Sequences and Series Test

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Sequences and Series Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The value of the product
$${ 6 }^{ \frac { 1 }{ 2 } }\times { 6 }^{ \frac { 1 }{ 4 } }\times { 6 }^{ \frac { 1 }{ 8 } }\times { 6 }^{ \frac { 1 }{ 16 } }\times ..$$ up to infinite terms is

A
$$6$$

B
$$36$$

C
$$216$$

D
$$512$$

Solution

$$6^{\tfrac12}\times 6^{\tfrac14}\times 6^{\tfrac18}........$$

$$=6^{\left(\tfrac12+\tfrac14+\tfrac18.......\right)}$$

$$=6^{\left(\tfrac{\tfrac12}{1-\tfrac12}\right)}$$

$$=6^1=6$$

Hence, the answer is $$6$$.
Question 2
1 / -0

Select the missing number from the given alternatives.
$$3$$ $$4$$ $$2$$ $$14$$
$$6$$ $$5$$ $$4$$ $$44$$
$$5$$ $$2$$ $$7$$ ?

A
$$58$$

B
$$14$$

C
$$49$$

D
$$4$$

Solution

$$3\times 2 + 4\times 2 = 6 + 8 = 14$$
$$6\times 4 + 5\times 4 = 24 + 20 = 44$$
Hence, $$5\times 7 + 2 \times 7 = 35 + 14 = 49$$.
Question 3
1 / -0

If $$12\times 16 = 188$$ and $$14\times 18 = 248$$, then find the value of $$16\times 20 = ?$$

A
$$320$$

B
$$360$$

C
$$316$$

D
$$318$$

Solution

$$12\times 16 = 192 - 4 = 188$$
$$14\times 18 = 252 - 4 = 248$$
$$16\times 20 = 320 - 4 = 316$$.
Question 4
1 / -0

If a and b are two unequal positive numbers, the:

A
$$\frac{2ab}{a+b}>\sqrt{ab}>\frac{a+b}{2}$$

B
$$\sqrt{ab}>\frac{2ab}{a+b}>\frac{a+b}{2}$$

C
$$\frac{2ab}{a+b}>\frac{a+b}{2}>\sqrt{ab}$$

D
$$\frac{a+b}{2}>\frac{2ab}{a+b}>\sqrt{ab}$$

E
$$\frac{a+b}{2}>\sqrt{ab}>\frac{2ab}{a+b}$$

Solution

a and b are two unequal positive numbers.
$$\therefore A.M>G.M>H.M\\ \cfrac { a+b }{ 2 } >\sqrt { ab } >\cfrac { 2ab }{ a+b } $$
Question 5
1 / -0

If $$y = x+x^2+x^3+...$$ up to infinite terms, where $$ x<1$$, then which one of the following is correct?

A
$$x=\dfrac { y }{ 1+y } $$

B
$$x=\dfrac { y }{ 1-y } $$

C
$$x=\dfrac {1+ y }{ y }$$

D
$$x=\dfrac { 1-y }{ y } $$

Solution

Given $$y=x+x^2+x^3+...\infty$$
$$y=x(1+x+x^2+x^3+.....\infty)$$
We know that $$x+x^2+x^3+x^4....\infty =y$$
$$y=x(1+y)$$
$$x=\dfrac{y}{1+y}$$
Question 6
1 / -0

The value of $$\dfrac { 1 }{ \log _{ 3 }{ e } } +\dfrac { 1 }{ \log _{ 3 }{ e^2 } } +\dfrac { 1 }{ \log _{ 3 }{ e^4 } } +...$$ up to infinite terms is

A
$$\log _{ e }{ 9 } $$

B
$$0$$

C
$$1$$

D
$$\log _{ e }{ 3 } $$

Solution

We can write $$\log_{3} e^2=2\log_{3} e$$

Similarly, $$\log_{3} e^4=4\log_{3} e$$ and so on

Let $$t=\dfrac{1}{\log_{ \ 3}e} $$

So, $$\dfrac{1}{\log_{ \ 3}e} + \dfrac{1}{\log_{ \ 3}e^2} +\dfrac{1}{\log_{ \ 3}e^4}..... =\dfrac{1}{\log_{ \ 3}e} + \dfrac{1}{2\log_{ \ 3}e} +\dfrac{1}{4\log_{ \ 3}e}..... $$
$$=t+\dfrac{t}{2}+\dfrac{t}{4}....$$
$$=t\left(1+\dfrac12+\dfrac14+\dfrac18+.....\right)$$
$$=t\times \dfrac{1}{1-\dfrac12}=2t$$

Now, $$2t=2\dfrac{1}{\log_{ \ 3}e} =2\dfrac{1}{\dfrac{\log e}{\log3}}=2\dfrac{\log 3}{\log e}=2\log_{e}3$$ $$=\log_{e}9$$

Therefore, the answer is $$\log_{e}9$$.
Question 7
1 / -0

If $$x =1-y+{ y }^{ 2 }-{ y }^{ 3 }+...$$ upto infinite terms, where |y| < 1, then which one of the following is correct?

