Sequences and Series Test

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Sequences and Series Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

If $$m * n = m+(m-1)+(m-2)+ ...... +(m-n)$$, evaluate $$7 * 5$$.

A
$$25$$

B
$$20$$

C
$$27$$

D
$$18$$

Solution

$$\Rightarrow$$ $$m * n=m+(m-1)+(m-2)+....+(m-n)$$
$$\Rightarrow$$ In $$7 * 5$$, $$m=7$$ and $$n=5$$
$$\Rightarrow$$ $$7 * 5=7+(7-1)+(7-2)+(7-3)+(7-4)+(7-5)$$
$$\Rightarrow$$ $$7 * 5=7+6+5+4+3+2$$
$$\therefore$$ $$7 * 5=27$$
Question 2
1 / -0

If $$a\odot b = 6\times a - 3\times b$$, evaluate $$(5\odot 3) \odot 20$$

A
$$2$$

B
$$41$$

C
$$66$$

D
$$1$$

Solution

$$\Rightarrow$$ $$a\odot b=6\times a-3\times b$$ [ Given ]
$$\Rightarrow$$ $$(5\odot 3)\odot 20$$
$$\Rightarrow$$ Let $$a=5$$ and $$b=3$$
$$\Rightarrow$$ $$(6\times 5-3\times 3)\odot 20$$
$$\Rightarrow$$ $$(30-9)\odot 20$$
$$\Rightarrow$$ $$21\odot 20$$
$$\Rightarrow$$ Let $$a=21$$ and $$b=20$$
$$\Rightarrow$$ $$6\times 21-3\times 20$$
$$\Rightarrow$$ $$126-60$$
$$\Rightarrow$$ $$66$$
$$\therefore$$ $$(5\odot 3)\odot 20=66$$
Question 3
1 / -0

What is the missing number in the sequence $$1, 5, 10, 16, 23, 31,$$ ____?

A
$$37$$

B
$$38$$

C
$$39$$

D
$$40$$

E
$$41$$

Solution

We can see that the difference between the consecutive numbers of the series is increasing by $$ 1 $$

$$ 5 - 1 = 4 $$
$$ 10 - 5 = 5 $$
$$ 16 - 10 = 6 $$
$$ 23 - 16 = 7 $$
$$ 31 - 23 = 8 $$

Hence, Next term $$ - 31 = 9 $$
=> Next term $$ = 31 + 9 = 40 $$
Question 4
1 / -0

$$6, 10, 18, 34, 66$$
The first number in the list above is $$6$$. Determine a rule for finding each successive number in the list.

A
Add $$4$$ to the preceding number.

B
Take $$\displaystyle \frac { 1 }{ 2 } $$ of the preceding number and then add $$7$$ to that result.

C
Double the preceding number and then subtract $$2$$ from that result.

D
Subtract $$2$$ from the preceding number and then double that result.

E
Triple the preceding number and then subtract $$8$$ from that result.

Solution

The series is $$6,10,18,34,66$$
In the given series:
$$6+6=12-2=10$$
$$10+10=20-2=18$$
$$18+18=36-2=34$$
$$34+34=68-2=66$$
Hence in this each successive number is obtained by double the preceding number and then subtract $$2$$ from the result.
Question 5
1 / -0

The value of the sum $$1.2.3+2.3.4+3.4.5+...$$ upto n terms =

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

B
$$\dfrac{1}{6}(n^2-1)(2n-1)(2n+3)$$

C
$$\dfrac{1}{8}(n^2+1)(n^2+5)$$

D
$$\dfrac{1}{4}n(n+1)(n+2)(n+3)$$

Solution
Question 6
1 / -0

$$m, 2m, 4m, . . . $$
The first term in the sequence above is $$m$$, and each term thereafter is equal to twice the previous term. If $$m$$ is an integer, which of the following could NOT be the sum of the first four terms of this sequence?

