3 is 99 

∴ a_n = 99 

⇒ A.P. is 12, 15,………,99 

We know, 

a_n = a + (n – 1)d 

Where a is first term or a and d is common difference and n is any natural number 

⇒ a₁ = 12, a₂ = 15 and a_n = 99 

Common difference, 

d₁ = a₂ – a₁ 

= 15 – 12 

= 3 

Now, 

a_n = a₁ + (n – 1)d 

⇒ a_n = 12 + (n – 1)3 

⇒ 99 = 12 + 3n – 3 

⇒ 99 = 3n + 9 

⇒ 99 – 9 = 3n 

⇒ 90 = 3n 

⇒ n = 30 

Hence, 

There are total 30 two-digit numbers which are divisible by 3.

Correct option is (B) 7

Given that \(n^{th}\) term of A.P. is 2n+5.

\(\therefore a_n=2n+5\)

\(\therefore\) First term is \(a_1=2\times1+5\)

\(=2+5=7\)

Correct option is B) 7

Given : 

6^th term of an A.P is 12 and 8^th terms of an A.P. is 22 

⇒ a₆ = 12 and a₈ = 22 

We know, 

a_n = a + (n – 1)d 

where a is first term or a₁ and d is common difference and n is any natural number 

When n = 6 : 

∴ a₆ = a + (6 – 1)d 

⇒ a₆ = a + 5d 

Similarly, 

When n = 8 : 

∴ a₈ = a + (8 – 1)d 

⇒ a₈ = a + 7d 

According to question : 

a₆ = 12 and a₈ = 22 

⇒ a + 5d = 12 ………………(i) 

And a + 7d = 22…………..(ii) 

Subtracting equation (i) from (ii) : 

a + 7d – (a + 5d) = 22 – 12 

⇒ a + 7d – a – 5d = 10 

⇒ 2d = 10

⇒ d = \(\frac{10}{2}\)

⇒ d = 5

Put the value of d in equation (i) : 

a + 5(5) = 12 

⇒ a + 25 = 12 

⇒ a = 12 – 25 

⇒ a = -13 

As, 

a_n = a + (n – 1)d 

a₂ = a + (2 – 1)d 

⇒ a₂ = a + d 

Now,

Put the value of a = 9 and d = 2 in an and a₂ 

⇒ a_n = a + (n – 1)d 

⇒ a_n = -13 + (n – 1)5 

⇒ a_n = -13 + 5n – 5 

⇒ a_n = -18 + 5n 

a₂ = a + d 

⇒ a₂ = -13 + 5 

⇒ a₂ = -8 

Hence, 

2^th term and n^th of the given A.P. are -8 and 5n – 18 respectively.

Given, 

6^th term of an A.P is 19 and 

17^th terms of an A.P. is 41 

⇒ a₆ = 19 and 

a₁₇ = 41 

We know, 

a_n = a + (n – 1)d 

Where a is first term or a1 and d is common difference and n is any natural number 

When n = 6 : 

∴ a₆ = a + (6 – 1)d 

⇒ a₆ = a + 5d 

Similarly, 

When n = 17 : 

∴ a₁₇ = a + (17 – 1)d 

⇒ a₁₇ = a + 16d 

According to question : 

a₆ = 19 and a₁₇ = 41 

⇒ a + 5d = 19 ………………(i) 

And a + 16d = 41…………..(ii) 

Subtracting equation (i) from (ii) : 

a + 16d – (a + 5d) = 41 – 19 

⇒ a + 16d – a – 5d = 22 

⇒ 11d = 22 

⇒ d = \(\frac{22}{11}\)

⇒ d = 2 

Put the value of d in equation (i) : 

a + 5(2) = 19 

⇒ a + 10 = 19 

⇒ a = 19 – 10 

⇒ a = 9 

As, 

a_n = a + (n – 1)d 

a_{40 =} a + (40 – 1)d 

⇒ a₄₀ = a + 39d 

Now,

Put the value of a = 9 and d = 2 in a₄₀ 

⇒ a₄₀ = 9 + 39(2) 

⇒ a₄₀ = 9 + 78 

⇒ a₄₀ = 87 

Hence, 

40^th term of the given A.P. is 87.