A
$$x =\frac { 1 }{ 1+y } $$

B
$$x =\frac { 1 }{ 1-y } $$

C
$$x = \frac { y }{ 1+y }$$

D
$$x=\frac { y }{ 1-y } $$

Solution
Question 8
1 / -0

If $$1.3 + 2.3^2 + 3.3^3 +...+ n.3^n $$= $$\dfrac { (2n-1)3^ a+b }{ 4 }$$ then $$a$$ and $$b$$ are respectively

A
$$n,2$$

B
$$n,3$$

C
$$n +1,2$$

D
$$n + 1,3$$

Solution

$$1.3 + 2.3^2 + 3.3^3 +...+ n.3^n $$

this can be wriiten as $$0.3^0+1.3 + 2.3^2 + 3.3^3 +...+ n.3^n $$

Sum of $$n$$ terms in AGP is $$S_n=\dfrac{a-[a+(n-1)d]r^n}{1-r}+\dfrac{dr(1-r^{n-1})}{(1-r)^2}$$

Here $$a=0,d=1,n=n ,r=3 $$

Substituting in the given formula we get,

$$S_n=\dfrac{(n-1)3^n}{2}+\dfrac{3(1-3^{n-1})}{4}$$

$$S_n=\dfrac{2(n-1)3^n+3-3^n}{4}$$

$$S_n=\dfrac{(2n-1)3^{n+1}+3}{4}$$

Thus, $$a$$ and $$b$$ are $$n+1$$ and $$3$$ respectively.
Question 9
1 / -0

General term of 4 + 6 + 9 + 13 + 18 + .......to n terms

A
$$\dfrac{1}{2}( n^2 \, + \,n\, + \,12)$$

B
$$\dfrac{1}{2}( n^2 \, + \,n\, + \,3)$$

C
$$\dfrac{1}{2}( n^2 \, + \,n\, + \,6)$$

D
$$\dfrac{1}{2}( n^2 \, + \,n\, + \,20)$$

Solution
Question 10
1 / -0

$$\displaystyle \sum _{ k=1 }^{ 6 }{ \left[ sin\frac { 2k\pi }{ 7 } -i\quad cos\quad \frac { 2k\pi }{ 7 } \right] } =$$

A
$$-1$$

B
$$0$$

C
$$-i$$

D
$$i$$

Solution

$$\displaystyle-i\sum _{ k=1 }^{ 6 }{ \left( cos\frac { 2k\pi }{ 7 } +i\quad sin\frac { 2k\pi }{ 7 } \right) } =$$
$$-i\quad \left[ 1+\alpha +{ \alpha }^{ 2 }+....+{ \alpha }^{ 6 }-1 \right] =\quad -i\left[ 0-1 \right] =i$$

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Re-Attempt Weekly Quiz Competition

$$3$$	$$4$$	$$2$$	$$14$$
$$6$$	$$5$$	$$4$$	$$44$$
$$5$$	$$2$$	$$7$$	?

Sequences and Series Test - 34

Result

Sequences and Series Test - 34

Score

-

Rank

-

SHARING IS CARING

Question 1

$$6$$

$$36$$

$$216$$

$$512$$

Solution

Question 2

$$58$$

$$14$$

$$49$$

$$4$$

Solution

Question 3

$$320$$

$$360$$

$$316$$

$$318$$

Solution

Question 4

$$\frac{2ab}{a+b}>\sqrt{ab}>\frac{a+b}{2}$$

$$\sqrt{ab}>\frac{2ab}{a+b}>\frac{a+b}{2}$$

$$\frac{2ab}{a+b}>\frac{a+b}{2}>\sqrt{ab}$$

$$\frac{a+b}{2}>\frac{2ab}{a+b}>\sqrt{ab}$$

$$\frac{a+b}{2}>\sqrt{ab}>\frac{2ab}{a+b}$$

Solution

Question 5

$$x=\dfrac { y }{ 1+y } $$

$$x=\dfrac { y }{ 1-y } $$

$$x=\dfrac {1+ y }{ y }$$

$$x=\dfrac { 1-y }{ y } $$

Solution

Question 6

$$\log _{ e }{ 9 } $$

$$0$$

$$1$$

$$\log _{ e }{ 3 } $$

Solution

Question 7

$$x =\frac { 1 }{ 1+y } $$

$$x =\frac { 1 }{ 1-y } $$

$$x = \frac { y }{ 1+y }$$

$$x=\frac { y }{ 1-y } $$

Solution

Question 8

$$n,2$$

$$n,3$$

$$n +1,2$$

$$n + 1,3$$

Solution

Question 9

$$\dfrac{1}{2}( n^2 \, + \,n\, + \,12)$$

$$\dfrac{1}{2}( n^2 \, + \,n\, + \,3)$$

$$\dfrac{1}{2}( n^2 \, + \,n\, + \,6)$$

$$\dfrac{1}{2}( n^2 \, + \,n\, + \,20)$$

Solution

Question 10

$$-1$$

$$0$$

$$-i$$

$$i$$

Solution

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