Arithmetic Progressions Test

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Arithmetic Progressions Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

The sum of the first $$100$$ natural numbers is

A
$$-5050$$

B
$$5005$$

C
$$5050$$

D
$$-5005$$

Solution

We know, sum of a arithmetic progression, with first term as $$a$$ and common difference $$d$$ is
$${S}_{n}=n[a+\frac{1}{2}(n-1)d]$$

Here, given, $$a=1$$, $$d=1$$ and $$n=100$$
$${S}_{100}=100[1+\displaystyle\dfrac{1}{2}(100-1)1]$$

$$ =100[1+\dfrac{99}{2}]$$

$$=100\times \displaystyle\frac{101}{2}$$

$$=5050$$
Question 2
1 / -0

Find the sum of all odd natural numbers from 1 to 150.

A
5625

B
5635

C
5645

D
5655

Solution

Clearly, the odd natural numbers from $$1$$ to $$150$$ are $$1,3,5,...,149$$.

This is an AP with first term $$a=1$$, common difference $$d=2$$ and last term $$a_n=l = 149$$.

Let there be $$n$$ terms in this AP.

Then,

$$a_n=149$$

$$\Rightarrow a_n = a+(n-1)d$$

$$149= 1+(n-1)\times2$$

$$149=1+ 2n-2$$

$$2n=150$$

$$\therefore n=75$$

$$\therefore$$ required sum $$=S_n=\dfrac n2[a+l]=\displaystyle \frac{75}2[1+149]=5625$$
Question 3
1 / -0

If the following sequence is an arithmetic progression, find its general term.
-5, 2, 9, 16, 23, 30, .......... is

A
$$7n-12$$

B
$$6n-12$$

C
$$5n-12$$

D
$$4n-12$$

Solution

We can observe the difference between two consecutives terms is equal:
$$2-(-5)=9-2= 16-9=23-16=7$$
Hence, the sequence is an arithmetic progression with first term $$a=-5$$ and common difference $$d= 7$$.
General term of an AP is $$t_n=a+(n-1)d$$
$$t_n=-5+(n-1)\times 7$$
$$t_n=7n-12$$
Question 4
1 / -0

Find $$S_{10}$$, if a = 6 and d = 3.

A
180

B
185

C
190

D
195

Solution

Given First term, $$a=6$$,
Common difference, $$d=3$$
Number of terms, $$n=10$$
Sum of $$n$$ terms is given as
$$S_n$$= $$\displaystyle \frac n2[2a+(n-1)d]$$
Substituting the values, we get
$$S_{10} =\displaystyle \frac {10}2[2\times 6+(10-1)\times3]$$
$$=5(39)$$
$$=195$$
Question 5
1 / -0

How many terms of the sequence $$18, 16, 14,....$$ should be taken so that their sum is zero?

A
$$19$$

B
$$17$$

C
$$18$$

D
$$16$$

Solution

Here $$a=18,d=-2$$.
Let there are $$n$$ terms so that the sum is zero.
Now, $$S_n=\dfrac n2[2a+(n-1)d]=0$$
$$\Rightarrow \dfrac n2[2\times18+(n-1)\times-2]=0$$
$$\Rightarrow 36+(n-1)\times-2=0$$
$$\Rightarrow n-1=18$$
$$\Rightarrow n=19$$
Therefore $$19$$ terms of the sequence has to be taken so that their sum is zero.
Question 6
1 / -0

How many terms are there in the A.P.: $$-1,\displaystyle\frac{-5}{6},\displaystyle-\frac{2}{3},\displaystyle\frac{-1}{2}, ....., \displaystyle\frac{10}{3}$$ ?

A
$$27$$

B
$$18$$

C
$$16$$

D
$$9$$

Solution

Given A.P. is $$-1,\ \dfrac{-5}{6},\ \dfrac{-2}{3},\ \dfrac{-1}{2},\ .....,\ \dfrac{10}{3}$$
Here, first term $$(a)=-1$$ and the common difference $$(d)=\displaystyle \dfrac{-5}6-(-1)=\dfrac{1}6.$$
Let $$n$$ be the number of terms of the given A.P.
We know that the $$n_{th}$$ term of an A.P. is:
$$a_n=a+(n-1)d\\$$
Here, $$a_n=\dfrac{10}3$$

$$\therefore \ a+(n-1)d=\dfrac{10}3$$

$$\Rightarrow -1+(n-1)\times\dfrac{1}6=\dfrac{10}3$$

$$\Rightarrow (n-1)\times\dfrac{1}6=\dfrac{13}3$$

$$\Rightarrow n-1=26$$

$$\Rightarrow n=27$$

Therefore, there are $$27$$ terms in the give A.P.
Question 7
1 / -0

Find the sum of $$7 + 10\dfrac{1}{2} + 14+...+84$$.

