Arithmetic Progressions Test

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Arithmetic Progressions Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

The sum of 24 terms of the following series : $$\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} +$$ ..................

A
$$300$$ $$\sqrt{2}$$

B
$$150$$ $$\sqrt{2}$$

C
$$250$$ $$\sqrt{2}$$

D
$$325$$ $$\sqrt{2}$$

Solution

We have $$\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} +$$ ..................

We can rewrite the series as $$1\sqrt{2}+2\sqrt{2}+3\sqrt{2} + 4 \sqrt{2}+........$$

Taking $$\sqrt{2}$$ common, we get:

$$\sqrt{2} \ [1+2+3+4+....... \text{upto} \ 24 \ \text{terms}] $$

The $$24^{th}$$ term will be $$24$$.

Hence, $$\sqrt{2} \ [1+2+3+4+....... +24] $$

We know that, the sum of $$n$$ terms of an AP is given by

$$S_{n}=\frac {n}{2}(a+l)$$

Here, $$a = 1, l = 24, n= 24 $$

$$\therefore \ S_{24}=\sqrt{2}\left[\frac{24}{2}(1+24)\right]$$

$$\Rightarrow \sqrt{2} \times \displaystyle \frac{24 \times 25}{2} = 300 \sqrt{2}$$

Hence, the sum of $$24$$ terms of the given series is $$300\sqrt{2}$$.
Question 2
1 / -0

In a flower bed, there are $$23$$ rose plants in the first row, $$21 $$ in the second, $$19$$ in the third, and so on. There are $$5$$ rose plants in the last row. The number of rows in the flower bed are......

A
$$11$$

B
$$41$$

C
$$10$$

D
$$9$$

Solution
Question 3
1 / -0

For an A.P., the seventh term is 19 and the thirteenth term is 37. Find its twenty second term.

A
60

B
54

C
64

D
78

Solution

For the given A.P., $$T_7=19$$ and $$ T_{13}=37$$
Now $$T_n = a+(n-1)d$$
$$\therefore T_7 = a + (7-1)d$$
$$\therefore 19=a+6d$$
$$\therefore a+6d = 19$$ ............ (1)
Again, $$T_{13} = a + (13-1)d$$
$$\therefore 37 = a + 12d$$
$$\therefore a+12 d =37$$ ............. (2)
Subtracting eq. (1) from eq. (2)
$$a + 12 d = 37$$
$$a + 6 d = 19$$
$$6d = 18$$
$$\therefore d=13$$
Substituting $$d=13$$ in $$a+6d =19$$,
$$a+6(3)=19$$
$$\therefore a+18 = 19$$
$$\therefore a =1$$
Thus, the first term of the A.P. is 1 and the common difference is 3.
Now, $$T_{22} =a + (22-1) d=1 + (22-1) (3)$$
$$=1 +(21) (3) =1+63=64$$
Thus, the twenty second term of the given A.P. is 64.
Question 4
1 / -0

The nth term of the A.P. $$3, 7, 11, 15, ....$$ is given by -

A
$$T_n = 4n-1$$

B
$$T_n=3n+4$$

C
$$T_n=3n-4$$

D
$$T_n = 2n+1$$

Solution

Given A.P. is $$3,7,11,15,....$$
Here $$a=3, d=4$$
We know $$T_n = a+(n-1) d $$
$$= 3 + (n-1) 4$$
$$ = 4n-1$$
Question 5
1 / -0

Strikers at a plant were ordered to return to work and were told they would be fined Rs. $$50$$ the first day they failed to do so, Rs. $$75$$ the second day, Rs. $$100$$ the third day, and so on. If the strikers stayed out for $$6$$ days, what was the fine for the sixth day?

A
$$200$$

B
$$150$$

C
$$125$$

D
$$175$$

Solution

Strikers on plant fined Rs.$$50$$ on first day,& $$75$$ on second ,Rs$$100$$ on third and so on,
difference in fine from day$$1$$ to day$$2=Rs.25$$
For $${ 6 }^{ th }$$ day,$$n=6,a=50$$ and $$d=25$$,
$$=a+(n-1)d\\ =50+(6-1)25\\ =50+(5\times 25)\\ =50+125\\ =175$$
Answer is$$(D)$$
Question 6
1 / -0

Write the sum of first five terms of the following Arithmetic Progressions where, the common difference $$d$$ and the first term $$a$$ are given:
$$a = 6, d = 6$$