Correct option is (D) 111

First number between 1 & 1000 which is divisible by 9 is 9

& last number between 1 & 1000 which is divisible by 9 is 999.

i.e., \(a_1=a=9\;\&\;a_n=999,d=9\)

\(a_n=a+(n-1)d\)

\(\Rightarrow999=9+(n-1)d\)

\(\Rightarrow9(n-1)=999-9=990\)

\(\Rightarrow n-1=\frac{990}9=110\)

\(\Rightarrow n=110+1=111\)

Hence, there are total 111 numbers which are divisible by 9 in between 1 & 1000.

Correct option is B) 111

Correct option is (C) 10

5, 15, 25, ........., 95 is the arithmetic progression of common difference 10 whose terms are odd numbers between 1 & 100 and divisible by 5.

\(\therefore a_1=a=5\;\&\;d=10,a_n=95\)

Now, \(a_n=a+(n-1)d\)

\(\Rightarrow95=5+(n-1)10\)

\(\Rightarrow10(n-1)=95-5=90\)

\(\Rightarrow n-1=\frac{90}{10}=9\)

\(\Rightarrow n=9+1=10\)

Hence, there are total 10 odd numbers between 1 & 100 that are divisible by 5.

Correct option is C) 10

Correct option is (C) 2048

Given G.P. is 2, 8, 32, .......

\(\therefore a_1=a=2,a_2=8\)

\(r=\frac{a_2}{a_1}=\frac82=4\)

\(\therefore a_6=ar^5=2.4^5\)

\(=2.1024=2048\)

Hence, \(6^{th}\) term of given G.P. is 2048.

Correct option is C) 2048

Given, an A.P of 60 terms

And, a = 7 and a₆₀ = 125

We know that a_n = a + (n – 1)d

⇒ a₆₀ = 7 + (60 – 1)d = 125

7 + 59d = 125

59d = 118

d = 2

So, the 32^nd term is given by

a₃₂ = 7 + (32 -1)2 = 7 + 62 = 69

⇒ a₃₂ = 69

Correct option is (B) 15

First two-digit number which is divisible by 6 is 12 and last two-digit number which is divisible by 6 is 96.

12, 18, 24, ........., 96 is the A.P. of two-digit numbers which are divisible by 6.

\(\therefore a=12,a_n=96,d=a_2-a_1=18-12=6\)

Now, \(a_n=a+(n-1)d\)

\(\Rightarrow96=12+(n-1)6\)

\(\Rightarrow6(n-1)=96-12=84\)

\(\Rightarrow n-1=\frac{84}6=14\)

\(\Rightarrow n=14+1=15\)

Hence, there are 15 two-digit numbers which are divisible by 6.

Correct option is B) 15

Let four numbers in A.P. are

a, a + d, a + 2d, a + 3d

According to question

a + (a + d) + (a + 2d) + (a + 3d) = 50

⇒ 4a + 6d = 50

⇒ 2(2a + 3d) = 50

⇒ 2a + 3d = 25 …..(i)

If larger number is 4 times the smaller number, then equation will be as follow,

a + 3d = 4 x a

⇒ a + 3d = 4a

⇒ 3d = 3a

⇒ d = a …(ii)

from equation (i) and (ii)

2a + 3a = 25 [∴ d = a]

a = 25

a = 5

∴ d = 5 [∵ eqⁿ(ii)]

∴ Numbers a = 5

a + d = 5 + 5 = 10

a + 2d = 5 + 2 × 5 = 15

a + 3d = 5 + 3 × 5 = 20

Hence, four numbers are 5, 10, 15, and 20

Correct option is (B) 128

The first number which is divisible by 7, between 100 & 1000 is 105 and last number which is divisible by 7 between 100 & 1000 is 994.

