Arithmetic Progressions Test

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Arithmetic Progressions Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The $$9th$$ term of an AP is $$499$$ and $$499th$$ terms is $$9.$$ The term which is equal to zero is

A
$$501 th$$

B
$$502 th$$

C
$$508 th$$

D
None of these

Solution

$$a_{9}=a+8d$$
$$a_9=499 (given)$$
$$\therefore a+8d=499.........(i)$$
$$ a_{499}=a+498d$$
$$a_{499}=9$$
$$\therefore a+498d=9.........(ii)$$
Subtract (i) from (ii)
$$\Rightarrow 490 d = - 490 $$
$$\Rightarrow d = -1 $$
substitute the value of $$d$$
$$\Rightarrow a + 8(-1) = 499$$
$$\Rightarrow a = 507 $$
$$\therefore$$ For $$ a_{n}=0$$
$$ \Rightarrow a+(n-1)d=0$$
$$ \Rightarrow 507+(n-1)(-1)=0$$
$$ \Rightarrow 507=n-1$$
$$ \Rightarrow n=508$$
So, $$508^{th}$$ term is equal to zero.
Question 2
1 / -0

$$8^{th}$$ term of series $$\displaystyle 2\sqrt{2}+\sqrt{2}+0+......$$ will be

A
$$\displaystyle -5\sqrt{2}$$

B
$$\displaystyle 5\sqrt{2}$$

C
$$\displaystyle 10\sqrt{2}$$

D
$$\displaystyle -10\sqrt{2}$$

Solution

Given series is : $$2\sqrt2+ \sqrt2+0+.......$$
Then, $$ a=2\sqrt{2},d=(\sqrt 2-2\sqrt 2)=-\sqrt{2}$$
$$ a_{8}=a+7d$$
$$\therefore 2\sqrt{2}+7\times (-\sqrt{2})$$
$$\therefore a_{8}=-5\sqrt{2}$$
Question 3
1 / -0

Find the sum of all whole numbers divisible by $$5$$ but less than $$100$$.

A
$$950$$

B
$$925$$

C
$$880$$

D
$$1050$$

Solution

All whole numbers divisible by $$5$$ but less then $$100$$ are
$$0,5,10,15,20........95$$
It is an A.P. with first term, common difference and last term as $$0, 5$$ and $$95$$ respectively.
$$a=0,d=5,a_n=95$$
$$a_n=a+(n-1)d$$
$$95=0+(n-1)5$$
$$5n=100$$
$$n=20$$
Hence total numbers are $$=20$$
Sum of numbers are
$$S_{n}=\dfrac{n}{2}[a+(n-1)d]$$

$$S_{20}=\dfrac{20}{2}[0+(20-1)5]$$

$$S_{20}=10\times 95\\\ \ \ \ \ =950$$
The sum of all natural number divisible by $$5$$ but less than $$100$$ is $$950$$.
Question 4
1 / -0

The $$4th$$ term from the end of the AP
$$-11, -8, -5, ....................49$$ is

A
$$37$$

B
$$40$$

C
$$43$$

D
$$58$$

Solution

Given series of AP:
$$- 11, -8, -5,.......49$$
New series : $$49,........-5,-8,-11$$
In new series of AP first term $$a = 49,$$ common difference $$d = (-8) - (-5) = - 3$$
$$\displaystyle a_{4}=a+3d=49+3(-3)=49-9=40$$
$$\therefore $$ Option B is correct.
Question 5
1 / -0

If the common difference of an AP is $$-6,$$ then what is $$\displaystyle a_{16}-a_{12}?$$

A
$$-24$$

B
$$24$$

C
$$-30$$

D
$$30$$

Solution

Given, $$d=-6$$
$$a_{16}=a+15d$$
$$a_{12}=a+11d$$
$$a_{16}-a_{12}=(a+15d)-(a+11d)=4d$$
$$\therefore a_{16}-a_{12}=4\times -6=-24$$
Question 6
1 / -0

What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

A
$$530, 820$$

B
$$179, 587$$

C
$$560, 758$$

D
$$164, 850$$

Solution

$$\textbf{Step1-Find sequence of an A.P according to question}$$
$$\text{The lowest 3 digit number is }$$$$101$$$$\text{ and the highest number is}$$ $$998$$.
$$\text{Therefore, the AP is }$$$$101,104,107, ....... , 998$$
$$\text{ where, the first term is }$$$$a=101$$,
$$\text{common difference is $$d=104 - 101 = 3$$ and }$$
$$\text{n$^{th}$ term is }$$$$a_n=998$$

$$\textbf{Step2-Find the total number of terms of sequence}$$
$$\text{Let us find the number of terms is n}$$
$$\text{We know that the n$^{th}$}$$ $$\text{term of AP is given by}$$
$$a_{n} = $$$$a+(n-1)d$$
$$\text{Now, substituting the values, we get:}$$
$$998=101+(n-1)3\\ \Rightarrow (n-1)3=998-101\\ \Rightarrow (n-1)3=897\\ \Rightarrow n-1=\dfrac { 897 }{ 3 } \\ \Rightarrow n-1=299\\ \Rightarrow n=299+1\\ \Rightarrow n=300$$

$$\textbf{Step3-Find sum of 300 terms of an A.P}$$
$$\text{We know }$$
$$S=\dfrac {n}{2}$$$$\text{(1st term + last term) }$$
$$\text{Therefore, we have:}$$
$$S=\dfrac { 300 }{ 2 } (101+998)$$
$$S=150\times 1099$$
$$S=164850$$

$$\textbf{Hence, the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.}$$
Question 7
1 / -0

