Arithmetic Progressions Test

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Arithmetic Progressions Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Write first four terms of the AP, when the first term $$a$$ and the common difference $$d$$ are given as follows:
$$a = -2, d =0$$

A
$$-2, -2, -2, -2$$

B
$$-2, -4, -6, -2$$

C
$$-2, -4, 4, -2$$

D
$$-2, 4, 4, -2$$

Solution

Here $$a=-2$$ and $$d=0$$.

Now, $$n^{th}$$ term $$=a_n=a+(n-1)d$$

$$\therefore t_1=a=-2$$

$$t_2=a+d=-2+0=-2$$

$$t_3=a+2d=-2+2\times0=-2$$

$$t_4=a+3d=-2+3\times0=-2$$

Thus, the first four terms of given AP are $$-2,-2,-2,-2$$.
Question 2
1 / -0

For the following AP, write the first term and the common difference
$$0.6, 1.7, 2.8, 3.9$$

A
First term: $$0.6$$ and Common difference: $$1.1$$

B
First term: $$1.7$$ and Common difference: $$2.8$$

C
First term: $$1.7$$ and Common difference: $$-1.1$$

D
First term: $$0.6$$ and Common difference: $$-2.8$$

Solution

Given series is $$0.6, 1.7, 2.8, 3.9,....$$
First term $$a=0.6$$
And common difference $$d=a_2-a_1$$
$$=1.7-0.6$$
$$=1.1$$
Question 3
1 / -0

Is it an AP?

$$1, 4, 7, 10, 13, 16, 19, 22, 25, ...$$

A
Yes

B
No

C
Ambiguous

D
Data insufficient

Solution

Given series is $$1,4,7,10,13,16,19,22,25,....$$
Since, $$4-1 = 7-4=10-7,.....$$
i.e $$3=3=3,......$$
Given the common difference is $$3$$.
Therefore, the given sequence is an AP.
Hence, option A is correct.
Question 4
1 / -0

Check if the sequence is an AP $$1, 3, 9, 27,....$$

A
Yes, it is an AP

B
No

C
Data Insufficient

D
Ambiguous

Solution

Given series is $$1,3,9,27,....$$
We have, $$3-1=2$$
$$9-3=6$$
$$27-9=18$$
This shows that the difference of a term and the preceding term is now always same. Hence, the given sequence is not an AP.
Question 5
1 / -0

$$11^{th}$$ term of the AP: $$-3,$$ $$-\dfrac{1}{2}, 2,...$$ is

A
$$28$$

B
$$22$$

C
$$-38$$

D
$$-48$$

Solution

Clearly, the given AP has first term $$a=-3$$ and common difference $$d=\displaystyle \frac{-1}2-(-3)=2-\dfrac{(-1)}{2}\Rightarrow\frac52$$.
Also, $$n=11$$
$$\therefore a_{11}=a+10d$$
$$=-3+10\times\displaystyle \frac52$$
$$=-3+25$$
$$=22$$
Hence, Option B is correct.
Question 6
1 / -0

Find the sums given below:
7 + 10$$\displaystyle\frac{1}{2}$$ + 14 +......+84

A
1066$$\displaystyle\frac{1}{2}$$

B
1096$$\displaystyle\frac{1}{2}$$

C
1046$$\displaystyle\frac{1}{2}$$

D
1006$$\displaystyle\frac{1}{2}$$

Solution

In given series of AP first term a=7, common difference d= $$10\frac 12-7=\frac {21}2-7=\frac 72$$ and last term l=84
Since, $$l=a+(n-1)d$$
$$\Rightarrow 84=7+(n-1)\frac 72 $$
$$\Rightarrow 84-7=(n-1)\frac 72 $$
$$\Rightarrow 77=(n-1)\frac 72 $$
$$\Rightarrow 77\times \frac 27=n-1 $$
$$\Rightarrow 22=n-1 $$
$$\Rightarrow n=23 $$
Since, $$ S_n=\frac n2(a+l) $$
$$\Rightarrow S_n=\frac {23}2(7+84) $$
$$\Rightarrow S_n=91 \times \frac {23}2 $$
$$\Rightarrow S_n=\frac {2093}2 $$
$$\Rightarrow S_n=1046\frac 12 $$
Question 7
1 / -0

