Sequences and Series Test

Result

Sequences and Series Test - 42

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Find: $$6,25,62,123,(?),341$$

A
$$216$$

B
$$214$$

C
$$215$$

D
$$218$$

Solution

$$6=2^{3}-2=6$$
$$25=3^{3}-2=25$$
$$62=4^{3}-2=62$$
$$123=5^{3}-2=123$$
$$214=6^{3}-2=214$$
$$341=7^3-2=341$$

Missing term is $$214$$
Question 2
1 / -0

Find the missing factor among the given figure ?

A
$$4$$

B
$$6$$

C
$$8$$

D
$$12$$

Solution
Question 3
1 / -0

The numbers $${\log _{180}}12,{\log _{2160}}12,{\log _{25920}}12$$ are in

A
AP

B
GP

C
HP

D
None of these

Solution
Question 4
1 / -0

If $$[x]$$ denotes the greates integer $$\le x$$, then $$\left[\dfrac{2}{3} \right] + \left[\dfrac{2}{3} + \dfrac{1}{99} \right] + \left[\dfrac{2}{3} + \dfrac{2}{99} \right] + .... + \left[\dfrac{2}{3} + \dfrac{98}{99} \right] =$$

A
$$99$$

B
$$98$$

C
$$66$$

D
$$65$$

Solution

$$\left [ \cfrac{2}{3} \right ]=0$$
$$\left [ \cfrac{2}{3}+\cfrac{1}{99} \right ]=0$$
$$\left [ \cfrac{2}{3}+\cfrac{2}{99} \right ]=0$$
$$...$$
$$...$$
$$\left [ \cfrac{2}{3}+\cfrac{33}{99} \right ]=\left [ \cfrac{2}{3}+\cfrac{1}{3} \right ]=\left [ 1 \right ]=1$$
$$\left [ \cfrac{2}{3}+\cfrac{34}{99} \right ]=\left [ 1+\cfrac{1}{99} \right ]=1$$
$$\left [ \cfrac{2}{3}+\cfrac{98}{99} \right ]=\left [ 1+\cfrac{65}{99} \right ]=1$$

So, the sum is
$$1+1+1+......... 66 $$ times
$$=66$$
Question 5
1 / -0

$$\sum _{ r=0 }^{ n }{ \dfrac { { 2 }^{ r+2 }\cdot ^{ n }C_{ r } }{ (r+1)(r+2) } } $$ is equal to

A
$$\dfrac{3^{n+2}-2n+5}{\left(n+1\right)\left(n+2\right)}$$

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

C
$$\dfrac{3^{n+2}-2n-5}{\left(n+1\right)\left(n+2\right)}$$

D
None of these
Question 6
1 / -0

If $$S={1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}....$$ upto $$n$$ terms and $$n$$ is even, then $$S$$ equals _____

A
$$\cfrac{n(n+1)}{2}$$

B
$$\cfrac{n(n-1)}{2}$$

C
$$\cfrac{-n(n+1)}{2}$$

D
$$\cfrac{-n(n-1)}{2}$$

Solution

$$S=1^{2}-2^{2}+3^{2}-4^{2}.........$$ upto $$n$$ term
$$=(1-2)(1+2)+(3-4)(3+4)+......+(n-1-n)(n-1+n)$$
$$=-(1+2+3+4+.....+n)$$
$$S=\dfrac{-n(n+1)}{2}$$
Question 7
1 / -0

The value of $$\sum\limits_{n = 2}^\infty {\left( {1 - \frac{1}{{{n^2}}}} \right)} $$ equals

A
$$-ln 3$$

B
$$0$$

C
$$-ln 2$$

D
$$-ln 5$$
Question 8
1 / -0

If $$x$$ and $$y$$ are the number of possibilities that $$A$$ can assume such that the unit digit of A and $$A^3$$ are same and the unit digit of $$A^2$$ and $$A^3$$ are same respectively ,then the value of $$x-y$$ is (where $$A$$ is a single digit number)

