Arithmetic Progressions Test

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Arithmetic Progressions Test - 22

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Question 1
1 / -0

If $$k, 2k -1$$ and $$2k + 1$$ are three consecutive terms as an $$A.P.$$ then find the value of $$k$$.

A
$$2$$

B
$$3$$

C
$$-3$$

D
$$5$$

E
$$4$$

Solution

$$k, 2k - 1 , 2k + 1 $$ are three terms in A.P. Then,
Common difference between first two terms = $$2k - 1 - k $$ =$$k - 1$$
Common difference between next two terms = $$2k + 1 - 2k + 1 $$ =$$2$$

The common difference between consecutive terms should be equal.
Hence, $$k - 1 = 2$$
$$k = 3$$
Question 2
1 / -0

The value of $$\dfrac{1}{97}+\dfrac{2}{97}+.....+\dfrac{96}{97}$$ is:

A
$$48$$

B
$$-48$$

C
$$1$$

D
None of the above

Solution

We know,
Sum of $$n$$ natural numbers $$=\dfrac{n(n+1)}{2}$$
$$\therefore \dfrac{1}{97}+\dfrac{2}{97}+.....+\dfrac{96}{97}$$

$$= \dfrac{1+2+3+....+96}{97}$$

$$= \dfrac{96\times 97}{2\times 97}$$

$$=48$$
Hence, the answer is $$48$$.
Question 3
1 / -0

If the $$n$$th term of an A.P is $$2n+3$$, then sum of first $$n$$ term of the A.P is

A
$$n(n+3)$$

B
$$n(n+4)$$

C
$$n(n-2)$$

D
$$n(n+1)$$

Solution

Here, $$a_n =2n+3 $$
$$\Rightarrow a_1 = 2(1)+3= 5$$
and $$a_2= 2(2)+3= 7$$
common difference $$d=7-5=2$$
sum of first n terms $${S}_{n}=\dfrac{n}{2}[2a+(n-1)d]$$
$$S_n=\dfrac{n}{2}[2\times 5+(n-1)2]$$
$$S_n=n[ 5+n-1]$$
$$S_n=n(n+4)$$
$$\therefore $$ Option B is correct.
Question 4
1 / -0

If an A.P is given by $$7,12,17,22$$, then $$n$$th term is

A
$$2n+5$$

B
$$4n+3$$

C
$$5n+2$$

D
$$3n+4$$

Solution

Given AP: $$7,12,17,22.....$$
Here first term $$a=7$$ and common difference $$d=12-7=5$$
Since, $$a_n=a+(n-1)d $$
$$\Rightarrow a_n=7+(n-1)5 $$
$$\Rightarrow a_n=5n+2 $$
$$n$$th term is $$5n+2$$
$$\therefore$$ Option C is correct.
Question 5
1 / -0

Which term of the sequence $$-7, -2, 3, 8,.........$$ is $$73$$?

A
$$17$$

B
$$18$$

C
$$16$$

D
none

Solution

Here first term $$a=-7$$ , common difference $$d= -2-(-7) = 5$$ and last term $$l= 73$$
Since, $$l=a+(n-1)d $$
$$\Rightarrow 73= -7+(n-1)5 $$
$$\Rightarrow 80= (n-1)5 $$
$$\Rightarrow 16= n-1 $$
$$\Rightarrow n=17 $$
$$\therefore $$ Option A is correct.
Question 6
1 / -0

Find common difference of the following
$$8, 15, 22, 29, .....$$

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

Solution

The given sequence is $$8, 15, 22, 29..... $$.

The common difference in this sequence is the difference between the consecutive terms.
So, the common difference would be $$2^{nd}$$ term - $$1^{st}$$ term $$= 15 - 8$$ i.e. $$7$$.

Difference between the $$3^{rd}$$ term and $$2^{nd}$$ term $$= 22 - 15$$ i.e. $$7$$.