Find the number of terms in each of the following AP's
$$7, 13, 19,....205$$

A
$$20$$ terms

B
$$28$$ terms

C
$$34$$ terms

D
$$40$$ terms

Solution

Clearly, the given Sequence is an AP with first term $$a=7$$ and common difference $$d=13-7=6$$.
Let $$205$$ be the $$nth$$ term of the given AP.
$$\Rightarrow a_n=205$$
$$\Rightarrow a+(n-1)d=205$$
$$\Rightarrow 7+(n-1)\times6=205$$
$$\Rightarrow (n-1)\times6=198$$
$$\Rightarrow n=34$$
Hence, there are $$34$$ terms in the given AP.
Question 8
1 / -0

Find the sum of $$2, 7, 12, ...$$to $$10$$ terms

A
$$160$$

B
$$245$$

C
$$290$$

D
$$300$$

Solution

Clearly, the given sequence is an AP with first term $$a=2$$, common difference $$d=5$$ and $$n=10$$.
Now,
$$S_{10}=\dfrac {10}2[2a+(10-1)d]$$
$$=5[2\times2+9\times5]$$
$$=245$$
Question 9
1 / -0

Find the sum of $$-37, -33, -29,..$$ to $$12$$ terms

A
$$-180$$

B
$$-210$$

C
$$-360$$

D
$$-390$$

Solution

The given series is $$-37, -33, -29, .........$$ to $$12$$ terms
The given sequence is an AP with first term $$a=-37$$, common difference $$d=4$$ and $$n=12$$.
$$\therefore S_{n}=\displaystyle \frac {n}2[2a+(n-1)d]$$
$$\therefore S_{12}=\displaystyle \frac {12}2[2a+(12-1)d]$$
$$=6[2\times-37+11\times4]$$
$$=-180$$
Question 10
1 / -0

For an A.P. $$a = 7, d = 3, n = 8$$, find $$a_8$$.

A
$$a_8 = 18$$

B
$$a_8 = 28$$

C
$$a_8 = 16$$

D
$$a_8 = 36$$

Solution

Here, $$a=7, d=3$$ and $$n=8$$.
By using $$a_n=a+(n-1)d$$
Therefore, $$a_8=a+7d$$
$$=7+7\times3$$
$$=28$$

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Arithmetic Progressions Test - 20

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Arithmetic Progressions Test - 20

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Question 1

$$-2, -2, -2, -2$$

$$-2, -4, -6, -2$$

$$-2, -4, 4, -2$$

$$-2, 4, 4, -2$$

Solution

Question 2

First term: $$0.6$$ and Common difference: $$1.1$$

First term: $$1.7$$ and Common difference: $$2.8$$

First term: $$1.7$$ and Common difference: $$-1.1$$

First term: $$0.6$$ and Common difference: $$-2.8$$

Solution

Question 3

Yes

No

Ambiguous

Data insufficient

Solution

Question 4

Yes, it is an AP

No

Data Insufficient

Ambiguous

Solution

Question 5

$$28$$

$$22$$

$$-38$$

$$-48$$

Solution

Question 6

1066$$\displaystyle\frac{1}{2}$$

1096$$\displaystyle\frac{1}{2}$$

1046$$\displaystyle\frac{1}{2}$$

1006$$\displaystyle\frac{1}{2}$$

Solution

Question 7

$$20$$ terms

$$28$$ terms

$$34$$ terms

$$40$$ terms

Solution

Question 8

$$160$$

$$245$$

$$290$$

$$300$$

Solution

Question 9

$$-180$$

$$-210$$

$$-360$$

$$-390$$

Solution

Question 10

$$a_8 = 18$$

$$a_8 = 28$$

$$a_8 = 16$$

$$a_8 = 36$$

Solution

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