The sum of all odd integers between $$2$$ and $$50$$ divisible by $$3$$ is

A
$$192$$

B
$$216$$

C
$$168$$

D
$$240$$

Solution

The odd intergers between 2 and 50 divisible by 3 are $$3, 9, 15,......45$$
This sequence forms an AP with first term $$a=3$$ and common difference $$d=9-3=6$$
Since, $$l=a+(n-1)d $$
$$\Rightarrow 45=3+(n-1)6 $$
$$\Rightarrow 42=(n-1)6 $$
$$\Rightarrow 7=n-1 $$
$$\Rightarrow n=8 $$
Number of terms in given sequence are 8.
$$ S_n=\dfrac n2(a+l) $$
$$\Rightarrow S_n=\dfrac{8}{2}(3+45)$$
$$\Rightarrow S_n=4(48)$$
$$\Rightarrow S_n= 192$$
$$\therefore $$ Option A is correct.
Question 8
1 / -0

If the $$5$$th term of an A.P is eight times the first term and $$8$$th term exceeds twice the $$4$$th term by $$3$$, then the common difference is

A
$$7$$

B
$$5$$

C
$$6$$

D
$$3$$

Solution

$$n^{th}$$ term of an AP is given as,

$$t_n=a+(n-1)d$$

Where $$a$$ and $$d$$ are first term and common difference respectively.

According to the given condition,

Thus,

$${ t }_{ 5 } = a + 4d$$

$${ t }_{ 1 } = a$$

$${ t }_{ 8 } = a + 7d$$

$$ { t }_{ 4 } = a + 3d$$

Given, $${ t }_{ 5 } = 8{ t }_{ 1 }$$

$$\Longrightarrow a+ 4d= 8a\\ \Longrightarrow \quad 4d= 7a\\ \Longrightarrow d= \dfrac { 7a }{ 4 } $$

Also given, $${ t }_{ 8 }$$ $$-2{ t }_{ 4 } = 3$$

$$\Longrightarrow a+ 7d- 2(a+ 3d)= 3\\ \Longrightarrow d - a= 3\\ $$

Substituting $$d= \dfrac { 7a }{ 4 } $$ in the above equation,

$$\dfrac{7a}{4}-a=3$$

$$\dfrac{3a}{4}=3$$

we get $$a = 4$$ and hence, $$d = 7$$.
Question 9
1 / -0

If $$9^{th}$$ term of an A.P. is $$47$$ and $$16^{th}$$ term is $$82$$, then the general term is

A
$$9n+2$$

B
$$6n-7$$

C
$$5n+2$$

D
none

Solution

$$a_9=a+8d=47$$........(1)
$$a_{16}=a+15d=82$$.....(2)
Subtract eq (1) by (2)
$$7d=35$$
$$\therefore d=5$$
$$a+8d=47$$
$$a+8\times 5=47$$
$$\therefore a=7$$
Hence general term is $$ a_n = a+(n-1)d = 5n+2 $$
Question 10
1 / -0

Find $${a}_{25}-{a}_{15}$$ for the A.P $$-6, -10, -14, -18,..........$$

A
$$-40$$

B
$$-48$$

C
$$40$$

D
$$-44$$

Solution

Given series is: $$-6,-10,-14,-18.....$$
$$a=-6, d=-4$$
As we know $$nth$$ term, $$a_n = a+(n-1)d$$
$$\therefore$$ $$a_{25}=a+24d$$
& $$a_{15}=a+14d$$
$$a_{25}-a_{15}=(a+24d)-(a+14d)=10d$$
In the following series $$d=-4$$
$$\therefore a_{25}-a_{15}=10\times (-4)=-40$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 43

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Arithmetic Progressions Test - 43

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Question 1

$$501 th$$

$$502 th$$

$$508 th$$

None of these

Solution

Question 2

$$\displaystyle -5\sqrt{2}$$

$$\displaystyle 5\sqrt{2}$$

$$\displaystyle 10\sqrt{2}$$

$$\displaystyle -10\sqrt{2}$$

Solution

Question 3

$$950$$

$$925$$

$$880$$

$$1050$$

Solution

Question 4

$$37$$

$$40$$

$$43$$

$$58$$

Solution

Question 5

$$-24$$

$$24$$

$$-30$$

$$30$$

Solution

Question 6

$$530, 820$$

$$179, 587$$

$$560, 758$$

$$164, 850$$

Solution

Question 7

$$192$$

$$216$$

$$168$$

$$240$$

Solution

Question 8

$$7$$

$$5$$

$$6$$

$$3$$

Solution

Question 9

$$9n+2$$

$$6n-7$$

$$5n+2$$

none

Solution

Question 10

$$-40$$

$$-48$$

$$40$$

$$-44$$

Solution

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