Arithmetic Progressions Test

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Arithmetic Progressions Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Find the sum of first $$32$$ multiples of $$4$$.

A
$$2,112$$

B
$$2,712$$

C
$$2,110$$

D
$$2,111$$

Solution

The multiples of $$4$$ are, $$4, 8, 12, 16,$$ ............
Therefore, $$a= 4, d = 4, n=32$$
We have $$S_{32} = \dfrac{n}{2} [2a + (n - 1)d]$$
$$= \dfrac{32}{2} [2 \times 4 + (32 - 1)4]$$
$$= 16 [8 + 31 \times 4]$$
$$= 16 [8 + 124]$$
Therefore, $$ S_{20} = 2,112$$
Question 2
1 / -0

Constant is subtracted from each term of an A.P. the resulting sequence is also an ______

A
Geometric Sequence

B
Arithmetic Progression

C
Both A and B

D
None of the above

Solution

If a constant is subtracted from each term of an A.P. the resulting sequence is also an arithmetic progression.

For example : Let the sequence be $$\{ a_1, a_2, a_3 ...... a_4 \}$$ be an arithmetic progression with common difference $$d$$.

Again subtract $$k$$ to the above sequence.

Then the resulting sequence is $$a_1 -k, a_2 - k, a_3 - k ...... $$ be an A.P.
Question 3
1 / -0

Find the difference of an A.P. $$a_n$$, if $$a_4 = 16$$ and $$a_2 = 8$$

A
$$3$$

B
$$4$$

C
$$2$$

D
$$1$$

Solution

Given, $$a_4=16, a_2=8$$
$$a_4 = a + 3d=16$$ ...... (1)
$$a_2 = a + d=8$$ ...... (2)
Multiplying equation (2) by $$3$$ and then subtract the equation, we get
$$16 = a + 3d$$
$$24 = 3a + 3d$$
(-) (-) (-)
----------------
$$-8 = - 2a$$
$$a = 4$$
Put $$a= 4$$ in equation (1), we get
$$16 = 4 + 3d$$
$$\Rightarrow 16 - 4 = 3d$$
$$\Rightarrow d = 4$$
Question 4
1 / -0

The last term of the AP $$21, 18, 15$$ ....... is $$-351$$. Find the nth term.

A
$$213$$

B
$$123$$

C
$$312$$

D
$$-231$$
Question 5
1 / -0

A train can travel $$200$$ m in the first hour, $$400$$ m the next hour, $$600$$ m the third hour and so on in an arithmetic sequence. What is the total distance the train travels in $$5$$ hours?

A
$$2,000$$

B
$$3,000$$

C
$$4,000$$

D
$$5,000$$

Solution

The sequence is $$200, 400, 600 .....$$
$$a = 200, d = 200$$
First find the common difference:
$$a_n = a_1 + (n - 1) d$$
$$a_5 = 200 + (5- 1) 200$$
$$= 200 + 4 \times 200$$
$$a_5 = 1,000$$
We know that, $$S_n = \dfrac{n}{2} $$ [First term $$+$$ Last term]
$$= \dfrac{5}{2} [200 + 1,000]$$
$$= 2.5 [1,200]$$
$$S_5 = 3,000$$
Hence, the train will travel $$3,000$$ m in $$5$$ hours.
Question 6
1 / -0

Find the sum of first $$20$$ multiples of $$13$$.

A
$$2,720$$

B
$$2,730$$

C
$$2,740$$

D
$$2,750$$

Solution

The multiples of $$13$$ are, $$13, 26, 39, .....$$
Therefore, $$a = 13, d = 13,n=20$$
We have $$S_{20} = \dfrac{n}{2} [2a + (n - 1)d]$$
$$= \dfrac{20}{2} [2 \times 13 + (20 - 1)13]$$
$$= 10 [26 + 19 \times 13]$$
$$ = 10 [26 + 247]$$
Therefore, $$S_{20} = 2,730$$
Question 7
1 / -0

Find the sum of the first 20 terms of the arithmetic series if $$a_{1} = 10$$ and $$a_{20}=100$$.

A
1000

B
1100

C
1200

D
1300

Solution

To find the sum of the first n terms of an arithmetic series use the formula,
$$S_{n}=\dfrac{n(a_{1}+a_{n})}{2}$$
$$S_{20}=\dfrac{20(10+100)}{2}$$
$$S_{20}=\dfrac{2200}{2}$$
$$S_{20}=1100$$
Question 8
1 / -0

What is the sum of $$t_n= (2n-5)$$ from n =10 to 150?

A
$$22650$$

B
$$21855$$

C
$$23250$$

D
$$21250$$

Solution

$$T_n = (2n-5)$$
$$T_{10} = 2\times10-5 = 15$$
$$T_{150} = 2\times15- 5 = 295$$

Sum of AP, if first terms and last term is given:
$$S = \dfrac n2(\text{first term } +\text{ last term})$$

Here, $$n = 150-10+1 = 141$$
$$S = \dfrac{141}2(15+295)$$

$$S = 141\times155 = 21855$$
Question 9
1 / -0

Calculate the $$15^{\text{th}}$$ term of the A.P. $$ -3, -4, -5, -6, -7....$$

A
$$-13$$

B
$$-15$$

C
$$-17$$

D
$$-19$$

Solution

Let the general form of A.P., $$a_n = a + (n-1)d$$
Common difference, $$d = -1$$
$$a_{15}=?$$
$$a_n = a + (n-1)d$$
$$a_{15} = -3 + (15-1)\times -1$$
$$a_{15} = -3 -14$$
$$a_{15} = -17$$
Question 10
1 / -0

Find the $$8^{th}$$ term of the A.P. : $$11, 14, 17, 20.....$$

A
$$11$$

B
$$-17$$

C
$$17$$

D
$$32$$

Solution

Given series is $$11,14,17,20,....$$
Let the general form of A.P., $$a_n = a + (n-1)d$$
Common difference, $$d = 3$$ and first term $$a=11$$
$$a_{8}=?$$
$$a_n = a + (n-1)d$$
$$a_{8} = 11 + (8-1)\times 3$$
$$a_{8} = 11 +21$$
$$a_{8} = 32$$

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Arithmetic Progressions Test - 27

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Arithmetic Progressions Test - 27

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Question 1

$$2,112$$

$$2,712$$

$$2,110$$

$$2,111$$

Solution

Question 2

Geometric Sequence

Arithmetic Progression

Both A and B

None of the above

Solution

Question 3

$$3$$

$$4$$

$$2$$

$$1$$

Solution

Question 4

$$213$$

$$123$$

$$312$$

$$-231$$

Question 5

$$2,000$$

$$3,000$$

$$4,000$$

$$5,000$$

Solution

Question 6

$$2,720$$

$$2,730$$

$$2,740$$

$$2,750$$

Solution

Question 7

1000

1100

1200

1300

Solution

Question 8

$$22650$$

$$21855$$

$$23250$$

$$21250$$

Solution

Question 9

$$-13$$

$$-15$$

$$-17$$

$$-19$$

Solution

Question 10

$$11$$

$$-17$$

$$17$$

$$32$$

Solution

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