Sequences and Series Test

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Sequences and Series Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\left(\displaystyle 1-\frac{1}{2}\right)\left(\displaystyle 1-\frac{1}{3}\right)\left(\displaystyle 1-\frac{1}{4}\right).........\left(\displaystyle 1-\frac{1}{10}\right)=$$ ___________.

A
$$\displaystyle\frac{10}{11}$$

B
$$\displaystyle\frac{1}{9}$$

C
$$\displaystyle\frac{1}{10}$$

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

Solution

$$\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)..........\left(1-\dfrac{1}{10}\right)$$

$$\Rightarrow$$ $$\dfrac{1}{2}\times \dfrac{2}{3}\times \dfrac{3}{4}......\times \dfrac{8}{9}\times \dfrac{9}{10}$$

$$\Rightarrow$$ $$\dfrac{1}{10}$$
Question 2
1 / -0

Sum $$1+3x+5{ x }^{ 2 }+7{ x }^{ 3 }+9{ x }^{ 4 }+.....$$ to infinity is $$\left( x<1 \right) $$

A
$$\dfrac { 1+x }{ { \left( 1-x \right) }^{ 2 } } $$

B
$$\dfrac { 1-x }{ { \left( 1+x \right) }^{ 2 } } $$

C
$$\dfrac { 1-x }{ { \left( 1+x \right) } } $$

D
$$\dfrac { 1+x }{ { \left( 1-x \right) } } $$

Solution

Let $$S=1+3x+5x^2+7x^3+9x^4+.......$$
$$xS = x + 3{x^2} + 5{x^3} + 7{x^4} + 9{x^5} + .......$$
$$S-xS = 1 + \left( {3x - x} \right) + \left( {5{x^2} - 3{x^2}} \right) + \left( {7{x^3} - 5{x^3}} \right) + \left( {9{x^4} - 7{x^4}} \right) + .......$$
$$S(1-x)=1+2x(1+x+x^2+x^3+....)$$
$$=1+2x \times \frac{1}{1-x}$$
$$=\dfrac{x+1}{1-x}$$
$$S=\dfrac{x+1}{(1-x)^2}$$
Question 3
1 / -0

$$1 \, + \, 2 \, \cdot \, 2 \, + \, 3 \, \cdot \, 2^2 \, + \, 4 \, \cdot \, 2^3 \, + \, ..... \, + \, 100 \, \cdot \, 2^{99}$$

A
$$S=50502^{100}$$

B
$$S=5050.2^{99}$$

C
$$S=5050.2^{101}$$

D
$$S=5050.2^{98}$$

Solution

Consider the given series.
$$1+2.2+3.2^2+4.2^3+5.2^4+.......+100.2^{99}$$

The series mentioned above is a $$A.G.P$$ series i.e arithmetic and geometric series.
The $$n^{th}$$ term for this series will be
$$=n.2^{n-1}$$

Now,
$$=\dfrac{n(n+1)}{2}.\dfrac{1.2^{n-1}}{2-1}$$

Since, $$n=100$$

Therefore,
$$=\dfrac{100(100+1)}{2}.\dfrac{1.2^{100-1}}{2-1}$$
$$=\dfrac{100(101)}{2}.{1.2^{99}}$$
$$=50(101).2^{99}$$
$$=5050.2^{99}$$

Hence, this is the answer.
Question 4
1 / -0

Select the missing number from the given matrix:
5 2 4
4 4 7
2 5 3
18 30 ?

