Arithmetic Progressions Test

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Arithmetic Progressions Test - 49

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Question 1
1 / -0

Calculate the sum of even numbers between $$12$$ and $$90$$ which are divisible by $$8$$.

A
$$500$$

B
$$510$$

C
$$520$$

D
$$620$$

Solution

The list of odd numbers between $$12$$ and $$90$$ which are divisible by $$8$$ are $$16, 24, ..... 88$$
The sum of first $$n$$ terms of arithmetic series formula can be written as,
$$S_n = \dfrac{n}{2} [2a + (n - 1)d]$$ ............ (1)
Where $$n =$$ number of terms = ?
$$a =$$ first odd number $$=16$$
$$d =$$ common difference of A.P. $$=8$$
$$T_n =$$ Last term $$=88$$
$$T_n = a + (n - 1)d$$
$$88 = 16 + (n - 1) 8$$
$$88 = 16 + 8n - 8$$
$$88 - 16 + 8 = 8n$$
$$ 80 = 8n$$
$$n = 10$$
Apply the given data in the formula (1),
$$S_{10} = \dfrac{10}{2} [2 \times 16 + (10 - 1)8]$$
$$= 5 [32 + (9)8]$$
$$= 5 [32 + 72]$$
$$= 5 [104]$$
$$= 520$$
So, the sum of even numbers between $$12$$ and $$90$$ which are divisible by $$8$$ is $$520$$.
Question 2
1 / -0

The first term and last term number is $$12$$ and $$30$$. The total number of term is $$50$$. Find its sum.

A
$$1,100$$

B
$$1,250$$

C
$$1,150$$

D
$$1,050$$

Solution

Given, $$n=50$$, first term $$=12$$ and last term $$=30$$
Sum of first $$n$$ numbers of an A.P. is
$$S_n = \dfrac{n}{2} [2a + (n - 1)d]$$ or $$\dfrac{n}{2}$$ [First term $$+$$ Last term]
$$= \dfrac{50}{2} [12 + 30]$$
$$= 25 [42]$$
$$S_{50} = 1,050$$
Question 3
1 / -0

Calculate the sum of whole numbers between 0 and 79 which are divisible by 11.

A
$$300$$

B
$$310$$

C
$$308$$

D
$$320$$

Solution

The numbers between $$0$$ and $$79$$ which are divisible by $$11$$ are $$11, 22, 33, 44, ...... 77$$.
The sum of first $$n$$ terms of arithmetic series formula can be written as,
$$S_n = \dfrac{n}{2} [2a + (n - 1)d]$$ ............ (1)
Where $$n =$$ number of terms = ?
$$a =$$ first odd number $$=11$$
$$d =$$ common difference of A.P. $$=11$$
$$T_n =$$ Last term $$=77$$
$$T_n = a + (n - 1)d$$
$$77 = 11 + (n - 1) 11$$
$$77 = 11 + 11n - 11$$
$$77 - 11 + 11 = 11n$$
$$ 77 = 11n$$
$$\Rightarrow n = 7$$
Apply the given data in the formula $$(1)$$, we get
$$S_{7} = \dfrac{7}{2} [2 \times 11 + (7 - 1)11]$$
$$= 3.5 [22 + (6)11]$$
$$= 3.5 [22 + 66]$$
$$= 3.5 [88]$$
$$= 308$$
So, the sum of whole numbers between $$0$$ and $$79$$ which are divisible by $$11$$ is $$308$$.
Question 4
1 / -0

Calculate the sum of even numbers between 100 and 150 which are divisible by 13.

A
$$494$$

B
$$410$$

C
$$420$$

D
$$430$$

Solution

The even numbers between $$100$$ and $$150$$ which are divisible by $$13$$ are $$104, 117, ............ 143$$
The sum of first $$n$$ terms of arithmetic series formula can be written as,
$$S_n = \dfrac{n}{2} [2a + (n - 1)d]$$ ............ (1)
Where $$n =$$ number of terms = ?
$$a =$$ first odd number $$=104$$
$$d =$$ common difference of A.P. $$=13$$
$$T_n =$$ Last term $$=143$$
$$T_n = a + (n - 1)d$$
$$143 = 104 + (n - 1) 13$$
$$143 = 104 + 13n - 13$$
$$143 - 104 + 13 = 13n$$
$$ 52 = 13n$$
$$n = 4$$
Apply the given data in the formula (1),
$$S_{4} = \dfrac{4}{2} [2 \times 104 + (4 - 1)13]$$
$$= 2 [208 + (3)13]$$
$$= 2 [208 + 39]$$
$$= 2 [247]$$
$$= 494$$
So, the sum of even numbers between $$100$$ and $$150$$ which are divisible by $$13$$ is $$494$$.
Question 5
1 / -0

Find the sum of even numbers between 1 and 25.