A
$$-26$$

B
$$-15$$

C
$$45$$

D
$$75$$

E
$$120$$

Solution

The first term in given Geometric series, $$a_1=m$$
The common ratio $$r$$ $$=\dfrac{4m}{2m}=2$$
No. of terms, $$n$$ $$=4$$
Applying sum of GP formula,
$$S_n=\dfrac{a(1-r^n)}{1-r}$$
$$=\dfrac{m(1-2^4)}{1-2}$$
$$=\dfrac{m(1-16)}{-1}=15m$$
If $$m$$ is an integer the sum of the first four terms of this sequence should be in multiple of $$15$$.
Hence, option A is correct which is not a multiple of $$15$$.
Question 7
1 / -0

The terms of a sequence are defined by $$a_{n} = 3a_{n - 1} - a_{n - 2}$$ for $$n > 2$$. Find the value of $$a_{5}$$ given that $$a_{1} =4$$ and $$a_{2} = 3$$.

A
$$12$$

B
$$23$$

C
$$25$$

D
$$31$$

E
$$36$$
Solution
- Given $${ a }_{ n }=3{ a }_{ n-1 }-{ a }_{ n-2 }$$ , $${a}_{1}=4$$ and $${a}_{2} = 3$$
- $${a}_{3} = 3{a}_{2}-{a}_{1} = 9-4=5$$
- $${a}_{4} = 3{a}_{3}-{a}_{2} = 15-3 = 12$$
- $${a}_{5}=3{a}_{4}-{a}_{3} = 36-5 = 31$$
Question 8
1 / -0

For all numbers a and b, let $$\displaystyle a\bigodot b$$ be defined by $$\displaystyle a\bigodot b=ab+a+b$$. Then for the numbers $$x$$, $$y$$ and $$z$$, which of the following is/are true?
I. $$\displaystyle x\bigodot y=y\bigodot x$$
II. $$\displaystyle \left( x-1 \right) \bigodot \left( x+1 \right) =\left( x\bigodot x \right) -1$$
III. $$\displaystyle x\bigodot \left( y+z \right) =\left( x\bigodot y \right) +\left( x\bigodot z \right) $$

A
I only

B
II only

C
III only

D
I and II only

E
I, II, and III

Solution

$$For(I)$$
$$x\odot y=y\odot x$$
$$x\odot y=xy+x+y\Rightarrow x\odot y=y\odot x$$
$$y\odot x=yx+y+x$$

$$For(II)$$
$$\left( x-1 \right) \odot \left( x+1 \right)$$
$$=(x-1)(x+1)+x-1+x+1$$
$$={ x }^{ 2 }-1+2x$$
$$\left( x\odot x \right) -1=x\times x+x+x-1$$
$${ x }^{ 2 }+2x-1$$
$$\therefore \left( x-1 \right) \odot \left( x+1 \right) =\left( x\odot x \right) -1$$

$$For(III)$$
$$x\odot \left( y+z \right) $$
$$=x\left( y+z \right) +x+y+z$$
$$=xy+xz+x+y+z$$
$$x\odot y=xy+x+y$$
$$x\odot z=xz+x+z$$
$$x\odot y+x\odot z=xy+xz+x+y+x+z$$
$$=xy+xz+2x+y+z$$
$$\Rightarrow x\odot \left( y+z \right) \neq x\odot y+x\odot z$$

$$\therefore (I)\& (II)$$ are true.
Question 9
1 / -0

The sum of $$1$$ st $$n$$ terms of the series
$$\cfrac { { 1 }^{ 2 } }{ 1 } +\cfrac { { 1 }^{ 2 }+{ 2 }^{ 2 } }{ 1+2 } +\cfrac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 1+2+3 } +........$$

A
$$\cfrac { n+2 }{ 3 } $$

B
$$\cfrac {n( n+2 )}{ 3 } $$

C
$$\cfrac {n( n-2 )}{ 3 } $$

D
$$\cfrac {n( n-2 )}{ 6 } $$

Solution

General term of the given series is, $$T_r=\dfrac{1^2+2^2+....+r^2}{1+2+.....+r}=\dfrac{\frac{r(r+1)(2r+1)}{6}}{\frac{r(r+1)}{2}}=\dfrac{2r+1}{3}$$

Hence required sum $$\displaystyle =\sum_{r=1}^nT_r=\dfrac{1}{3}\sum_{r=1}^n(2r+1)$$
$$\displaystyle=\dfrac{1}{3} \left(2\sum_{r=1}^n r+\sum_{r=1}^n 1 \right)=\dfrac{1}{3}\left(2\cdot \dfrac{n(n+1)}{2} +n\right)=\dfrac{1}{3}(n^2+n+n)=\dfrac{n(n+2)}{3}$$
Question 10
1 / -0

The sum of the series,
$$\displaystyle \frac{1}{2.3}\cdot 2+ \frac{2}{3.4}\cdot 2^2+\frac{3}{4.5}\cdot 2^3+ ......$$ to n terms is _____.

