Sequences and Series Test 25

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Sequences and Series Test 25

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Question 1
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 2
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 3
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 4
1 / -0

In the $$xy$$-plane, the vertices of a triangle are $$(-1,3), (6,3)$$ and $$(-1,-4)$$. The area of the triangle is ___ square units.

A
$$10$$

B
$$17.5$$

C
$$24.5$$

D
$$35$$

Solution

If $$(x_1,y_1)$$,$$(x_2,y_2)$$ and $$(x_3,y_3)$$ are the vertices of a triangle then its area is given by,
$$A=\left| \dfrac { 1 }{ 2 } (x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 })) \right| $$
Therefore, with the vertices $$(-1,3)$$, $$(6,3)$$ and $$(-1,-4)$$.
Area of triangle is given by,
$$A=\left| \dfrac { 1 }{ 2 } (-1(3-(-4))+6(-4-3)+(-1)(3-3)) \right|$$
$$ =\left| \dfrac { 1 }{ 2 } (-7-42) \right| $$
$$=\left| \dfrac { 1 }{ 2 } (-49) \right| $$
$$=\left| -24.5 \right|$$
$$ =24.5$$ square units.
Question 5
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 6
1 / -0

Let $$\boxed { n }$$ be defined as $$\frac{(n+2)!}{(n-1)!}$$, what is the value of $$\frac{\boxed{7}}{\boxed {3}}$$ ?

A
4.4

B
8.4

C
12.4

D
16.4

E
20.4

Solution

Given, box $$n$$ is equal to $$\dfrac{(n+2)!}{(n-1)!}=(n+2)(n+1)n$$
We get box $$7$$ is equal to $$9 \times 8 \times 7 =504$$
We get box $$3$$ is equal to $$5 \times 4 \times 3 = 60$$
Therefore, If we divide those two, we get $$\dfrac{504}{60} = 8.4$$.
Question 7
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 8
1 / -0

In fig., the area of triangle ABC (in sq. units) is:

A
$$15$$

B
$$10$$

C
$$7.5$$

D
$$2.5$$

Solution

Given: Coordinates of Point $$A (1,3) ,B (-1,0)$$ and $$C (4,0)$$
Construction: Drop a perpendicular from $$A$$ on $$x-$$ axis, which meets x-axis at $$D\equiv(1,0)$$
Now in $$\Delta ADC, AD = 3, DC = 3$$
Area of $$\Delta ADC = \dfrac12\times DC\times AD$$
$$= \dfrac12\times3\times3 = \dfrac92 \ cm^2$$

Now in $$\Delta ADB, AD = 3, DB = 2$$
Area of $$\Delta ADB = \dfrac12\times DB\times AD$$
$$= \dfrac12\times2\times3 = 3 \ cm^2$$

Area of $$\Delta ABC =$$ Ara of $$\Delta ADC + $$ Area of $$\Delta ABD$$
$$ = \dfrac92 + 3 = \dfrac{15}2 = 7.5\ cm^2$$
Question 9
1 / -0

Area of the triangle formed by the points $$\left( 0,0 \right) ,\left( 2,0 \right) $$ and $$\left( 0,2 \right) $$ is

A
$$1$$ sq. unit

B
$$2$$ sq. units

C
$$4$$ sq. units

D
$$8$$ sq. units

Solution

Given: $$A=(x_1,y_1)=(0,0)$$

$$B=(x_2,y_2)=(2,0)$$ and

  $$C=(x_3,y_3)=(0,2)$$

Area of triangle $$=\dfrac {1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)++x_3(y_1-y_2)]$$

$$=\dfrac {1}{2}[0(0-2)+2(2-0)+0(0-0)]$$

         $$=\dfrac {1}{2}[2(2)]$$

         $$=2$$ sq. units.
Question 10
1 / -0

Consider an incomplete pyramid of balls on a square base having $$18$$ layers; and having $$13$$ balls on each side of the top layer. Then the total number $$N$$ of balls in that pyramid satisfies

A
$$9000 < N < 10000$$

B
$$8000 < N < 9000$$

C
$$7000 < N < 8000$$

D
$$10000 < N < 12000$$

Solution

The top layer has $$(13\times 13)$$ balls the layer below it will have $$(14\times 14)$$ balls

We have $$18$$ layers

So the total number of balls

$$N=(13\times 13)+(14\times 14)+.......(30\times 30)\\ N={ 13 }^{ 2 }+{ 14 }^{ 2 }+.....{ 30 }^{ 2 }$$

Sum of squares of first $$n$$ natural numbers is $$\dfrac { n(n+1)(2n+1) }{ 6 } \\ $$

$$\therefore N=$$ sum of first $$30$$ $$-$$ sum of first $$12$$

$$N=\dfrac { 30\times 31\times 61 }{ 6 } -\dfrac { 12\times 13\times 25 }{ 6 } \\ N=8805\\ $$

$$\Rightarrow 8000<N<9000$$
Hence, option B is correct.

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Sequences and Series Test 25

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Sequences and Series Test 25

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Question 1

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 2

$$-26$$

$$-15$$

$$45$$

$$75$$

$$120$$

Solution

Question 3

I only

II only

III only

I and II only

I, II, and III

Solution

Question 4

$$10$$

$$17.5$$

$$24.5$$

$$35$$

Solution

Question 5

$$25$$

$$20$$

$$27$$

$$18$$

Solution

Question 6

4.4

8.4

12.4

16.4

20.4

Solution

Question 7

$$2$$

$$41$$

$$66$$

$$1$$

Solution

Question 8

$$15$$

$$10$$

$$7.5$$

$$2.5$$

Solution

Question 9

$$1$$ sq. unit

$$2$$ sq. units

$$4$$ sq. units

$$8$$ sq. units

Solution

Question 10

$$9000 < N < 10000$$

$$8000 < N < 9000$$

$$7000 < N < 8000$$

$$10000 < N < 12000$$

Solution

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