A
$$155$$

B
$$156$$

C
$$157$$

D
$$158$$

Solution

The sum of first n terms of arithmetic series formula can be written as,
$$S_n = \dfrac{n}{2} [2a + (n - 1)d]$$ ............ (1)
The list of even numbers between 1 and 25 is 2, 4, 6, 7, 8, .......... 24
Where n = number of terms = ?
a = first even number
$$a = 2$$
d = common difference of A.P.
$$ d=2$$
$$T_n =$$ Last term
$$T_n = 24 $$
$$T_n = a + (n - 1)d$$
$$24 = 2 + (n - 1) 2$$
$$24 = 2 + 2n - 2$$
$$24 - 2 + 2 = 2n$$
$$ 24 = 2n$$
$$n = 12$$
Apply the given data in the formula (1),
$$S_{12} = \dfrac{12}{2} [2 \times 2 + (12 - 1)2]$$
$$= 6 [4 + (11)2]$$
$$= 6 [4 + 2]$$
$$= 6 [26]$$
$$= 156$$
So, the sum of even numbers between 1 and 25 is 156.
Question 6
1 / -0

If $$a_3 = 9$$ and $$a_7 = 21$$. Find $$S_5$$

A
40

B
45

C
50

D
55

Solution

Using the formula,
$$a_n = a + (n - 1) d$$
So, $$a_3 = a + (3 - 1) d$$
$$a = a + 2d$$ ......... (1)
$$a_7 = a + (7 - 1) d$$
$$21 = a + 6d $$ ........ (2)
Subtracting equation (1) & (2)
$$a = a + 2d$$ ........ (1)
$$21 = a = bd$$ ....... (2)
(-) (-) (-)
------------------
$$- 20 = -4d$$
$$d = 5$$
Put the value of d in equation (1)
$$a = a + 2(5)$$
$$a = - 1$$
The sum of first n term is given by $$S_n = \dfrac{n}{2} [2a + (n - 1)d]$$
$$S_5 = \dfrac{5}{2} [2(-1) + (5 - 1) 5]$$
$$= \dfrac{5}{2} (-2 + 20)$$
$$= 2.5 (18)$$
$$S_5 = 45$$
Question 7
1 / -0

Check whether $$80$$ is a term of the AP. $$0, 0.5, 1, 1.5,\dots$$

A
True

B
False

C
Data Insufficient

D
Can't Say

Solution

$$0, 0.5,1,1.5,\dots$$

Here,
$$a = 0, d = 0.5$$
Let $$80$$ be the $$n^{th}$$ term of this AP.

Since,
$$a_n = a + (n - 1) d$$
$$\Rightarrow 80 = 0 + (n - 1) 0.5$$
$$\Rightarrow 80 = 0.5 n - 0.5$$
$$\Rightarrow 80 + 0.5 = 0.5 n$$
$$\Rightarrow n = 161$$

Clearly, $$n$$ is an integer.
Therefore, $$80$$ is the $$161^{th}$$ term of this AP.
Question 8
1 / -0

Find the $$12^{th}$$ term from the end of the A.P. $$2, 6, 10 ......... 58$$

A
$$22$$

B
$$14$$

C
$$16$$

D
$$28$$

Solution

The given A.P. can be written in reverse order as $$58 .......... 10, 6, 2$$
Here $$a = 58, d = -4, n = 12$$
$$a_{12} = a + (12- 1) d$$
$$= 58 + 11 \times (-4)$$
$$ = 58 - 44$$
$$a_{12} = 14$$
$$\therefore$$ $$14$$ is the $$12^{th}$$ term from the end.
Question 9
1 / -0

Find the common difference for the arithmetic sequence whose formula is $$a_n = 12n - 7$$

A
$$10$$

B
$$11$$

C
$$12$$

D
$$13$$

Solution

Assume $$n = 1, 2, 3, $$ .......
Put those values in $$a_n = 12n - 7$$
$$a_1 = 12 (1) - 7 = 5$$
$$a_2 = 12 (2) - 7 = 17$$
$$a_3 = 12 (3) - 7 = 29$$
$$a_4 = 12 (4) - 7 = 41$$
Therefore, the common difference, $$d$$ is $$12$$
Question 10
1 / -0

Find the common difference, if $$a_2 = 10$$ and $$a_5 = 20$$.

A
$$3.33..........$$

B
$$3.35..........$$

C
$$3.65..........$$

D
$$3.75..........$$

Solution

The nth term of A.P. is $$a_n = a_1 + (n - 1) d$$
$$\therefore a_2 = a + (n -1) d$$
$$10 = a + (2- 1) d$$
$$10 = a + d $$ ..... (1)
$$a_5 = a + (n - 1) d$$
$$20 = a + 4d$$ ..... (2)
Equating equation (1) & (2)
$$ 10 = a + d$$
$$20 = a + 4d$$
(-) (-) (-)
-----------------
$$-10 = - 3d$$
$$d = \dfrac{+10}{+3}$$
$$d = 3.33 ........$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 49

Result

Arithmetic Progressions Test - 49

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SHARING IS CARING

Question 1

$$500$$

$$510$$

$$520$$

$$620$$

Solution

Question 2

$$1,100$$

$$1,250$$

$$1,150$$

$$1,050$$

Solution

Question 3

$$300$$

$$310$$

$$308$$

$$320$$

Solution

Question 4

$$494$$

$$410$$

$$420$$

$$430$$

Solution

Question 5

$$155$$

$$156$$

$$157$$

$$158$$

Solution

Question 6

40

45

50

55

Solution

Question 7

True

False

Data Insufficient

Can't Say

Solution

Question 8

$$22$$

$$14$$

$$16$$

$$28$$

Solution

Question 9

$$10$$

$$11$$

$$12$$

$$13$$

Solution

Question 10

$$3.33..........$$

$$3.35..........$$

$$3.65..........$$

$$3.75..........$$

Solution

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