\(\therefore a_1=105,d=7\;\&\;a_n=994\)

\(\Rightarrow a+(n-1)d=994\)            \((\because a_n=a+(n-1)d)\)

\(\Rightarrow105+(n-1)7=994\)

\(\Rightarrow7(n-1)=994-105=889\)

\(\Rightarrow n-1=\frac{889}7=127\)

\(\Rightarrow n=1+127=128\)

Hence, total 128 numbers between 100 & 1000 which are divisible by 7.

Correct option is B) 128

Let in four numbers,

First number = a – 3d

Second number = a – d

Third number = a + d

Fourth number = a + 3d

Sum of four number is 20

∴ 20 = (a – 3d) + (a – d) + (a + d) + (a + 3d)

⇒ 20 = 4a

⇒ a = 5

Sum of squares of four numbers is 120.

∴ (a – 3d)² + (a – d)² + (a + d)² + (a + 3d)² = 120

⇒ [a² + 9d² – 6ad + a² + d² – 2ad + a² + d² + 2ad + a² + 9d² + 6ad] = 120

⇒ 4 (a² + 5d²) = 120

⇒ a² + 5d² = 30 [∵ a = 5]

⇒ 5² + 5d² = 30

⇒ 25 + 5d² = 30

⇒ 5d² = 30 – 25

⇒ 5d² = 5

⇒ d² = 1

⇒ d = ±1

Thus, a = 5 and d = ± 1

∴ Four numbers are 2, 4, 6, 8 or 8, 6, 4, 2

Odd numbers divisible by 3, between 1 and 1000 are 3, 9, 15, 21 …….. 999.

Clearly series 3, 9, 15, 15,21 …… 999 is A.P.

whose first term (a) = 3 and common difference (d) = 6.

Let us assume that this series contains n terms.

∴ a_n = 999

⇒ a + (n – 1)d = 999

⇒ 3 + (n – 1) × 6 = 999

⇒ 6n – 3 = 999

⇒ 6n = 1002

⇒ n = 1002/6

⇒ n = 167

∴ Required sum

S_n = n/2(a + l)

S₁₆₇ = 167/2(3 + 999)

= 167/2 × 1002

= 167 × 501

= 83667

Hence, required sum = 83667

Correct answer is (c) 4920

The positive integers divisible by 6 are 6, 12, 18,....

This is an AP with a = 6 and d = 6.

Also, n = 40 (Given)

Using the formula, \(S_n=\frac{n}{2}[2a+(n-1)d],\) we get

\(S_{40}=\frac{40}{2}[2\times6+(40-1)\times6]\)

= 20(12 + 234)

= 20 x 246

= 4920

Thus, the required sum is 4920

(i) Given A.P. : 9, 17, 25, …

First term (a) = 9, common difference (d) 

= 17 – 9 = 8

Let no. of terms be n

S_n = 636

⇒ n/2[2a + (n – 1)d] = 636

⇒ n/2[2 × 9+ (n – 1)8] = 636

⇒ n/2[18 + 8n – 8] = 636

⇒ n/2[8n + 10] = 636

⇒ n(4n + 5) = 636

⇒ 4n² + 5n = 636

⇒ 4n² + 5n – 636 = 0

⇒ 4n² + 53 n – 48n – 636 = 0

⇒ n(4n + 53) – 12(4n + 53) = 0

⇒ (4n + 53)(n – 12) = 0

⇒ n – 12 = 0 or 4n + 53 = 0

⇒ n = 12 or – 53/4

∵ n cannot be  negative

So, ignore n = – 53/4

∴ n = 12

Hence, sum of 12 terms of given A.P. is 636.

(ii) Given A.P. 63, 60, 57 …..

