Arithmetic Progressions Test

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Arithmetic Progressions Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

The sum of $$2, 7, 12, ..............$$ to $$10$$ terms is :

A
$$240$$

B
$$248$$

C
$$245$$

D
$$250$$

Solution

Given series is: $$\displaystyle 2+7+12+........ $$ to $$10$$ terms
Here, $$\displaystyle a=2,d=7-2=5,n=10$$
$$\displaystyle \Rightarrow { S }_{ n }=\frac { n }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} $$
$$\displaystyle \Rightarrow { S }_{ 10}=\frac { 10 }{ 2 } \left\{ 4+45 \right\} $$
$$=5\times 49$$
$$=245$$
Question 2
1 / -0

Which term in the $$\text{A.P.}~~ 5,13,21,...$$ is $$ 181 $$?

A
$${21}^{\text{st}}$$

B
$${22}^{\text{nd}}$$

C
$${23}^{\text{rd}}$$

D
$${24}^{\text{th}}$$

Solution

Given sequence is $$5, 13, 21, ...$$
Given: $$a = 5, ~d = 8, ~n^{th}\text{ term} = 181$$
To find: $$n$$
$$\because~~ { t }_{ n }~ = a +(n- 1)d\\ \Longrightarrow 181 =5 + (n - 1)8\\ \Longrightarrow 181 = 5+ 8n - 8\\ \Longrightarrow 181= 8n - 3\\ \Longrightarrow 8n= 184\\\implies ~n =23$$.

$$\therefore$$ $$181$$ is the $$23^{\text{rd}}$$ term.
Question 3
1 / -0

If the sum of $$n$$ terms of AP. is $$476, l = 20, a = 36$$, then $$n$$ is equal to :

A
$$14$$

B
$$15$$

C
$$16$$

D
$$17$$

Solution

Given, $$\displaystyle { S }_{ n }=476,l=20,a=36,n=?$$
We know $$\displaystyle { S }_{ n }=\frac { n }{ 2 } \left[ a+l \right] $$
$$\Rightarrow \displaystyle 476=\frac { n }{ 2 } \left[ 36+20 \right]$$
$$ =\dfrac { 56n }{ 2 } $$
$$=28n$$
$$\Rightarrow \therefore n=\dfrac { 476 }{ 28 } =17$$
Question 4
1 / -0

If $$\displaystyle a=3,n=20$$ and $$\displaystyle { S }_{ n }=300$$, then $$l$$ is :

A
$$30$$

B
$$25$$

C
$$27$$

D
$$14$$

Solution

Given, $$a=3, n=20, S_n=300$$
We need to find $$l$$
$$\displaystyle { S }_{ n }=\frac { n }{ 2 } \left( a+l \right) $$
$$\displaystyle \therefore 300=\frac { 20 }{ 2 } \left( 3+l \right) $$
$$\displaystyle \therefore l=27$$
Question 5
1 / -0

The sum of $$n$$ natural number is :

A
$$\displaystyle \frac { n\left( n+1 \right) }{ 2 } $$

B
$$\displaystyle \frac { n(n-1) }{ 2 } $$

C
$$\displaystyle \left( n+2 \right) $$

D
$$\displaystyle \frac { n+1 }{ 2 } $$

Solution

As the first $$n$$ natural numbers are $$1,2,3,4,5,6,.........n$$
They form an AP with $$a=1$$ and $$d=1$$
The sum of AP for $$n$$ numbers $$=\dfrac{n}{2}(2a+(n-1)d)$$
$$=\dfrac{n}{2}(2+n-1)$$
$$=\dfrac{n(n+1)}{2}$$
Question 6
1 / -0

The $$3^{rd}$$ term of an A.P. is $$-40$$ and $$13^{th}$$ term is zero, then $$d$$ is equal to :

A
$$-4$$

B
$$4$$

C
$$0$$

D
$$-1$$

Solution

Given $$3^{rd}$$ term $$=-40$$, $$13^{th}$$ term $$=0$$
Now using, $$a_n=a+(n-1)d$$
$$\therefore \displaystyle { a }_{ 3 }=a+2d=-40$$ ......(i)
and $$\displaystyle { a }_{ 13 }=a+12d=0$$ ......(ii)
Subtracting equations (i) and (ii), we get
$$\displaystyle d=4$$
Question 7
1 / -0

If numbers $$a, b$$ and $$c$$ are in AP, then

A
$$\displaystyle b-a=c-b$$

B
$$\displaystyle b+a=c+b$$

C
$$\displaystyle a-c=b-d$$

D
None of these

Solution

Since, given three numbers are in AP, then the common difference will be same.
$$\therefore$$ Common difference = $$ b - a = c - b$$
Question 8
1 / -0

The $$4^{th}$$ term of an AP is $$14$$ and its $$12^{th}$$ term is $$70$$. What is its first term?

A
$$-10$$

B
$$-7$$

C
$$7$$

D
$$10$$

Solution

As we know for an AP with first term $$a$$ and common difference $$d$$, genral term is $$a_n = a+(n-1)d$$
$$\therefore \displaystyle a+3d=14$$ and
$$\displaystyle a+ 11d=70$$
On subtracting both the above equations, we get
$$\displaystyle 8d=56$$
or $$\displaystyle d=7$$
$$\displaystyle \therefore \quad a=-7$$
Question 9
1 / -0

If $$p, (p - 2)$$ and $$3 p$$ are in AP, then the value of $$p$$ is

A
$$-3$$

B
$$-2$$

C
$$3$$

D
$$2$$

Solution

Since, given three numbers are in AP, then the common difference between two consecutive terms will be same.
$$\therefore$$ $$\displaystyle (p-2)-p=3p-(p-2)$$
$$\Rightarrow -2=2p+2$$
or $$\displaystyle p=-2$$
Question 10
1 / -0

The common difference of the $$A.P:\;\;5, 3, 1, -1,\dots$$ is

A
$$-2$$

B
$$2$$

C
$$-1$$

D
$$3$$

Solution

Common difference of an $$A.P$$ is equal to difference between the consecutive terms of the sequence,

$$d= 3 - 5$$
$$= -2$$

Hence, option $$A$$ is the correct answer.

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 24

Result

Arithmetic Progressions Test - 24

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SHARING IS CARING

Question 1

$$240$$

$$248$$

$$245$$

$$250$$

Solution

Question 2

$${21}^{\text{st}}$$

$${22}^{\text{nd}}$$

$${23}^{\text{rd}}$$

$${24}^{\text{th}}$$

Solution

Question 3

$$14$$

$$15$$

$$16$$

$$17$$

Solution

Question 4

$$30$$

$$25$$

$$27$$

$$14$$

Solution

Question 5

$$\displaystyle \frac { n\left( n+1 \right) }{ 2 } $$

$$\displaystyle \frac { n(n-1) }{ 2 } $$

$$\displaystyle \left( n+2 \right) $$

$$\displaystyle \frac { n+1 }{ 2 } $$

Solution

Question 6

$$-4$$

$$4$$

$$0$$

$$-1$$

Solution

Question 7

$$\displaystyle b-a=c-b$$

$$\displaystyle b+a=c+b$$

$$\displaystyle a-c=b-d$$

None of these

Solution

Question 8

$$-10$$

$$-7$$

$$7$$

$$10$$

Solution

Question 9

$$-3$$

$$-2$$

$$3$$

$$2$$

Solution

Question 10

$$-2$$

$$2$$

$$-1$$

$$3$$

Solution

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