Arithmetic Progressions Test

Result

Arithmetic Progressions Test - 1

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Question 1
1 / -0

If a + 1, 3a, 4a + 2 are in AP, then the value of a is

A
3

B
2

C
1

D
0

Solution

a + 1, 3a, 4a + 2 are in AP.
If x, y, z are in AP, then 2y = x + z.
Therefore, 2(3a) = a + 1 + 4a + 2
6a = 5a + 3
6a - 5a = 3
a = 3
Question 2
1 / -0

If k + 2, 4k – 6 and 3k – 2 are in AP, then the value of k is

A
0

B
1

C
2

D
3

Solution

k + 2, 4k – 6, 3k – 2 are in AP.
2(4k - 6) = (k + 2) + (3k - 2)
8k – 12 = k + 2 + 3k – 2
8k – 12 = 4k
4k = 12
k = 3
Question 3
1 / -0

In an AP, the common difference of two terms is 3 and the first term is 7. Find the 14^th term.

A
43

B
44

C
45

D
46

Solution

The 14^th term of the respective AP will be calculated as follows:
T₁₄ = a₁ + (14 - 1)d
14^th term = 7 + (13)3 = 46
Question 4
1 / -0

What is the n^th term of the given sequence?

- 2, 0, 2,........

A
- 2n

B
4n - 6

C
2n - 4

D
3n - 8

Solution

Here first term = a and difference = d = 2
n^th term = a + (n - 1)d
= -2 + (n - 1)2
= 2n - 4
Question 5
1 / -0

Find the 19^th term of the given squence:

10, 14, 19, 25 ...

A
235

B
236

C
248

D
259

Solution

T₁ = 10
T₂ = 14 = 10 + 4 = T₁ + 4
T₃ = 19 = 14 + 5 = T₂ + 5 = T₁ + 4 + 5
T₄ = 25 = 19 + 6 = T₃ + 6 = T₂ + 5 + 6 = T₁ + 4 + 5 + 6
Continuing in the same fashion,
T₁₉= T₁ + (4 + 5 + 6 +...18^th terms) = T₁ + S
S = Sum of an AP with 1^stterm = 4
and common difference = 1
S =
= 9(8 + 17) = 225
So, T₁₉ = 10 + 225 = 235
Question 6
1 / -0

What is the 17^th term of the sequence 3, 6, 11, 18, 27 ...?

A
222

B
224

C
291

D
298

Solution

n^th term of the given series = n² + 2
17^th term = (17)² + 2
= 289 + 2 = 291
Question 7
1 / -0

What is the 9^th term of the sequence -3, -12, -27, -48, ...?

A
-143

B
-153

C
-243

D
None of these

Solution

n^th term = -3n²
So, 9^th term = -3 × 81 = -243
Question 8
1 / -0

Samuel received Rs. 10 on the first day of his job. For every day after that, he received Rs. 5 more than the previous day. How much money had he received in total after the 15^th day?

A
Rs. 100

B
Rs. 675

C
Rs. 575

D
Rs. 475

Solution

10, 15, 20, _____
Amount after the 15^th day =
= = Rs. 675
Question 9
1 / -0

A type of bacteria doubles every 2 hours. If there are 500 bacteria at the beginning, how many bacteria will there be after 24 hours?

A
2048000

B
2149000

C
2047050

D
2046000

Solution

As bacteria gets doubled every 2 hours, so sequence every 2 hours will be as follows:
500, 1000, 2000, 4000, 8000, 16000, 32000, 64000, 128000, 256000, 512000, 1024000, 2048000
So, count of bacteria after 24 hours will be 2048000.
Question 10
1 / -0

Ruby has 36 stamps in her collection. She adds 7 stamps to her collection every month. How many stamps will she have at the end of 7^th month?

A
85

B
95

C
105

D
115

Solution

Initially, Rub has 36 stamps.
At the end of 7^th month, she will have 36 + 7 × 7 = 85 stamps
Question 11
1 / -0

Cherry has 70 marbles with her. She buys 5 marbles every week. How many marbles will she have at the end of the seventh week?

A
100

B
105

C
110

D
115

Solution

7^th term = dn + (a – d)
= 5 × 7 + (75 – 5)
= 35 + 70 = 105
Question 12
1 / -0

Mr. Jones has £2500 in his account. He withdraws £200 every month. How much money would be there in his account at the end of the tenth month?

A
£500

B
£600

C
£700

D
£800

Solution

The sequence is 2500, 2300, 2100....
Here, a = 2500, d = - 200 and n = 11
11^th term = - 200 × 10 + (2500)
= - 2000 + 2500
= £500
Question 13
1 / -0

Tom's teacher gave him a sequence, which was the times table of 2. After that, he thought to change the sequence and placed -2 and 0 in front of the previous one. The new sequence becomes -2, 0, 2, .... What will be the 8^th term in the new sequence?

A
10

B
11

C
12

D
16

Solution

-2, 0, 2 ...
8^th term = a + (n - 1)d
= -2 + (8 - 1)(2)
= -2 + 14 = 12
Question 14
1 / -0

Tom had 4 chocolates. He decided to buy 3 more chocolates each day than that of the previous day. How many chocolates will he buy on the 11^th day?

A
16

B
29

C
30

D
34

Solution

4, 7, 10………..
11^th term = 3(11) + (4 - 3) = 33 + 1 = 34
Hence, Tom bought 34 chocolates on the 11^th day.
Question 15
1 / -0

Priyanka has 30 chocolates with her. She buys 5 chocolates every week. How many chocolates will she have at the end of the 5^th week if she didn't eat any or give any to anybody else?

A
45

B
55

C
65

D
75

Solution

6^th term = a + (n - 1)d
= 30 + (6 - 1)5
= 30 + 25 = 55

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 1

Result

Arithmetic Progressions Test - 1

Score

-

Rank

-

SHARING IS CARING

Question 1

3

2

1

0

Solution

Question 2

0

1

2

3

Solution

Question 3

43

44

45

46

Solution

Question 4

- 2n

4n - 6

2n - 4

3n - 8

Solution

Question 5

235

236

248

259

Solution

Question 6

222

224

291

298

Solution

Question 7

-143

-153

-243

None of these

Solution

Question 8

Rs. 100

Rs. 675

Rs. 575

Rs. 475

Solution

Question 9

2048000

2149000

2047050

2046000

Solution

Question 10

85

95

105

115

Solution

Question 11

100

105

110

115

Solution

Question 12

£500

£600

£700

£800

Solution