First term(a) = 63

Common difference (d) = 60 – 63 = -3

Let number of terms be n

S_n = 693

We know that

S_n = n/2[2a + (n – 1)d]

⇒ 693 = n/2[2 × 63 + (n – 1) (-3)]

⇒ 693 = n/2[126 – 3n + 3]

⇒ 1386 = n(129 – 3n)

⇒ 1386 = 129n – 3n²

⇒ 3n² – 129n + 1386 = 0

⇒ n² – 43n + 462 = 0

⇒ n² – 21n – 22n + 462 = 0

⇒ = n(n – 21) – 22(n – 21) = 0

⇒ (n – 21)(n – 22) = 0

⇒ n – 21 = 0 or n – 22 = 0

n = 21 or n = 22

By taking 21 or 22 terms of given A.P. we will get sum 693.

Here, s_p = ap² + bp 

S₁= a + b 

S₂= ax² + bx² 

= 4a + 2b 

S₂ = a₁ + a₂ 

4a + 2b = a₁ + a₂ 

A₂ = 4a + 2b – (a + b) 

A₂ = 3a + b 

Now, d = common difference = a₂ – a₁ 

= 3a + b – (a + b)

= 2a

Answer is (B) 6

a= 1, l = 11, S_n = 36

S_n = n/2(a + l)

36 = n/2(1 + 11)

⇒ 36 = n/2 × 12

⇒ 36 = 6n

⇒ n = 6

Correct option is (B) 133

(A) \(\frac{100}7=14\frac27\) not divisible by 7

(B) \(\frac{133}7=19\) is divisible by 7

(C) \(\frac{137}7=19\frac47\) not divisible by 7

(D) \(\frac{143}7=20\frac37\) not divisible by 7

Correct option is B) 133

Number of terms (n) = 60

First term (a) = 7

Last term (a_n) = 125

Formula, a_n = a + (n – 1)d

⇒ 125 = 7 + (60 – 1)d

⇒ 125 – 7 = 59 d

⇒ 118 = 59 d

⇒ d = 118/59

⇒ d = 2

Thus, 32^nd term

a₃₂ = a + (32 – 1)d

= 7 + 31 × 2

= 7 + 62 = 69

Hence, 32^nd term of A.P. is 69.

Let first term of A.P. is a and common difference is d.

Given a₃ = 16

a + (3 – 1)d = 16

⇒ a + 2d = 16 …(i)

According to question, a₇ = 12 – a₅

⇒ a₇ – a₅ = 12

[a + (7 – 1)d] – [a+(5 – 1)d] = 12

⇒ a₇ + 6d – a – 4d = 12

⇒ a + 6d – a – 4d = 12

⇒ 2d = 12

d = 12/2 = 6

Substituting value of d in equation (i)

a + 2(6) = 16

∴ a = 16 – 12 = 4

AP. a, a + d, a + 2d,…

= 4, 4 + 6, 4 + 2 × 6,…

= 4, 10, 16,…

Hence, required A.P. is 4, 10, 16, 22,…

 A.P. is \(\frac{1}{2q}, \frac{1-2q}{2q}, \frac{1-4q}{2q},...., \)

d= a₂ – a₁

d = \(\frac{1-2q}{2q}- \frac{1}{2q}\)

d = \(\frac{1-2q + 1}{2q}\)

d = -1

Given A.P.

1, 2, 3, 4, 5 …….. 1000

First term (a) = 1

Common difference (d) = 2 – 1 = 1

Last term (l) = 1000

Formula, S_n = n/2(a + l)

S₁₀₀₀ = 1000/2 (1 + 1000)

= 500 × 1001 = 500500

Hence, sum of first 1000 positive integers.

= 500500

Correct answer is (c) 400

The first 20 odd natural numbers are 1, 3, 5,.., 39.

These numbers are in AP.

Here. a = 1, l = 39 and n = 20

∴ Sum of first 20 odd natural numbers

= \(\frac{20}{2}(1+39)\)      [\(S_n=\frac{n}{2}(a+l)\)]

= 10 x 40

= 400

Here A.P is 1,3, 5, …… 

a= 1, d= 2 and n = 20 

S_n = n/2(2a + (n–1) d) 

S_n = 20/2 (2 x 1 + (19) 2) 

S_n = 10 (2 + 38) 

S_n= 400

Given A.P.

