9

Q1. If A₁, A₂, A₃,..., A_n are n numbers between a and b, such that a, A₁, A₂, A₃,..., A_n, b are in A.P., then

• A₁, A₂, A₃,...., A_n are arithmetic means between a and b
• A₁, A₂, A₃,...., A_n all are equal
• A₁, A₂, A₃,....., A_n are in geometric progression
• A₁, A₂, A₃,....., A_n are in harmonic progression

Solution

If A₁, A₂, A₃,..., A_n are n numbers between a and b, such that a, A₁, A₂, A₃,..., A_n, b are in A.P., then A₁, A₂, A₃,...., A_n are arithmetic means between a and b.

Q2. The sum of the series 1³ - 2³ + 3³ - 4³ + ....... + 9³ is:

• 1225
• 925
• 425
• -1525

Solution

S_n = 1³ - 2³ + 3³ - 4³ + ... + 9³ = 1³ + 3³ + 5³ + ... + 9³ -  (2³ + 4³ + ... + 8³) = S₁ - S₂  (let)           For S₁, t_n = (2n - 1)³ = 8n³ - 12n² + 6n - 1 Here n = 5 Hence, S₁ = 2  25  36 - 60  11+ 3 30 - 5                 = 1800 - 660 + 90 - 5                 = 1890 - 665 = 1225 For S₂, t_n= 8n³; = 2  16  25 = 800. (for n = 4) Required sum = 1225 - 800 = 425

Q3. Three numbers are in G.P. if each is multiplied by 2 , the resulting sequence is

• A G.P. with the same common ratio as the original G.P
• A G.P. with double the common ratio as the original G.P
• A G.P. with the common ratio 2 more than the common ratio of   the original G.P
• A G.P. with half the  common ratio as the original G.P

Solution

Let the G.P. be a,ar,a r² ..  Then the new G.P. is 2a,2ar,2ar²..   or (2a),(2a)r,(2a)r² So we see that the common ratio is same as that of the original G.P.

Q4. What is the 10^th term of the sequence defined by a_n = (n + 1) (2 + n) (3 - n) ?

• 924
• -924
• 792
• -792

Solution

a₁₀ = (10 + 1) (2 + 10) (3 - 10) = (11)(12)(-7) = - 924

Q5. The A.M. between two numbers is 34 and their G.M. is 16.The numbers are

• 62 and 6
• 32 and 8
• 64 and 4
• 60 and 18

Solution

Let the two numbers be a and b According to question,a+b=2(34)=68Andab=1616=256Solving we get,The numbers are 64 and 4

Q6. The terms of an A.P. are doubled, then the resulting sequence is

• An A.P. with common difference half the common difference of the original A.P
• An A.P. with common difference double the common difference of the original A.P
• An A.P. with common difference equal to the common difference of the  original A.P
• An A.P. with common difference thrice  the common difference of the  original A.P

Solution

Let the terms of the A.P. bea,(a + d), (a + 2d) After doubling, the terms become 2a,(2a + 2d), (2a + 4d) So, the common difference of doubled A.P. = double the common difference of the original A.P.

Q7. Sum to 20 terms of the series 1.3² + 2.5² + 3.7² +... is:

• 168090
• 178090
• 188090
• 198090

Solution

T_n = [nth term of 1, 2, 3, ...] .[ nth term of 3, 5, 7, ....]²

Q8. The third term of a G.P. is 3, the product of first five terms of this progression is:

• 27
• 81
• 243
• The information is not sufficient.

Solution

Let a₁ be the first term and r be the common ratio of a G.P.

Q9. The first term of a G.P. is 10 and the sum to infinity is 5, the common ratio is

• -1
• 1
• 2
• -2

Solution

Q10. The first negative term of the A.P. 62,57,52.... is the

• 10 th term
• 18 th term
• 14 th term
• 12 th term

Solution

Q11. The sum of the series 2 + 6 + 18 + ....+ 4374 is:

• 8748
• 6876
• 6560
• 8798

Solution

Here, a = 2, r = 3 and l = 4374 Sum =

Q12. The sum of the first hundred even natural numbers divisible by 5 is

• 55000
• 50500
• 50000
• 55005

Solution

Q13. The arithmetic mean between 6 and - 12 is:

• 6
• -6
• 9
• -3

Solution

The arithmetic mean between 6 and -12 =

Q14. If n numbers are inserted between 15 and 60 such that the ratio of the first to the last is 1 : 3, then the value of n is:

• 3
• 4
• 11
• 15

Solution

Number of terms = n + 2,

Q15. The ratio of the sum of n terms of two A.P's is (7n + 1) : (4n + 27). The ratio of their 12^th terms is:

• 155 : 115
• 162 : 119
• 169 : 123
• 176 : 127

Solution

Let a₁, a₂, d₁, d₂ be the first terms and common differences of two A.P's respectively.

Q16. Write the given sequence in the form of a function: -3,-1,1,3,5,7

• 3 + 2 (n - 1)
• -3 + 2 (n + 1)
• -3 + 2 (n + 1)
• -3 + 2 (n - 1)

Solution

Q17. For a G.P. the ratio of the 7^th and the third terms is 16. The sum of 9 terms is 2555. What is the first term?