A
$$\displaystyle\frac{2093}{2}$$

B
$$\displaystyle\frac{1093}{2}$$

C
$$\displaystyle\frac{2193}{2}$$

D
$$\displaystyle\frac{3093}{2}$$

Solution

Clearly, terms of the given series form an AP with first term and last term $$a=7$$ & $$ l= a_n = 84$$ respectively and common difference $$d=\displaystyle \frac{21}2-7=\frac72$$.
Let the given series has $$n$$ terms.
Therefore,
$$a_n=84=a+(n-1)d$$
$$84= 7+(n-1)\times\frac72$$
$$ (n-1)\times\frac72=77$$
$$\therefore n=23$$
Here we use the following formula to calculate the sum of terms:
$$S_n =\dfrac {n}{2} (a+l) $$
$$\therefore$$Required sum $$=S_{23}=\displaystyle \dfrac{23}2(7+84)$$
$$\Rightarrow S_{23}=\displaystyle \frac{23\times91}2=\dfrac{2093}2$$
Question 8
1 / -0

The fourth term of an A.P. is $$4$$. Then the sum of the first $$7$$ terms is ?

A
$$4$$

B
$$28$$

C
$$16$$

D
$$40$$

Solution

Let a and d be the first term and common difference of A.P. Then,
$$a_4=4\Rightarrow a+3d=4$$ ...(1)
Now, $$S_n=\frac{n}{2}[2a+(n-1)d]$$
$$ \therefore S_7=\frac72[2a+(7-1)d]$$
$$\Rightarrow S_7=\frac72(2a+6d)$$
$$\Rightarrow S_7=7(a+3d)=7\times4=28$$ ( $$\because of (1)$$)
Therefore option B is correct
Question 9
1 / -0

Write the arithmetic progression when first term $$a = -1.5$$ and common difference $$d = -0.5$$.

A
$$-1.5, -2, -2.5, -3,.....$$

B
$$-1.5, -1, -0.5, 0,.....$$

C
$$-1.5, -2, -2.5, -5,.....$$

D
$$-1.5, 2, 2.5, 3,.....$$

Solution

Here $$a=-1.5$$ and $$d=-0.5$$.
The $$n^{th}$$ term of the AP is $$a_n=a+(n-1)d$$
$$\therefore a_1=a=-1.5$$
$$a_2=a+d\\=-1.5+(-0.5)\\=-2$$
$$a_3=a+2d\\=-1.5+2\times(-0.5)\\=-2.5$$
$$a_4=a+3d\\=-1.5+3\times(-0.5)\\=-3$$ and so on.
Therefore, the required AP is
$$-1.5,-2,-2.5,-3,...$$
Question 10
1 / -0

Find the $$18^{th}$$ term of the A.P., $$\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}$$.....

A
$$36\sqrt{2}$$

B
$$35\sqrt{2}$$

C
$$34\sqrt{2}$$

D
$$33\sqrt{2}$$

Solution

Here $$a=\sqrt2$$ and $$d=3\sqrt2-\sqrt2=2\sqrt2$$.
The nth term of the AP is $$a_n=a+(n-1)d$$
$$\therefore a_{18}=a+17d$$
$$=\sqrt2+17\times2\sqrt2$$
$$=35\sqrt2$$

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Arithmetic Progressions Test - 18

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Arithmetic Progressions Test - 18

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SHARING IS CARING

Question 1

$$-5050$$

$$5005$$

$$5050$$

$$-5005$$

Solution

Question 2

5625

5635

5645

5655

Solution

Question 3

$$7n-12$$

$$6n-12$$

$$5n-12$$

$$4n-12$$

Solution

Question 4

180

185

190

195

Solution

Question 5

$$19$$

$$17$$

$$18$$

$$16$$

Solution

Question 6

$$27$$

$$18$$

$$16$$

$$9$$

Solution

Question 7

$$\displaystyle\frac{2093}{2}$$

$$\displaystyle\frac{1093}{2}$$

$$\displaystyle\frac{2193}{2}$$

$$\displaystyle\frac{3093}{2}$$

Solution

Question 8

$$4$$

$$28$$

$$16$$

$$40$$

Solution

Question 9

$$-1.5, -2, -2.5, -3,.....$$

$$-1.5, -1, -0.5, 0,.....$$

$$-1.5, -2, -2.5, -5,.....$$

$$-1.5, 2, 2.5, 3,.....$$

Solution

Question 10

$$36\sqrt{2}$$

$$35\sqrt{2}$$

$$34\sqrt{2}$$

$$33\sqrt{2}$$

Solution

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