A
$$43$$

B
$$42$$

C
$$33$$

D
$$32$$

Solution

In first column $$(5+4)\times 2=18$$ In second column $$(2+4)\times 5=30$$

so in third column ans

$$=(4+7)\times3$$

$$=11\times 3=33.$$
Question 5
1 / -0

The two sequence of number $${1,4,16,64,.......}$$ and $${3,12,48,192,........}$$ are mixed as follows:$${1,3,4,12,16,48,64,192,.......}$$. one of the numbers in the mixed in the mixed series is $$1048576$$, then number immediately preceding is-

A
$$3.4^9$$

B
$$3.4^{10}$$

C
$$4^9$$

D
$$4^{10}$$

Solution

$$1,4,16,64,....\equiv 1,4,4^2,4^3,..\equiv 4^{n-1}$$

$$3,12,48,192,... \equiv 3\times1,3\times4,3\times4^2,3\times4^3...\equiv 3\times4^{n-1}$$

$$1,3,4,12,16,48,64,192,.. \equiv 1,3\times1,4,3\times4,4^2,3\times4^2,4^3,3\times4^3,..$$

$$1048576=4^{10} is \space 4^{11-1}$$
Number preceding $$3\times4^{10-1}=3.4^9$$
Question 6
1 / -0

If $$\dfrac{a-x}{px}=\dfrac{a-y}{qy}=\dfrac{a-z}{rz}$$ and $$p, q, r,$$ be in A.P. then $$x, y, z$$ are in?

A
A.P.

B
G.P.

C
H.P.

D
A.G.P.

Solution

Given $$\dfrac{a-x}{px}=\dfrac{a-y}{qy}=\dfrac{a-z}{rz}=k$$(let)

$$\therefore p=\dfrac{a-x}{xk}, q=\dfrac{a-y}{yk}$$ and $$r=\dfrac{a-z}{zk}$$

Given p,q,r are in A.P. $$\therefore 2q=p+r$$
$$\Rightarrow 2\times(\dfrac{a-y}{yk})=(\dfrac{a-x}{xk})+(\dfrac{a-z}{zk})$$
$$\Rightarrow 2\times(\dfrac{a-y}{y})=(\dfrac{a-x}{x})+(\dfrac{a-z}{z})$$
$$\Rightarrow 2\times(\dfrac{a}{y}-1)=\dfrac{a}{x}-1+\dfrac{a}{z}-1$$
$$\Rightarrow \dfrac{2a}{y}-2=\dfrac{a}{x}+\dfrac{a}{z}-2$$
$$\Rightarrow \dfrac{2a}{y}=\dfrac{a}{x}+\dfrac{a}{z}$$
$$\Rightarrow \dfrac{2}{y}=\dfrac{1}{x}+\dfrac{1}{z}$$
$$\therefore x,y,z $$ are in H.P.
Question 7
1 / -0

What is the next term of the AP $$\sqrt{2}, \sqrt{8}, \sqrt{18},$$ ____?

A
$$\sqrt{24}$$

B
$$\sqrt{28}$$

C
$$\sqrt{30}$$

D
$$\sqrt{32}$$

Solution

We are given a series : $$ \sqrt{2},\sqrt{8},\sqrt{18} $$
$$ a = \sqrt{2} $$
$$ a + d = \sqrt{8} = \sqrt{4.2} = 2\sqrt{2} $$
so, $$ d = 2\sqrt{2} - \sqrt{2} = \sqrt{2} $$
so, $$ 4^{th} $$ term will be
$$ a+3d = \sqrt{2} + 3\sqrt{2} = 4\sqrt{2} $$
$$ a+3d= \sqrt{4^{2}.2} = \sqrt{16.2} = \sqrt{32} $$
so, the next term will be $$ \sqrt{32} $$
Question 8
1 / -0

$$A$$ is twice as fast as $$B$$ is thrice as fast as $$C$$. The journey covered by $$C$$ in $$42$$ minutes, what will be covered by $$A$$ is

A
$$21$$ Min

B
$$64$$ Min

C
$$17$$ Min

D
$$40$$ Min

Solution

time taken ny C to complete the journey $$=$$ 42 minutes
time taken by A to complete the journey $$=\dfrac{42}{2}$$ minutes ($$\because $$ A is twice as fast as C)
$$=21 $$ minutes.
Question 9
1 / -0