A
$$\dfrac{2^{n+1}}{n+2}+1$$

B
$$\dfrac{2^{n+1}}{n+2}-1$$

C
$$\dfrac{2^{n+1}}{n+2}+2$$

D
$$\dfrac{2^{n+1}}{n+2}-2$$

Solution

Let $$S = \displaystyle \frac{1}{2.3}\cdot 2+ \frac{2}{3.4}\cdot 2^2+\frac{3}{4.5}\cdot 2^3+ ......$$

General term is given by, $$T_r = \dfrac{r}{(r+1)(r+2)}\cdot 2^r=\dfrac{r\cdot 2^r}{(r+1)(r+2)}=\dfrac{2(r+1)-(r+2)}{(r+1)(r+2)}\cdot 2^r=\dfrac{(r+1)2^{r+1}-(r+2)2^r}{(r+1)(r+2)}$$

$$\Rightarrow T_r = \dfrac{2^{r+1}}{r+2}-\dfrac{2^r}{r+1}$$, split the terms

Therefore the required sum $$=\displaystyle \sum_{r=1}^nT_r=T_1+T_2+T_3+...........+T_n$$

$$\quad =\left( \dfrac{2^2}{3}-\dfrac{2}{2}\right)+\left( \dfrac{2^3}{4}-\dfrac{2^2}{3}\right)+\left(\dfrac{2^4}{5}-\dfrac{2^3}{4} \right)+..............+\left(\dfrac{2^{n+1}}{n+2} -\dfrac{2^n}{n+1}\right)$$

$$=-\dfrac{2}{2}+\dfrac{2^{n+1}}{n+2}=\dfrac{2^{n+1}}{n+2}-1$$, all the remaining terms will get cancelled

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Sequences and Series Test - 27

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Sequences and Series Test - 27

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SHARING IS CARING

Question 1

$$25$$

$$20$$

$$27$$

$$18$$

Solution

Question 2

$$2$$

$$41$$

$$66$$

$$1$$

Solution

Question 3

$$37$$

$$38$$

$$39$$

$$40$$

$$41$$

Solution

Question 4

Add $$4$$ to the preceding number.

Take $$\displaystyle \frac { 1 }{ 2 } $$ of the preceding number and then add $$7$$ to that result.

Double the preceding number and then subtract $$2$$ from that result.

Subtract $$2$$ from the preceding number and then double that result.

Triple the preceding number and then subtract $$8$$ from that result.

Solution

Question 5

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

$$\dfrac{1}{6}(n^2-1)(2n-1)(2n+3)$$

$$\dfrac{1}{8}(n^2+1)(n^2+5)$$

$$\dfrac{1}{4}n(n+1)(n+2)(n+3)$$

Solution

Question 6

$$-26$$

$$-15$$

$$45$$

$$75$$

$$120$$

Solution

Question 7

$$12$$

$$23$$

$$25$$

$$31$$

$$36$$

Solution

Question 8

I only

II only

III only

I and II only

I, II, and III

Solution

Question 9

$$\cfrac { n+2 }{ 3 } $$

$$\cfrac {n( n+2 )}{ 3 } $$

$$\cfrac {n( n-2 )}{ 3 } $$

$$\cfrac {n( n-2 )}{ 6 } $$

Solution

Question 10

$$\dfrac{2^{n+1}}{n+2}+1$$

$$\dfrac{2^{n+1}}{n+2}-1$$

$$\dfrac{2^{n+1}}{n+2}+2$$

$$\dfrac{2^{n+1}}{n+2}-2$$

Solution

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