Arithmetic Progressions Test

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Arithmetic Progressions Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle t_{54}$$ an A.P is -61 and $$\displaystyle t_{4}=64$$ find $$\displaystyle t_{10}$$

A
59

B
49

C
36

D
64

Solution

$$t_{54}=a+53d=-64...........(1)$$
$$t_4=a+3d=61.........(2)$$
Subtracting eq 1 and 2
$$50d=-125$$
$$d=-2.5$$
Substitute the value of d in eq 2
$$a=64-(3\times -2.5)$$
$$a=64+7.5=71.5$$

$$t_{10}=a+9d$$
$$\Rightarrow 71.5+(9\times -2.5)$$
$$\Rightarrow 71.5-22.5=49$$
Question 2
1 / -0

Find the number of terms in the sequence$$:\ 4,\ 12,\ 20,\ .\ .\ .\ 108$$

A
$$12$$

B
$$19$$

C
$$13$$

D
$$14$$

Solution

Given sequence: $$4,\ 12,\ 20,\ .\ .\ .\ 108$$

Here first term $$a=4$$,
common difference $$d=12-4=8$$
and last term $$ a_n=108$$

Since, $$a_n=a+(n-1)d $$
$$\Rightarrow 108=4+(n-1)8 $$
$$\Rightarrow 104=8n-8 $$
$$\Rightarrow 112=8n $$
$$\Rightarrow n=14 $$

So, the number of terms in the given sequence is $$14$$
Question 3
1 / -0

The sum of the first $$40$$ natural numbers is

A
$$210$$

B
$$820$$

C
$$610$$

D
$$710$$

Solution

Sum of '$$n$$' natural numbers $$=\dfrac { n(n+ 1) }{ 2 } $$

$$\therefore$$ sum of $$40$$ natural numbers $$= \dfrac {40(40+ 1) }{ 2 } = 820$$.
hence, option B is correct.
Question 4
1 / -0

If $$ 5, k, 11$$ are in $$AP,$$ then the value of $$k$$ is

A
$$6$$

B
$$8$$

C
$$7$$

D
$$9$$

Solution

If $$5, k$$ and $$11$$ are in AP, then
$$k - 5 = 11 - k$$
$$\therefore 2k= 16$$
$$\therefore k= 8$$
Question 5
1 / -0

30th term of the A.P. : 10, 7, 4,.........., is :

A
97

B
77

C
-77

D
-87

Solution

nth term of an AP is given by $$\qquad \qquad \\ { t }_{ n }\quad =\quad a\quad +\quad (n\quad -\quad 1)d$$

In the given sequence, 'a' = 10, 'd' = -3, 'n' = 30.
So, 30th term = $$\qquad \qquad \\ \qquad { t }_{ 30 }\quad =\quad 10\quad +\quad (30\quad -\quad 1)(-3)\\ \Longrightarrow { t }_{ 30 }\quad =\quad 10\quad -\quad 87\quad =\quad -77$$.
Question 6
1 / -0

The sum of the first $$22$$ terms of the A.P. $$8, 3, -2, ..........$$ is

A
$$-875$$

B
$$-979$$

C
$$-717$$

D
$$1072$$

Solution

Given, $$\displaystyle 8, 3, -2,.........$$ are in A.P.
Thus, first term and common difference of the A.P. is $$8$$ and $$-5$$ respectively.
$$\displaystyle a=8,d=3-8=-5,n=22$$
$$\displaystyle { S }_{ n }=\frac { n }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} $$
$$\displaystyle =\frac { 22 }{ 2 } \left\{ 16+\left( 21 \right) \left( -5 \right) \right\} $$
$$\displaystyle =11\left\{ 16-105 \right\} $$
$$=-979$$
Question 7
1 / -0

If $$\displaystyle { a }_{ n }=2n$$ of an A.P., then common difference is :

