Arithmetic Progressions Test

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Arithmetic Progressions Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

If 100 times the $$100^{\mathrm{t}\mathrm{h}}$$ term of an AP with non zero common difference equals the 50 times its $$50^{\mathrm{t}\mathrm{h}}$$ term, then the $$150^{\mathrm{t}\mathrm{h}}$$ term of this AP is:

A
$$-150$$

B
$$150$$ times its $$50^{th}$$ term

C
$$150$$

D
$$0$$

Solution

Let say $$a$$ be the first term and $$d$$ be the common difference of the AP.

Given: $$100(a+99d)=50(a+49d)$$ [using $$n^{th}$$ term $$a_{n}= a+(n-1)d $$]

$$\Rightarrow 2a+198d=a+49d$$

$$\Rightarrow a+149d=0$$

Also, $$T_{150}= a+ 149d=0$$

Hence, the option 'D' is correct.
Question 2
1 / -0

Sum of first $$15$$ terms of $$2 + 5 + 8 + ...$$ is?

A
$$44$$

B
$$42$$

C
$$345$$

D
$$386$$

Solution

Difference between terms $$=5-2=8-5=3$$
We can observe that difference between two consecutive terms is equal, hence the terms are in A.P.
Sum of fist n terms in an A.P. is given by the formula: $$S_n=\dfrac{n}{2}\times[2a+(n-1)d] $$
Given $$n=15,a= 2, d=5-2=3 ,$$
$$S_n=\dfrac{15}{2}\times[2\times2+(15-1)3] $$
$$S_n=\dfrac{15}{2}\times 46$$
$$S_n=345 $$
Question 3
1 / -0

The list of numbers $$10, 6, 2, -2,\cdots$$ is

A
an $$A.P.$$ with $$d = 16$$

B
an $$A.P.$$ with $$d = -4$$

C
an $$A.P.$$ with $$d = 4$$

D
not an $$A.P.$$

Solution

Given series: $$10, 6, 2, -2...$$

Let, $$a_1 = 10$$,
$$a_2 = 6$$,
$$a_3 = 2$$,
$$a_4 = -2$$

$$d_1 = a_2 - a _1 = 6-10 = -4$$
$$d_2 = a_3 - a _2 = 2-6 = -4$$
$$d_3 = a_4 - a _3 = -2-2 = -4$$

The differences are equal, hence common difference of this A.P. is $$d = -4$$
Question 4
1 / -0

What is the sum of all natural numbers from 1 to 100?

A
5050

B
50

C
4550

D
5150

Solution

We know,
Sum of n natural numbers $$=\dfrac{n(n+1)}{2}$$
$$\therefore$$ Sum from 1 to 100 $$=\dfrac{100.101}{2}=5050$$
$$(Ans \to A)$$
Question 5
1 / -0

The common difference of the AP $$\dfrac{1}{P}, \dfrac{1-P}{P}, \dfrac{1-2 P}{P}$$, .......... is:

A
P

B
-1

C
-P

D
1

Solution

Let $$a_{1}=\dfrac{1}{P}$$ , $$a_{2}=\dfrac{1-P}{P}$$ , $$a_{3}=\dfrac{1-2P}{P}$$

$$\therefore$$ Common difference = $$a_{2}-a_{1} or \hspace{2 mm} a_{3}-a_{2}; $$

= $$\dfrac{(1-P)-(1)}{P} ; \hspace{2 mm} \dfrac{(1-2P)-(1-P)}{P}$$
= $$-1$$
Question 6
1 / -0

If the first term of an A.P. is $$-1$$ and common difference is $$-3$$, then its $$12^{th}$$ term is

A
$$34$$

B
$$32$$

C
$$-32$$

D
$$-34$$

Solution

Given that,
The first term of an A.P. $$(a)=-1$$,
The common difference $$(d)=-3$$
To find out: $$12^{th}$$ term of the A.P.

