Arithmetic Progressions Test

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Arithmetic Progressions Test - 52

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Question 1
1 / -0

Find which the term of the A.P. 20, 30, 40.... is 100.

A
$$11$$

B
$$12$$

C
$$13$$

D
$$9$$

Solution

Let the general form of A.P., $$a_n = a + (n-1)d$$
Common difference, $$d = 10$$
$$a_{n}=100$$
$$a_n = a + (n-1)d$$
$$100 = 20 + (n-1)\times 10$$
$$100-20 = 10n-10$$
$$80+10 = 10n$$
$$90 = 10n$$
$$n = 9$$
Question 2
1 / -0

Find the sum of the following AP : $$2 , 4 ,6 , 8 ,...... (15 $$ terms).

A
$$210$$

B
$$220$$

C
$$230$$

D
$$240$$

Solution

First term of the given arithmetic series = 2
Second term of the given arithmetic series = 4
Third term of the given arithmetic series = 6
Fourth term of the given arithmetic series = 8
Now, Second term - First term = 4 - 2 = 2
Third term - Second term = 6 - 4 = 2
Therefore, common difference of the given arithmetic series is 2.
The number of terms of the given A. P. series (n) = 15
We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is
$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$
$$S_{15}=\frac{15}{2}[2\times 2+(15-1)2]$$
$$S_{15}=7.5[4+28]$$
$$S_{15}=240$$
Question 3
1 / -0

Which term of the A.P. 18, 16, 14.... is -20?

A
$$10$$

B
$$20$$

C
$$30$$

D
$$40$$

Solution

Let the required term $$=n$$
The general form of A.P., $$a_n = a + (n-1)d$$
Common difference $$ d = -2$$
$$a_{n}=-20$$
$$a_n = a + (n-1)d$$
$$-20 = 18 + (n-1)\times -2$$
$$-20-18 = -2n+2$$
$$-38-2 = -2n$$
$$-40 = -2n$$
$$n = 20$$
Question 4
1 / -0

Find the sum of the first 20 terms if the first term is -12 and the common difference is -5 in an arithmetic series.

A
$$-1190$$

B
$$-1290$$

C
$$-1390$$

D
$$-1490$$

Solution

First term =$$ a = -12$$
$$n = 20$$
$$d = -5$$
$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$
$$S_{20}=\frac{20}{2}[2\times -12+(20-1)(-5)]$$
$$S_{20}=10[-24-95]$$
$$S_{20}=-1190$$
Question 5
1 / -0

In an arithmetic series, find the sum of the first 10 terms if the first term is 3 and the common difference is 4.

A
$$110$$

B
$$210$$

C
$$310$$

D
$$410$$

Solution

First term = a = 3
n = 10
d = 4
$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$
$$S_{10}=\frac{10}{2}[2\times 3+(10-1)4]$$
$$S_{10}=5[6+36]$$
$$S_{10}=210$$
Question 6
1 / -0

What is the sum of the first $$n$$ terms if the first term is $$2$$, the common difference is $$5$$ and the $$n^{th}$$ term is $$122$$ in an arithmetic series?

A
$$1250$$

B
$$1350$$

C
$$1450$$

D
$$1550$$

Solution

First term: $$a = 2$$
last term, $$l = 122$$
$$d = 5$$
$$a_{n}=a+(n-1)d$$
$$122=2+(n-1)5$$
$$122-2+5=5n$$
$$125=5n$$
$$n = 25$$
$$S_{n}=\dfrac{n}{2}[a+l]$$
$$S_{25}=\dfrac{25}{2}[2+122]$$
$$S_{25}=12.5[124]$$
$$S_{25}=1550$$
Question 7
1 / -0

In an arithmetic series, $$a_1 = -14$$ and $$a_{5}=50$$. Find the sum of the first 5 terms.

A
$$60$$

B
$$70$$

C
$$80$$

D
$$90$$

Solution

First term = a = -14
last term, l = 50
$$a_{n}=50$$
n = 5
$$S_{n}=\cfrac{n}{2}[a+l]$$
$$S_{5}=\cfrac{5}{2}[-14+50]$$
$$S_{5}=2.5[36]$$
$$S_{5}=90$$
Question 8
1 / -0

If the first term of an arithmetic sequence is $$-14$$ and the common difference is $$2$$, find the sum of the first $$8$$ terms.

A
$$-56$$

B
$$-46$$

C
$$-36$$

D
$$-26$$

Solution

Let $$S_{n}$$ be the sum of $$n$$ terms of AP.
First term: $$a = -14$$
$$S_{8}=?$$
$$d = 2$$
$$n = 8$$
We know that,
$$S_{n}=\dfrac{n}{2}[2a+(n-1)d]$$
$$S_{8}=\dfrac{8}{2}[2\times (-14)+(8-1)2]$$
$$S_{8}=4[-28+14]$$
$$S_{8}=-56$$
Question 9
1 / -0

$$a_1=1.5, a_{20}=58.5$$ in an arithmetic series, find $$S_{15}$$.

A
$$137.5$$

B
$$237.5$$

C
$$337.5$$

D
$$437.5$$

Solution

$$a_n = a+(n-1)d$$
$$\Rightarrow a_{20} = 1.5+(20-1)d$$
$$\Rightarrow 58.5-1.5 = 19d$$
$$\Rightarrow 57= 19d$$
$$\Rightarrow d = 3$$

$$S_n = \dfrac{n}{2}[2a+(n-1)d]$$
$$\Rightarrow S_{15} = \dfrac{15}{2}[2\times 1.5+(15-1)3]$$
$$\Rightarrow S_{15} = 7.5[3+42]$$
$$\Rightarrow S_{15} = 7.5[45]$$
$$\Rightarrow S_{15} = 337.5$$
Question 10
1 / -0

Given A.P is $$\dfrac{1}{3q},\dfrac{1-6q}{3q}, \dfrac{1-12q}{3q}$$. Find the common difference

A
$$1$$

B
$$-1$$

C
$$2$$

D
$$-2$$

Solution

Given A.P is $$\dfrac{1}{3q},\dfrac{1-6q}{3q}, \dfrac{1-12q}{3q}$$
Then, $$ \dfrac{1}{3q},\dfrac{1}{3q}-\dfrac{6q}{3q},\dfrac{1}{3q}-\dfrac{12q}{3q}$$ will also be in AP.

Which results in,
$$ \dfrac{1}{3q},\dfrac{1}{3q}-2,\dfrac{1}{3q}-4$$

It can be seen clearly that the series is decreasing by the quantity (2)
Hence, the common difference is $$-2$$

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Arithmetic Progressions Test - 52

Result

Arithmetic Progressions Test - 52

Score

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Rank

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SHARING IS CARING

Question 1

$$11$$

$$12$$

$$13$$

$$9$$

Solution

Question 2

$$210$$

$$220$$

$$230$$

$$240$$

Solution

Question 3

$$10$$

$$20$$

$$30$$

$$40$$

Solution

Question 4

$$-1190$$

$$-1290$$

$$-1390$$

$$-1490$$

Solution

Question 5

$$110$$

$$210$$

$$310$$

$$410$$

Solution

Question 6

$$1250$$

$$1350$$

$$1450$$

$$1550$$

Solution

Question 7

$$60$$

$$70$$

$$80$$

$$90$$

Solution

Question 8

$$-56$$

$$-46$$

$$-36$$

$$-26$$

Solution

Question 9

$$137.5$$

$$237.5$$

$$337.5$$

$$437.5$$

Solution

Question 10

$$1$$

$$-1$$

$$2$$

$$-2$$

Solution

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