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Given, $$\displaystyle { a }_{ n }=2n$$
If, $$n=1$$, $$a_1=2$$,
$$n=2$$, $$a_2=4$$
$$n=3$$, $$a_3=6$$
$$\therefore \displaystyle { a }_{ 1 }=2,{ a }_{ 2 }=4,{ a }_{ 3 }=6$$
$$\displaystyle \therefore $$ Series is $$2, 4 ,6, .....$$
$$\therefore$$ common difference, $$\displaystyle $$ $$\displaystyle d=4-2=2$$
Question 8
1 / -0

If 7 th term of AP is 34 and 13th term is 64 then its 18th term is :

A
87

B
88

C
89

D
90

Solution

Given, 7th term = 34 and 13th term = 64.

$$a\quad +\quad 6d\quad =\quad 34\\ a\quad +\quad 12d\quad =\quad 64$$.
From the above two equations, we can deduce, 'd' = 5.

$$\qquad -6d\quad =\quad -30\\ \Longrightarrow \quad d\quad =\quad 5$$. [subtracting both equations]

Substituting 'd' = 5 in either equation, will give 'a'.
$$\qquad \qquad a\quad +\quad 6d\quad =\quad 34\\ \Longrightarrow \qquad a\quad +\quad 6(5)\quad =\quad 34\\ \Longrightarrow \qquad a\quad +\quad 30\quad =\quad 34\quad or\quad a\quad =\quad 4$$.

First term 'a' = 4.

$$\qquad \qquad { t }_{ 18 }\quad =\quad a\quad +\quad 17d\\ \Longrightarrow \qquad { t }_{ 18 }\quad =\quad 4\quad +\quad 17(5)\\ \Longrightarrow \qquad { t }_{ 18 }\quad =\quad 89$$.

18th term of the sequence = 89.
Question 9
1 / -0

The sum of 2 + 4 + 6 + 8 + .......... + 80 is :

A
$$1540$$

B
$$1640$$

C
$$1740$$

D
$$1440$$

Solution

Given sequence is: $$2+4+6+.......+80$$
Here, $$\displaystyle a=2,d=4-2=2,$$$$\displaystyle { a }_{ n }=80$$
$$\displaystyle { a }_{ n }=a+\left( n-1 \right) d$$
$$\displaystyle \Rightarrow 80=2+\left( n-1 \right) 2$$
$$\displaystyle \Rightarrow 80=2n\Rightarrow n=40$$
$$\Rightarrow \displaystyle { S }_{ 40 }=\frac { 40 }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} $$
$$\displaystyle =20\left\{ 4+78 \right\}$$
$$ =20\times 82$$
$$=1640$$
Question 10
1 / -0

The sum of the first $$1000$$ positive integer is

A
$$500500$$

B
$$650650$$

C
$$300300$$

D
$$700700$$

Solution

Sum of '$$n$$' natural numbers $$=$$ $$ \dfrac { n(n+ 1) }{ 2 } \\ $$.
So, sum of first $$1000$$ natural numbers $$=$$ $$ \dfrac { 1000(1000+ 1) }{ 2 } = 500(1001)= 500500\\ $$.
Hence, option A is correct.

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Arithmetic Progressions Test - 23

Result

Arithmetic Progressions Test - 23

Score

-

Rank

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SHARING IS CARING

Question 1

59

49

36

64

Solution

Question 2

$$12$$

$$19$$

$$13$$

$$14$$

Solution

Question 3

$$210$$

$$820$$

$$610$$

$$710$$

Solution

Question 4

$$6$$

$$8$$

$$7$$

$$9$$

Solution

Question 5

97

77

-77

-87

Solution

Question 6

$$-875$$

$$-979$$

$$-717$$

$$1072$$

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 8

87

88

89

90

Solution

Question 9

$$1540$$

$$1640$$

$$1740$$

$$1440$$

Solution

Question 10

$$500500$$

$$650650$$

$$300300$$

$$700700$$

Solution

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