Thus common difference between the terms in this sequence is $$7$$.
Question 7
1 / -0

The sum of first $$20$$ natural numbers is

A
$$55$$

B
$$210$$

C
$$110$$

D
$$215$$

Solution

The sum of first '$$n$$' natural numbers is given by $$ \dfrac { n(n + 1) }{ 2 } $$.
So, the sum of first $$20$$ natural numbers is $$ \dfrac { 20(20 + 1) }{ 2 } $$.
$$= \dfrac { 20(21) }{ 2 } $$
$$ = 210$$.
Question 8
1 / -0

Find the sum of all odd natural numbers less than $$200$$.

A
$$1000$$

B
$$10100$$

C
$$9900$$

D
$$10000$$

Solution

The odd natural numbers less than 200 are $$1,\ 3,\ 5\, .\ .\ .\ 199$$

The numbers form an AP

Where, first term $$a=1$$
and common difference $$d= 2$$
and last term $$a_n = 199$$

We know that $$n^{th}$$ term of an AP $$=a+(n-1)d$$

$$\Rightarrow\;199= 1+(n-1)\times2$$

$$\Rightarrow\;\displaystyle\frac{198}{2}=n-1$$

$$\Rightarrow\;n=99+1$$

$$\Rightarrow\;n=100$$

We know
Sum of $$n$$ terms of an AP $$ = \dfrac{n}{2} (2a+(n-1)d)$$

Let the sum of odd natural numbers less than $$200$$ be $$S_{100}$$

$$\Rightarrow\;S_{100}=\displaystyle\frac{100}{2}[1+199]$$

$$\Rightarrow\;S_{100}=50\times200$$

$$=10000$$
Question 9
1 / -0

Which of the following will not be a term of the sequence $$11,20,29,.....$$

A
$$326$$

B
$$173$$

C
$$388$$

D
none

Solution

As $${a}_{n}=a+(n-1)d$$
$$a=11, d=9$$
Take each option and see if $$(n-1)$$ is integer or not
Since, $$(n-1)=\cfrac{{a}_{n}-a}{d}$$, in other way if $${a}_{n}-a$$ is divisible by $$d$$ or not.
$${a}_{n}=326, \cfrac{326-11}{9}=\cfrac{315}{9}=35$$, divisible
$${a}_{n}=173, \cfrac{173-11}{9}=\cfrac{162}{9}=18$$, divisible
$${a}_{n}=388, \cfrac{388-11}{9}=\cfrac{377}{9}$$, not divisible
Hence option (3) is correct, as $$388$$ is not a term of the sequence.
Question 10
1 / -0

Find the 12th term of the AP. whose first term is 9 and common difference is 10.

A
110

B
119

C
128

D
130

Solution

The nth term in an AP is given by $$\qquad \qquad { t }_{ n }\quad =\quad a\quad +\quad (n - 1)d$$, where 'a' is the first term, 'n' the number of terms and 'd' is the common difference.
So, the 12th term would be $$\qquad \qquad { t }_{ 12 }\quad =\quad 9\quad +\quad (12 - 1)10\\ \Longrightarrow \qquad { t }_{ 12 }\quad =\quad 9\quad +\quad 110\\ \Longrightarrow \qquad { t }_{ 12 }\quad =\quad 119$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 22

Result

Arithmetic Progressions Test - 22

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SHARING IS CARING

Question 1

$$2$$

$$3$$

$$-3$$

$$5$$

$$4$$

Solution

Question 2

$$48$$

$$-48$$

$$1$$

None of the above

Solution

Question 3

$$n(n+3)$$

$$n(n+4)$$

$$n(n-2)$$

$$n(n+1)$$

Solution

Question 4

$$2n+5$$

$$4n+3$$

$$5n+2$$

$$3n+4$$

Solution

Question 5

$$17$$

$$18$$

$$16$$

none

Solution

Question 6

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 7

$$55$$

$$210$$

$$110$$

$$215$$

Solution

Question 8

$$1000$$

$$10100$$

$$9900$$

$$10000$$

Solution

Question 9

$$326$$

$$173$$

$$388$$

none

Solution

Question 10

110

119

128

130

Solution

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