• 7
• 5
• 4
• 9

Solution

Q18. The sum of three numbers in A.P. is 9 and the sum of their squares is 59.What are the numbers?

• -1,3,5
• -1,4,6
• -3,5,7
• -1,3,7

Solution

Q19. Write the nth term of the sequence : 4,9,14,19,...

• 4n - 1
• 4n + 1
• 5n - 1
• 5n + 1

Solution

The first term is 5x1 -1 = 4 , : The second term is 5x2 -1 = 9 , : The third term is 5x3 - 1 and so on . So, : The nth term is = 5n - 1

Q20. The G.M. of 5 and 8 is

• 20√2
• 2√10 
• 20
• 40

Solution

The G.M. of two numbers a and b is equal to . Hence. G.M. of 5 and 8 =

Q21. 8 numbers between 6 and 33 such that the sequence is an AP are

• 9,12,15,..30
• 6,9,12, ...,33
• 9,11,13,..28
• 12,15,18,..,30

Solution

Q22. If a, b, c are in A.P., then

• b is the arithmetic mean of a and c.
• 2a = b + c
• a - b = c - b
• b - a = b  - c

Solution

If a, b, c are in A.P., thenCommon difference = Thus, b is the arithmetic mean of a and c.

Q23. How many terms of the G.P. 4 + 16 + 64 + ... will make the sum 5460?

• 6
• 7
• 8
• 9

Solution

4 + 16 + 64 + ...  Let the sum of n terms of n terms be 5460. 6^th term of a given GP will make the sum 5460.

Q24. What is the 10th A.M between 2 and 57 if 10 A.M s are inserted between these numbers?

• 54
• 55
• 52
• 53

Solution

Q25. Three terms in A.P. are such that their sum is 45.What is the middle term?

• 15
• 10
• 25
• 30

Solution

Take the three terms as a - d,a and a+d where a is the first term and d is the common difference. Then  (a - d) + a + (a + d) = 45 So, a = 15

Q26. Four A.Ms are inserted between 12 and 42. The common difference between the terms of the resulting A.P. is

• 7
• 6
• 10
• 8

Solution

Q27. The sum of the squares of the first 15 natural numbers is

• 3600
• 1240
• 1400
• 3500

Solution

We know that the sum of squares of first n natural numbers is /6 So, in this case, substituting n=15 we get the answer.

Q28. The sequence whose terms follow the certain pattern is called a

• real sequence
• finite sequence
• series
• progression

Solution

The sequence whose terms follow the certain pattern is called a progression.

Q29. The G.M. between the numbers: 56 and 14 is:

• 35
• 28
• 70
• 14

Solution

G.M. =

Q30. The number of terms of the series 51,49,47,...37 is

• 10
• 8
• 5
• 7

Solution

Q31. How many terms of the series 24,20,16,...are required so that their sum is 72?

• 12
• 8 or 6
• 9 or 4
• 11

Solution

Q32. A man saved Rs 21700 in 14 years. In each year after the first he saved Rs 100 more than he did in the preceding year. How much did he save in the first year?

• Rs 600
• Rs 900
• Rs 1300
• Rs 2100

Solution

Let the amount saved be Rs 'a' in the first year.Thus, he saved Rs 900 in the first year.

Q33. If a, b, c are in A.P., then (b+c)²-a², (a+c)² -b², (a+b)² -c² are in....

• A.P.
• G.P.
• H.P.
• A.G.P.

Solution

Q34. The digits of a positive integer having three digits are in AP and sum of their digits is 21. The number obtained by reversing the digits is 396 less than the original number. Find the original number.

• 876
• 678
• 579
• 975

Solution

Q35. The sum of 9 terms of G.P. 3, 6, 12, ... is:

• 1533
• 3069
• 1536
• 511

Solution

Here, a = 3, r = 2

Q36. The three numbers between 1 and 256 such that the sequence is in GP are

• 8, 16, 64
• 2, 8, 32
• 8, 16, 32
• 4, 16, 64

Solution

When we insert 3 numbers x,y,z between the given numbers,  i.e  1,x,y,z,,256 becomes a G.P. with common ratio r say 256 is the 5^th term of the GP whose first term is 1 and common ratio r  256 = (r)^5-1  (4)⁴ = r⁴ So, r = 4 Hence, the required numbers = ar, ar², ar³ = 1(4), 1(16), 1(64) = 4, 16, 64

Q37. A sequence is a function whose domain is the set of

• Real numbers
• Natural Numbers
• Whole numbers
• Integers

Solution

A sequence is a function whose domain is the set of natural numbers.

Q38. The ratio of the A.M. between two numbers whose sum is 100 to the G.M. is 5:4The numbers are

• 60,40
• 80,20
• 70,30
• 90,10

Solution

Q39. The numbers whose arithmetic mean is 34 and geometric mean is 16 are:

• 14 and 54
• 4 and 64
• 8 and 60
• 12 and 56

Solution

Let numbers be a and b. Then the numbers are 64 & 4 or 4 & 64.

Q40. Sequence containing finite number of terms is called a ____ sequence, otherwise _____.

• infinite, finite
• finite, infinite
• finite, finite
• infinite, infinite