A
$$4$$

B
$$2$$

C
$$3$$

D
$$5$$

Solution

Here, according to the question
if $$A=1$$, then $${ A }^{ 3 }={ 1 }$$,
We see that unit digit number of $$A$$ and $${ A }^{ 3 }$$ are same.
for, $$A=2\Rightarrow { A }^{ 3 }=8,$$, unit digits are not same
$$A=3\Rightarrow { A }^{ 3 }=27,$$ unit digits are not same
$$A=4\Rightarrow { A }^{ 3 }=64,$$ unit digits are same
$$A=5\Rightarrow { A }^{ 3 }=125,$$ unit digits are same
$$A=6\Rightarrow { A }^{ 3 }=216,$$ unit digits are not same
$$A=7\Rightarrow { A }^{ 3 }=343,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 3 }=512,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 3 }=512,$$ unit digits are same
So, there are five possible solutions where unit digits of $$A$$ and $${ A }^{ 3 }$$ are same.
$$\therefore x=5$$
Again, if we take $$A=1,{ A }^{ 2 }=1,{ A }^{ 3 }={ 1 }$$, unit digits are same
$$A=2\Rightarrow { A }^{ 2 }=4,{ A }^{ 3 }=8,$$ unit digits are not same
$$A=3\Rightarrow { A }^{ 2 }=9,{ A }^{ 3 }=27,$$ unit digits are not same
$$A=4\Rightarrow { A }^{ 2 }=16,{ A }^{ 3 }=64,$$ unit digits are not same
$$A=5\Rightarrow { A }^{ 2 }=25,{ A }^{ 3 }=125,$$ unit digits are same
$$A=6\Rightarrow { A }^{ 2 }=36,{ A }^{ 3 }=216,$$ unit digits are same
$$A=7\Rightarrow { A }^{ 2 }=49,{ A }^{ 3 }=343,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 2 }=64,{ A }^{ 3 }=512,$$ unit digits are not same
$$A=9\Rightarrow { A }^{ 2 }=81,{ A }^{ 3 }=729$$ unit digits are not same
Here, we can see that there are three possible solutions where unit digits of $${ A }^{ 2 },{ A }^{ 3 }$$ are same.
So, $$y=3$$
Hence, $$x-y=5-3=2.$$
Question 9
1 / -0

$$\begin{bmatrix} 12 & 47 & 21 \\ 10 & 52 & 4 \\ 64 & ? & 24 \end{bmatrix}$$

A
$$40$$

B
$$83$$

C
$$62$$

D
$$16$$

Solution

$$\begin{bmatrix} 12 & 47 & 21\\ 10 & 52 & 4\\ 64 & ? & 24\end{bmatrix}$$
We can see that dividing third number in each row with second digit of middle number and multiplying with first digit gives first member.
Similarly, out of $$4$$ options, dividing $$24$$ by $$3$$ and multiplying by $$8$$ gives $$64$$.
$$\Rightarrow$$ Option $$(B)$$.
Question 10
1 / -0

If $$(1+ax)^{n }=1+8x+24x^{2}+...;$$ then $$\cfrac {a-n}{a+n}$$ is equal to

A
$$3$$

B
$$\dfrac {-1}{3}$$

C
$$-3$$

D
$$\dfrac {1}{3}$$

Solution

$$(1+ax)^{ n }=1+n\left( ax \right) +^{ n }{ C }_{ 2 }{ \left( ax \right) }^{ 2 }+^{ n }{ C }_{ 3 }{ \left( ax \right) }^{ 3 }....$$
According to question
$$(1+ax)^{ n }=1+8x+24x^{ 2 }+...$$
$$\Rightarrow an=8$$
$${ a }^{ 2 }=\cfrac { (n)(n-1) }{ 2 } =24$$
$$\Rightarrow { a }^{ 2 }\left( n \right) \left( n-1 \right) =48$$
$$\Rightarrow a\left( n-1 \right) =6$$
$$\Rightarrow \cfrac { n }{ n-1 } =\cfrac { 8 }{ 6 } =\cfrac { 4 }{ 3 } $$
$$\Rightarrow n=4,\Rightarrow a=2$$
$$\therefore \cfrac { a-n }{ a+n } =\cfrac { 2-4 }{ 2+4 } =\cfrac { -2 }{ 6 } =\cfrac { -1 }{ 3 } $$