Arithmetic Progressions Test

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Arithmetic Progressions Test - 39

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Question 1
1 / -0

Find $$n$$ if the given value of $$x$$ is the $$n^{th}$$ term of the AP: $$5\dfrac{1}{2}, 11, 16\dfrac{1}{2},22,... $$ ; $$x=550$$.

A
$$100$$

B
$$110$$

C
$$200$$

D
$$220$$

Solution

Here, first term $$a=5\displaystyle \frac12=\frac{11}2$$ and the common difference $$d=11-\displaystyle \frac{11}2=\frac{11}2.$$
Given, $$a_n=x=550$$
$$\Rightarrow a+(n-1)d=550$$
$$\Rightarrow \dfrac{11}2+(n-1)\times\dfrac{11}2=550$$
$$\Rightarrow 1+(n-1)=550\times\dfrac2{11}$$
$$\Rightarrow n=100$$
Question 2
1 / -0

Which term of the AP: $$3, 10, 17,..$$ will be $$84$$ more than its $$13^{th}$$ term?

A
$$25^{th}$$

B
$$24^{th}$$

C
$$35^{th}$$

D
$$34^{th}$$

Solution

Clearly, the given sequence is an AP with first term $$a=3$$ and common difference $$d=10-3=7$$.
Given $$n^{th}$$ term is $$84$$ more than its $${13}^{th}$$ term.
$$\Rightarrow t_n=t_{13}+84$$
$$\Rightarrow a+(n-1)d=(a+12d)+84$$
$$\Rightarrow (n-1)d=12d+84$$
$$\Rightarrow (n-1)7=12\times7+84$$
$$\Rightarrow (n-1)7=7(12+12)$$
$$\Rightarrow n-1=24$$
$$\Rightarrow n=25$$
Therefore $${25}^{th}$$ term is the required term.
Question 3
1 / -0

How many terms are there in the A.P., $$7, 13, 19, ...205$$ ?

A
$$32$$

B
$$33$$

C
$$34$$

D
$$35$$

Solution

Given sequence is $$7,13,19,...,205$$

The first term $$a=7$$ and

the common difference $$d=13-7=6$$

the last term is $$205$$

Let the last term be the $$n^{th}$$ term

We know that the $$n^{th}$$ term of the arithmetic progression is given by $$a+(n-1)d$$

Therefore, $$a+(n-1)d=205$$

$$\implies 7+(n-1)\times(6)=205$$

$$\implies 7-6+6n=205$$

$$\implies 1+6n=205$$

$$\implies 6n=205-1$$

$$\implies n=\dfrac{204}{6}$$

$$\implies n=34$$

Therefore, the number of terms in the given sequence is $$34$$
Question 4
1 / -0

If the sum of $$n$$ terms of an AP is $$2n$$$$^2$$ $$+ 5n$$, then its $$n$$th term is

A
$$4n - 3$$

B
$$3n - 4$$

C
$$4n + 3$$

D
$$3n + 4$$

Solution

Given $${ S }_{ n }=2{ n }^{ 2 }+5n$$

$$ { S }_{ n-1 }=2{ (n-1) }^{ 2 }+5(n-1)$$

$$ =2({ n }^{ 2 }-2n+1)+5n-5$$

$$ =2{ n }^{ 2 }-4n+2+5n-5$$

$$ =2{ n }^{ 2 }+n-3$$

Now, $${ t }_{ n }={ S }_{ n }-{ S }_{ n-1 }$$

   $$=2{ n }^{ 2 }+5n-(2{ n }^{ 2 }+n-3)$$

     $$=2{ n }^{ 2 }+5n-2{ n }^{ 2 }-n+3$$

     $$=4n+3$$
Question 5
1 / -0

Find the sum of all even integers between $$101$$ and $$999.$$

A
$$246950$$

B
$$249650$$

C
$$256950$$

D
$$276950$$

Solution

$$\textbf{Step -1: Find the number of terms.}$$
$$\text{Even integers between }101\text{ and }999 \text{ are:}$$
$$102,104,106,\ldots998.$$
$$\text{It is an A.P. sequence whose first term}(a)=102,$$
$$\text{common difference}(d)=2$$
$$\text{and last term}(l)=998.$$
$$\text{Let this sequence has }n \text{ terms.}$$
$$\therefore a_n=l=a+(n-1)d$$
$$\Rightarrow 998=102+(n-1)2$$
$$\Rightarrow 2n-2=896$$
$$\Rightarrow 2n=898$$
$$\Rightarrow n=449$$
$$\textbf{Step -2: Find the required sum.}$$
$$\because \text{Sum of an A.P.}(S_n)=\dfrac{n}{2}(a+l).$$
$$\therefore\text{The sum of all even integers between }101\text{ and }998,$$
$$=\dfrac{449}{2}(102+998)$$
$$=\dfrac{449\times1100}{2}$$
$$=449\times550$$
$$=246950$$
$$\textbf{Hence,The correct option is A.}$$
Question 6
1 / -0

Which term of the A.P., $$84, 80, 76,..$$ is $$0?$$

A
$$18^{th}$$ term

B
$$20^{th}$$ term

C
$$22^{nd}$$ term

D
$$24^{th}$$ term

Solution

The given sequence is an AP in which first term $$a=84$$ and common difference $$d=-4$$.
We have to find value of n for which $$a_n$$ is $$0$$.
Then,
$$a+(n-1)d=0$$
$$\Rightarrow 84+(n-1)\times-4=0$$
$$\Rightarrow 88-4n=0$$
$$\Rightarrow n=22$$
Hence, $${22}^{nd}$$ term is $$0$$.
Question 7
1 / -0

Find the sum of first $$n$$ odd natural numbers.