A
$$88$$

B
$$89$$

C
$$90$$

D
$$91$$

Solution

Here the first term $$a=6$$ and common difference $$d=6$$.
Now $$a_n=a+(n-1)d$$
$$\therefore a_1=a=6$$
$$a_2=a+d=6+6=12$$
$$a_3=a+2d=6+2\times6=18$$
$$a_4=a+3d=6+3\times6=24$$
$$a_5=a+4d=6+4\times6=30$$
$$\therefore$$ series in A.P is $$6,12,18,24,30,....$$
Thus, the sum of first five terms $$=6+12+18+24+30=90$$
Hence, the answer is $$90$$.
Or
We can use the formula for sum of A.P $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
$$\therefore S_5$$ =$$\dfrac{5}{2} \times [2(6)+(5-1)(6)]=90$$
Question 7
1 / -0

Find the twenty fifth term of the A.P. : 12, 16, 20, 24, ..........

A
108

B
112

C
116

D
120

Solution

Let $$a$$ be the first term and $$d$$ be the common difference of given AP.
Hence $$a=12$$ and $$d=16-12=4$$.
$$t_n=a+(n-1)d$$
$$\Rightarrow t_{25}=12+24\times4$$
$$\Rightarrow t_{25}=12+96$$
$$\Rightarrow t_{25}=108$$
Question 8
1 / -0

Find the sum of first 20 natural numbers.

A
210

B
220

C
230

D
240

Solution

We have to find the sum n terms of the AP 1,2,3,...
Here
$$a=1,d=1$$.
Therefore, Required sum $$S_n=\frac n2[2a+(n-1)d]=\frac n2[2\times1+(n-1)\times1]$$
$$\Rightarrow S_n=\displaystyle \frac n2[2+n-1]$$

$$\Rightarrow S_n=\displaystyle \frac{n(n+1)}2$$ ...(1)

Put $$n=20$$ in (1), we get
$$S_{20}=\displaystyle \frac{20(20+1)}2=210$$
Question 9
1 / -0

Find $$t_{11}$$ from the given A.P. 4, 9, 14, ...

A
54

B
55

C
51

D
21

Solution

Given series is $$4,9,14,....$$
Clearly, the given sequence is an AP with first term $$a=4$$ and the common difference $$d=5$$.
We know, $$a_n=a+(n-1)d$$
Now, $$t_{11}=a+10d$$
$$\therefore t_{11}=4+10\times5=54$$
Question 10
1 / -0

Write the sum of the first five terms of the Arithmetic Progression for which the common difference $$d$$ and the first term $$a$$ are given below:
$$a = 5, d = 2$$

A
$$40$$

B
$$45$$

C
$$50$$

D
$$55$$

Solution

Given: the first term $$a=5$$ and common difference $$d=2$$.
Now $$a_n=a+(n-1)d$$
$$\therefore a_1=a=5$$
$$a_2=a+d=5+2=7$$
$$a_3=a+2d=5+2\times2=9$$
$$a_4=a+3d=5+3\times2=11$$
$$a_5=a+4d=5+4\times2=13$$
$$\therefore $$ The series in A.P. is $$5,7,9,11,13,.....$$
Sum of the first five terms is $$=5+7+9+11+13=45$$
Hence, the answer is $$45$$.
Or
We can also use the formula for sum of A.P => $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
$$\therefore S_5$$ =$$\dfrac{5}{2} \times [2(5)+(5-1)(2)]=45$$

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Arithmetic Progressions Test - 17

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Arithmetic Progressions Test - 17

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SHARING IS CARING

Question 1

$$300$$ $$\sqrt{2}$$

$$150$$ $$\sqrt{2}$$

$$250$$ $$\sqrt{2}$$

$$325$$ $$\sqrt{2}$$

Solution

Question 2

$$11$$

$$41$$

$$10$$

$$9$$

Solution

Question 3

60

54

64

78

Solution

Question 4

$$T_n = 4n-1$$

$$T_n=3n+4$$

$$T_n=3n-4$$

$$T_n = 2n+1$$

Solution

Question 5

$$200$$

$$150$$

$$125$$

$$175$$

Solution

Question 6

$$88$$

$$89$$

$$90$$

$$91$$

Solution

Question 7

108

112

116

120

Solution

Question 8

210

220

230

240

Solution

Question 9

54

55

51

21

Solution

Question 10

$$40$$

$$45$$

$$50$$

$$55$$

Solution

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