The sum of the infinite series $$1+\left ( 1+\dfrac{1}{5} \right )\left ( \dfrac{1}{2} \right )+\left ( 1+\dfrac{1}{5}+\dfrac{1}{5^2} \right )\left ( \dfrac{1}{2^2} \right )+.......$$

A
$$\dfrac{20}{9}$$

B
$$\dfrac{10}{9}$$

C
$$\dfrac{5}{9}$$

D
$$\dfrac{5}{3}$$

Solution

Let $$\displaystyle S= 1+(1+\frac{1}{5})(\frac{1}{2})+(1+\frac{1}{5}+\frac{1}{5^{2}})(\frac{1}{2^{2}})+...$$

$$\displaystyle \Rightarrow \frac{S}{2}= \frac{1}{2}+(1+\frac{1}{5})(\frac{1}{2^{2}})+(1+\frac{1}{5}+\frac{1}{5^{2}})(\frac{1}{2^{3}})+...$$

$$\displaystyle S-\frac{S}{2}= 1+\frac{1}{5}(\frac{1}{2})+\frac{1}{5^{2}}(\frac{1}{2^{2}})+(\frac{1}{5^{3}})(\frac{1}{2^{3}})+...$$

$$\displaystyle\Rightarrow \frac{S}{2}= 1+\frac{1}{10}+\frac{1}{(10)^{2}}+\frac{1}{(10)^{3}}+...$$

$$\dfrac{S}{2}= \dfrac{1}{1-\dfrac{1}{10}}= \dfrac{10}{9}$$ [infinite G.P sum $$S=\dfrac{a}{1-r}]$$

$$S= \dfrac{20}{9}$$

Option A is correct
Question 10
1 / -0

The sum of the series $$\sum_{r=0}6{88} \sec r^o\cdot \sec(r+1)^o$$ is equal to

A
$$cosec\, 1^o\cdot \tan 1^o$$

B
$$cosec\, 1^o\cdot \cot 1^o$$

C
$$\sec\, 1^o\cdot \tan 1^o$$

D
$$\sec\, 1^o\cdot \cot 1^o$$

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Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 15

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Sequences and Series Test - 15

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SHARING IS CARING

Question 1

$$\displaystyle\frac{10}{11}$$

$$\displaystyle\frac{1}{9}$$

$$\displaystyle\frac{1}{10}$$

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

Solution

Question 2

$$\dfrac { 1+x }{ { \left( 1-x \right) }^{ 2 } } $$

$$\dfrac { 1-x }{ { \left( 1+x \right) }^{ 2 } } $$

$$\dfrac { 1-x }{ { \left( 1+x \right) } } $$

$$\dfrac { 1+x }{ { \left( 1-x \right) } } $$

Solution

Question 3

$$S=50502^{100}$$

$$S=5050.2^{99}$$

$$S=5050.2^{101}$$

$$S=5050.2^{98}$$

Solution

Question 4

$$43$$

$$42$$

$$33$$

$$32$$

Solution

Question 5

$$3.4^9$$

$$3.4^{10}$$

$$4^9$$

$$4^{10}$$

Solution

Question 6

A.P.

G.P.

H.P.

A.G.P.

Solution

Question 7

$$\sqrt{24}$$

$$\sqrt{28}$$

$$\sqrt{30}$$

$$\sqrt{32}$$

Solution

Question 8

$$21$$ Min

$$64$$ Min

$$17$$ Min

$$40$$ Min

Solution

Question 9

$$\dfrac{20}{9}$$

$$\dfrac{10}{9}$$

$$\dfrac{5}{9}$$

$$\dfrac{5}{3}$$

Solution

Question 10

$$cosec\, 1^o\cdot \tan 1^o$$

$$cosec\, 1^o\cdot \cot 1^o$$

$$\sec\, 1^o\cdot \tan 1^o$$

$$\sec\, 1^o\cdot \cot 1^o$$

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