Sequences and Series Test

Result

Sequences and Series Test - 4

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Weekly Quiz Competition

Question 1
2 / -0.83

The three arithmetic mean between –2 and 10 are:

A
-1, 3 and 7

B
1, 4 and 7

C
-1, 3 and 8

D
2, 5 and 8

Solution
Question 2
2 / -0.83

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

A
b - a = b - c

B
b is the arithmetic mean of a and c.

C
2a = b + c

D
a - b = c - b

Solution

If a, b, c are in A.P., then b is the arithmetic mean of a and c.
a b and c are in AP.
∴b - a = c - b
⇒2b = a + c ........ (1)
The arithmetic mean of a and c
= (a+c)/2
Using equation (1), we get
= 2b/2
= b
∴The arithmetic mean of a and c = b
Thus, the arithmetic mean of a and c is b.
Question 3
2 / -0.83

If we want to insert 8 numbers between the numbers 4 and 31 such that the resulting sequence is an AP.The difference between the consecutive numbers will be

A
4

B
3

C
2

D
5

Solution

a = 4, l = 31, n = 8
(l-a)/(n+1) = (31-4)/(8+1)
= 27/9
= 3
Question 4
2 / -0.83

For what value of n, is the arithmetic mean (AM) of a and b?

A
1

B
1/2

C
-1/2

D
zero

Solution

(a ⁿ⁺¹+ b ⁿ⁺¹)/(a ⁿ+ b ⁿ) is mean between a &b
Mean of a &b = ( a + b)/2
=>(a ⁿ⁺¹+ b ⁿ⁺¹)/(a ⁿ+ b ⁿ) = ( a + b)/2
=>2a ⁿ⁺¹+ 2b ⁿ⁺¹ = a ⁿ⁺¹ + b ⁿ⁺¹ + ba ⁿ+ b ⁿa
=>a ⁿ⁺¹+ b ⁿ⁺¹= ba ⁿ+ b ⁿa
=>a ⁿ(a - b) = b ⁿ(a - b)
=>a ⁿ= b ⁿ
=>(a/b)ⁿ= 1
=>n = 0 or a = b
Question 5
2 / -0.83

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.

A
876

B
579

C
975

D
678

Solution

Let the digits at ones, tens and hundreds place be (a −d)a and (a+d) respectively. The, the number is
(a+d)×100+a ×10+(a −d) = 111a+99d
The number obtained by reversing the digits is
(a −d)×+a ×10+(a+d) = 111a −99d
It is given that the sum of the digits is 21.
(a −d)+a+(a+d) = 21 ...(i)
Also it is given that the number obtained by reversing the digits is 594 less than the original number.
∴111a −99d = 111a+99d −396 ...(ii)
⟹3a = 21 and 198d = 396
⟹a = 7 and d = -2
Original number = (a −d)×+a ×10+(a+d)
= 100(9) + 10(7) + 5
= 975
Question 6
2 / -0.83

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?

A
Rs 2100

B
Rs 1300

C
Rs 900

D
Rs 600

Solution

d = 100
S_n = 21700
n = 14
S_n = n/2[2a + (n-1)d]
21700 = 14/2[2a + (14-1)100]
= 21700 = 7[2a + 1300]
= 3100 = 2a + 1300
⇒3100 - 1300 = 2a
⇒1800 = 2a
⇒a = 900
Question 7
2 / -0.83

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

A
54

B
53

C
52

D
55

Solution

2 and 57 have 10 terms between them so including them there would be 12 terms
a_n = 57, a = 2, n = 12
a_n = a + (n-1)d
=>57 = 2 + (12 - 1)d
=>55 = 11d
d = 5
T10 = a + 10d
=>2 + 10(5)
= 52
Question 8
2 / -0.83

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:

A
3

B
4

C
11

D
15

Solution
Question 9
2 / -0.83

The arithmetic mean between 6 and –12 is:

A
6

B
-3

C
-6

D
9

Solution

6, x, -12
x = (6-12)/2
x = -6/2
x = -3
Question 10
2 / -0.83

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 n^th term from beginning is:

A
b

B
A_n-1

C
A_n

D
A₁

Solution

A₁ ,A₂ ,......, An are inserted between a and b then the series will become
a, A₁ ,A₂ ,A₃ ,......, An,b. Now a becomes the first term, A₁ will be second, A₂ will become third term
An will become A(n+1)th term
therefore A(n-1) will become nth term.

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Sequences and Series Test - 4

Result

Sequences and Series Test - 4

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Question 1

-1, 3 and 7

1, 4 and 7

-1, 3 and 8

2, 5 and 8

Solution

Question 2

b - a = b - c

b is the arithmetic mean of a and c.

2a = b + c

a - b = c - b

Solution

Question 3

4

3

2

5

Solution

Question 4

1

1/2

-1/2

zero

Solution

Question 5

876

579

975

678

Solution

Question 6

Rs 2100

Rs 1300

Rs 900

Rs 600

Solution

Question 7

54

53

52

55

Solution

Question 8

3

4

11

15

Solution

Question 9

6

-3

-6

9

Solution

Question 10

b

An-1

An

A1

Solution

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A_n-1

A_n

A₁