We know that, $$ n^{th}$$ term of any AP is given by:
$$T_n=a+(n-1)d $$
Hence,
$$T_{12}=(-1)+(12-1)(-3)$$
$$\Rightarrow T_{12}=(-1)+(11)(-3)=-1-33$$
$$\therefore \ T_{12}= -34$$

Hence, the $$12^{th}$$ term of the given A.P. is $$-34$$.
Question 7
1 / -0

Calculate $$10^{th}$$ term of an AP: $$4, 6, 8, ........ $$

A
$$18$$

B
$$20$$

C
$$22$$

D
$$26$$

Solution

Let $$S=4+6+8+...\infty$$
The above series is an arithmetic progression with common difference of $$2$$.
We know that, $$a_{n}=a_{1}+(n-1)d$$
$$\therefore a_{10}=4+(10-1)(2)$$
$$=4+9(2)$$
$$=18+4$$
$$=22$$
Question 8
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,
$$a = 2, d = 2.5$$

A
$$32$$

B
$$33$$

C
$$34$$

D
$$35$$

Solution

Here the first term $$a=2$$ and common difference $$d=2.5$$.
Now $$a_n=a+(n-1)d$$
$$\therefore a_1=a=2$$
$$a_2=a+d=2+2.5=4.5$$
$$a_3=a+2d=2+2\times2.5=7$$
$$a_4=a+3d=2+3\times2.5=9.5$$
$$a_5=a+4d=2+4\times2.5=12$$
$$\therefore $$ series in A.P. is $$2, 4.5, 7, 9.5, 12,....$$
Then, the sum of first five terms is $$=2+4.5+7+9.5+12=35$$
Or
We can use the formula for sum of A.P $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
$$\therefore S_5$$ =$$\dfrac{5}{2} \times [2(2)+(5-1)(2.5)]=35$$
Question 9
1 / -0

A property valued at Rs. 30000 will depreciate Rs. 1380 the first year, Rs. 1340 the second year, Rs. 1300 the third year, and so on. What will the depreciation during the eight year?

A
Rs. 1150

B
Rs. 1200

C
Rs. 1100

D
Rs. 1250

Solution

Every year depreciation decrease by 40, so the given problem can be solved through the AP: 1380, 1340, 1300, ........
So $$a = 1380, d = 40, n = 8$$
$$\therefore$$ Depreciation during eight year
$$=1380 + (8-1) \times (-40) = Rs. 1100$$
Question 10
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:
$$a = 4, d = 0$$

A
$$20$$

B
$$22$$

C
$$24$$

D
$$26$$

Solution

Here the first term $$a=4$$ and common difference $$d=0$$.
Now $$a_n=a+(n-1)d$$
$$\therefore a_1=a=4$$
$$a_2=a+d=4+0=4$$
$$a_3=a+2d=4+2\times0=4$$
$$a_4=a+3d=4+3\times0=4$$
$$a_5=a+4d=4+4\times0=4$$
$$\therefore$$ series in A.P. is $$4,4,4,4,4,....$$
Thus, the sum of first five terms is $$4+4+4+4+4=20$$
Hence, the answer is $$20$$.
Or
We can use the formula for sum of A.P $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
$$\therefore S_5$$$$=$$$$\dfrac{5}{2} \times [2(4)+(5-1)(0)]$$
$$=20$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 16

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Arithmetic Progressions Test - 16

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Question 1

$$-150$$

$$150$$ times its $$50^{th}$$ term

$$150$$

$$0$$

Solution

Question 2

$$44$$

$$42$$

$$345$$

$$386$$

Solution

Question 3

an $$A.P.$$ with $$d = 16$$

an $$A.P.$$ with $$d = -4$$

an $$A.P.$$ with $$d = 4$$

not an $$A.P.$$

Solution

Question 4

5050

50

4550

5150

Solution

Question 5

P

-1

-P

1

Solution

Question 6

$$34$$

$$32$$

$$-32$$

$$-34$$

Solution

Question 7

$$18$$

$$20$$

$$22$$

$$26$$

Solution

Question 8

$$32$$

$$33$$

$$34$$

$$35$$

Solution

Question 9

Rs. 1150

Rs. 1200

Rs. 1100

Rs. 1250

Solution

Question 10

$$20$$

$$22$$

$$24$$

$$26$$

Solution

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