5, 11, 17, 23, ……..

First term (a) = 5

Common difference (d) = 11 – 5 = 6

n^th term (a_n) = 299

Formula a_n = a + (n – 1)d

⇒ 299 = 5 + (n – 1) × 6

⇒ 299 – 5 = (n – 1) 6

⇒ 294 = (n – 1) 6

⇒ n – 1 = 294/6

⇒ n – 1 = 49

⇒ n = 49 + 1

⇒ n = 50

n is whole no.

∴ Hence, 299 is 50^th term of sequence 5, 11, 17, 23 ….

Given,

First term (a) = 7, last term (a_n) = 49 and sum of n terms (S_n) = 420

Now, we know that

a_n = a + (n – 1)d

⟹ 49 = 7 + (n – 1)d

⟹ 43 = nd – d

⟹ nd – d = 42  ….. (1)

Next,

S_n = \(\frac{n}{2}\)(2(7) + (n − 1)d)

⟹ 840 = n[14 + nd – d]

⟹ 840 = n[14 + 42] [using (1)]

⟹ 840 = 54n

⟹ n = 15 …. (2)

So, by substituting (2) in (1), we have

nd – d = 42

⟹ 15d – d = 42

⟹ 14d = 42

⟹ d = 3

Therefore, the common difference of the given A.P. is 3.

(i) Given A.P. 21, 18, 15, ….

First term (a) = 21

Common difference

(d) = 18 – 21 = -3

∵ a_n = a + (n – 1)d

According to question

-81 = 21 + (n – 1)(-3)

⇒ -81-21 = (n – 1) × -3

⇒ -102 = (n – 1) × -3

⇒ (n – 1) = -102/3

⇒ n – 1 = 34

⇒ n = 34 + 1 = 35

Hence, 35^th term of given series is -81.

(ii) Given A.P. 84, 80, 76…..

First term (a) = 84

Common difference (d) = 80 – 84 = -4

∵ a_n = a + (n – 1)d

According to question,

0 = 84 + (n – 1)(-4)

⇒ -84 = (n – 1) × -4

⇒ (n – 1) = -84/-4

⇒ n – 1 = 21

⇒ n = 21 + 1 = 22

Hence. 22^nd term of given A.P. is zero.

(iii) Given A.P. 5, 11, 17, 23 …..

First term (a) = 5

Common difference

(d) = 11 – 5 = 6

∵ a_n = a + (n – 1)d

According to question.,

⇒ 301 = 5 + (n – 1)(6)

⇒ 301 – 5 = 6(n—1)

⇒ 6(n – 1) = 296

⇒ (n – 1) = 296/6

⇒ n – 1 = 49.33

⇒ n = 49.33 + 1 = 50.33

∴ Value of n cannot be fraction, it means n is not a whole number. Thus no term can be 301 in given series A.P.

(iv) Given A.P. : 11, 8, 5, 2 …

First term (a) = 11 and common difference (d) = 8 – 11 = – 3

Let n^th term, a_n = -150

⇒ a + (n – 1)d = -150

⇒ 11 + (n – 1) × (-3) = -150

⇒ -3(n – 1) = -150 – 11 = -161

⇒ (n – 1) = -161/-3

= 53.6 (approx)

∴ n = 53.6 + 1 = 54.6

⇒ n is not a whole no.

Hence, -150 is no  term is given A.P.

Given,

The first term of the A.P (a) = 22

The nth term of the A.P (l) = -11

And, sum of all the terms S_n = 66

Let the common difference of the A.P. be d.

So, finding the number of terms by

66 = (\(\frac{n}{2}\))[22 + (−11)]

66 = (\(\frac{n}{2}\))[22 − 11]

(66)(2) = n(11)

6 × 2 = n

n = 12

Now, for finding d

We know that, l = a + (n – 1)d

– 11 = 22 + (12 – 1)d

-11 = 22 + 11d

11d = – 33

d = – 3

Hence, the number of terms is n = 12 and the common difference d = -3