Sequences and Series Test 27

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Sequences and Series Test 27

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Question 1
1 / -0

If $$\alpha ,\beta , \gamma $$ are three consecutive integers. If these integers are raised to first, second and third positive powers respectively, and added then they form a perfect square, the square root of which is equal to the sum of these integers. Also, $$\alpha < \beta < \gamma $$. Then, $$\gamma$$ is equals to:

A
$$3$$

B
$$14$$

C
$$5$$

D
$$11$$

Solution

Let the numbers be $$n-1,n,n+1$$
As per the given information, we have
$$(n-1)+{ n }^{ 2 }+{ (n+1) }^{ 3 }={ (n-1+n+n+1) }^{ 2 }$$
$$\Rightarrow { n }^{ 3 }-5{ n }^{ 2 }+4n=0$$
$$\Rightarrow n=0,1,4$$
For $$n=0,$$ the numbers are: $$-1,0,1$$ this is out as all the numbers should be positive
For $$n=1,$$ the numbers are: $$0,1,2$$ this is out as all the numbers should be positive (‘0’ can’t be taken as positive)
For $$n=4,$$ the numbers are: $$3,4,5$$
We have got largest number which is $$5$$
Hence, the value of $$\gamma$$ is $$5$$.
Question 2
1 / -0

Find the $$4^{th}$$ term of $${ \left( 9x-\cfrac { 1 }{ 3\sqrt { x } } \right) }^{ 18 }$$.

A
$$16500$$

B
$$18564$$

C
$$16540$$

D
$$32600$$

Solution

We know that the $$r^{th}$$ term in expansion of $$(a+b)^n$$ is given by $$^nC_{r-1}a^{n+1-r}b^{r-1}$$
Here $$a=9x, \ n=18 \ \And \ b=-\cfrac{1}{3\sqrt{x}}$$
$$\therefore$$ the fourth term in expansion of $$\left (9x-\cfrac{1}{3\sqrt{x}}\right)^{18}$$ is $$^{18}C_{12} {x}^{6}\left (-\cfrac{1}{3\sqrt{x}}\right)^{12} =18564$$
Question 3
1 / -0

Find the area (in square units) of the triangle whose vertices are $$(a, b+c), (a, b-c) $$ and $$(-a, c). $$

A
$$2ac$$

B
$$2bc$$

C
$$b(a+c)$$

D
$$c(a-h)$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times |[ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] |$$
Given the vertices of triangle,
$$A (a,b+c) ,B(a,b-c) ,C(-a,c)$$
Therefore, area is given by
$$=\dfrac { 1 }{ 2 } |\left[ a\left[ b-c-c \right] +a\left[ c-b-c \right] +(-a)\left[ b+c-b+c \right] \right]| \\ =\dfrac { 1 }{ 2 } |\left[ a(b-2c)+a(-b)-a(2c) \right]| \\ =\dfrac { 1 }{ 2 } |\left[ ab-2ac-ab-2ac \right]| =\left| \dfrac { -4ac }{ 2 } \right| =2ac$$
Question 4
1 / -0

Find the missing number in the circle:

A
$$66$$

B
$$72$$

C
$$71$$

D
$$78$$

Solution

$$3\times 2 - 1, 5\times 2 - 1, 9\times 2 - 1, 17\times 2 - 1 + 39 = 72$$.

$$\therefore$$ The solution is $$72$$.
Question 5
1 / -0

13, 74, 290, 650,.......

A
$$1248$$

B
$$1370$$

C
$$1346$$

D
$$1452$$

E
$$1625$$

Solution

Here we observe the first term 13 can be written as: $$2^2+3^2$$
Similarly other numbers can be written as:
$$74: 5^2+7^2$$
$$290: 11^2+13^2$$
$$650: 17^2+19^2$$
Here, we observe that all above term of series can be written as sum of squares of two prime numbers which comes next to it's previous pair.
Like: $$(2,3);(5,7);(11,13);(17,19)$$ So, other pair of prime number will be $$ (23,29).$$
Hence next term of given series will be:
$$\Rightarrow 23^2+29^2=1370$$
So, correct answer is $$1370$$.
Question 6
1 / -0

An alphabet contains a $$A^{'s}$$ and b $$ B^{'s}$$ . (In all a+b letters ). The number of words each containing all the $$ A^{'s}$$ and any number of $$ B^{'s}$$, is

A
$$^{a+b}C_b$$

B
$$^{a+b+1}C_a$$

C
$$^{a+b+1}C_b$$

D
none of these

Solution
Question 7
1 / -0

The maximum number of intersection points of n circles and n straight lines , among themselves is 80.The value of n is

A
$$7$$

B
$$6$$

C
$$5$$

D
$$8$$

Solution
Question 8
1 / -0

Figures $$1$$ and $$2$$ are related in a particular manner. Establish the same relationship between figures $$3$$ and $$4$$ by choosing a figure from amongst the options.

A

B

C

D

Solution

Hexagon is rotating in anti-clockwise direction two times to get the next figure. (1 to 2).
Same for getting the figure (3 to 4) rotate the figure in an anti-clockwise direction (two times) we will get the desired figure.
Question 9
1 / -0

Select the INCORRECT match

A
$$3249-MMMCCXLIX$$

B
$$1667-MDCLXVII$$

C
$$207-CCXVII$$

D
$$499-CDXCIX$$

Solution

(a) $$MMMCCXLIX=3000+200+40+9=3249$$
(b) $$MDCLXVII=1000+500+100+60+7=1667$$
(c) $$CCXVII=217=200+17$$
(d) $$CDXCIX=400+90+9=499$$
Hence the correct match is option C.
Question 10
1 / -0

If D (3, -1), E (2, 6) and F (5, 7) are the vettices of the sides of $$\Delta DEF$$, the area of triangle DEF is sq. units.

A
$$11$$

B
$$22$$

C
$$48$$

D
$$50$$

Solution

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

Area of triangle $$DEF=|\dfrac{1}{2}(3(6-7)+2(7+1)+5(-1-6))|$$

$$=|\dfrac{1}{2}(-3+16-35)|$$

$$=11$$ sq. Units.