A
$$n^3$$

B
$$n(n+1)$$

C
$$n(n+1)^2$$

D
$$n^2$$

Solution

We have to find sum of the series : $$1+3+5+...$$ to $$n$$ terms
Clearly, terms of the given series form an AP with first term $$a=1$$ and common difference $$d=2$$.
Therefore, $$S_n=\dfrac n2[2a+(n-1)d]$$
$$\Rightarrow S_n=\dfrac n2[2\times1+(n-1)\times2]$$
$$\Rightarrow S_n=\dfrac n2[2+2n-2]$$
$$\Rightarrow S_n=n^2$$
Question 8
1 / -0

Find the $$n^{th}$$ term of the sequence $$m -1, m - 3, m - 5,.....$$

A
$$m - n + 1$$

B
$$m + 2n + 1$$

C
$$m - 2n + 1$$

D
$$m - 2n$$

Solution

Clearly, the given sequence is an AP with first term $$a=m-1$$ and common difference $$d=(m-3)-(m-1)=-2$$.
$$\therefore$$ nth term $$a_n=a+(n-1)d$$
$$\Rightarrow a_n=(m-1)+(n-1)\times-2$$
$$=m-1-2n+2$$
$$=m-2n+1$$
Question 9
1 / -0

Find the $$12^{th}$$ term from the end of the arithmetic progression $$3, 5, 7, 9,...201?$$

A
$$179$$

B
$$175$$

C
$$172$$

D
$$174$$

Solution

Clearly, the given sequence is an AP with first term $$a=3$$ and common difference $$d=2$$.
Let there are $$n$$ terms in this AP. Therefore,
$$a_n=201\Rightarrow a+(n-1)d=201$$
$$\Rightarrow 3+(n-1)2=201$$
$$\Rightarrow (n-1)2=198$$
$$\Rightarrow n-1=99$$
$$\Rightarrow n=100$$
Now $$12^{th}$$ term from end=$$(100-12+1)^{th}$$ term from beginning.
$$\Rightarrow 12^{th}$$ term from end=$$89^{th}$$ term from beginning=$$3+(89-1)\times2=3+176=179.$$
Question 10
1 / -0

Check if the series is an $$AP.$$ Find the common difference $$d$$. Also, find the next three terms.

$$-10, -6, -2 , 2.....$$.

A
It is an $$AP$$ and $$d=4$$, other terms $$6,10,14$$

B
It is an $$AP$$ and $$d=\dfrac{3}{5}$$, other terms $$5,10,15$$

C
It is not an $$AP$$, other terms $$3,4,5$$

D
None of these

Solution

Given series is $$-10, -6,-2,2,.......$$
Clearly, the given sequence is an AP with first term $$a=-10$$ and common difference $$d=-6-(-10)=4$$
$$n^{th}$$ term $$=a_n=a+(n-1)d$$
$$\Rightarrow a_5=a+4d=-10+4\times4=6$$
$$\Rightarrow a_6=a+5d=-10+5\times4=10$$
$$\Rightarrow a_7=a+6d=-10+6\times4=14$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 39

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Arithmetic Progressions Test - 39

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SHARING IS CARING

Question 1

$$100$$

$$110$$

$$200$$

$$220$$

Solution

Question 2

$$25^{th}$$

$$24^{th}$$

$$35^{th}$$

$$34^{th}$$

Solution

Question 3

$$32$$

$$33$$

$$34$$

$$35$$

Solution

Question 4

$$4n - 3$$

$$3n - 4$$

$$4n + 3$$

$$3n + 4$$

Solution

Question 5

$$246950$$

$$249650$$

$$256950$$

$$276950$$

Solution

Question 6

$$18^{th}$$ term

$$20^{th}$$ term

$$22^{nd}$$ term

$$24^{th}$$ term

Solution

Question 7

$$n^3$$

$$n(n+1)$$

$$n(n+1)^2$$

$$n^2$$

Solution

Question 8

$$m - n + 1$$

$$m + 2n + 1$$

$$m - 2n + 1$$

$$m - 2n$$

Solution

Question 9

$$179$$

$$175$$

$$172$$

$$174$$

Solution

Question 10

It is an $$AP$$ and $$d=4$$, other terms $$6,10,14$$

It is an $$AP$$ and $$d=\dfrac{3}{5}$$, other terms $$5,10,15$$

It is not an $$AP$$, other terms $$3,4,5$$

None of these

Solution

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