Sequences and Series Test

Result

Sequences and Series Test - 49

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 the sum of first $$15$$ terms of the sequence whose $${n}^{th}$$ term is $$3+4n$$.

A
$$525$$

B
$$425$$

C
$$495$$

D
$$None\,of\,thes$$

Solution

$$n^{th}$$ term = 3+4 n
$$1^{st}$$ term = 3+4(1)=7
$$2^{nd}$$ term = 3+4(2)=11
$$3^{rd}$$ term 3+4(3)=15
It is an AP with a=7, d=4
$$S_{15}=\frac{15}{2}[2(7)+14(4)]$$
$$= 15[7+28]$$
$$= 525$$
Question 2
1 / -0

Let $$\sum _{ r=1 }^{ n }{ { r }^{ 4 } } =f\left( n \right) ,$$ then $$\sum _{ r=1 }^{ n }{ { \left( 2r-1 \right) }^{ 4 } } $$ is equal to

A
$$f\left( 2n \right) -16f\left( n \right) $$

B
$$f\left( 2n \right) +7f\left( n \right) $$

C
$$f\left( 2n+1 \right) 8f\left( n \right) $$

D
none of these

Solution
Question 3
1 / -0

solve

A
$$225$$

B
$$175$$

C
$$335$$

D
$$350$$

Solution

Pattens is as follows:-

$$ (20 - 9)^{2} = 121 , $$ $$(24 - 11)^{2} = 169 $$

So $$ (28 - 13)^{2} = 15^{2} = 225$$

Hence missing place = 225
Question 4
1 / -0

$$\displaystyle\sum _{ r=1 }^{ n }{ \frac { r }{ { r }^{ 4 }+{ r }^{ 2 }+1 } } $$ is equal to

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

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

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

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

Solution
Question 5
1 / -0

$$C_1 + 2C_2 + 3C_3 + ...... + nC_n$$ is equal to

A
$$2^{n-1}$$

B
$$2^{n +1}$$

C
$$n. 2^{n -1}$$

D
$$n. 2^{n +1}$$

Solution

Given that,

$$ {{C}_{1}}+2.{{C}_{2}}+3.{{C}_{3}}+.............n{{C}_{n}} $$

$$ =n+2\times \dfrac{n\left( n-1 \right)}{2!}+3\times \dfrac{n\left( n-2 \right)\left( n-3 \right)}{3!}.......n\times 1 $$

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

$$ =n[1+\dfrac{\left( n-1 \right)}{1}+\dfrac{\left( n-2 \right)\left( n-3 \right)}{2}.......1 $$

Put, n-1=N

$$ =n[1+\dfrac{N}{1}+\dfrac{\left( N+1 \right)\left( N-1 \right)}{2}.......1 $$

$$ =n\left( {}^{N}{{C}_{0}}+{}^{N}{{C}_{1}}+{}^{N}{{C}_{2}}...............{}^{N}{{C}_{N}} \right) $$

$$ =n{{.2}^{N}} $$

$$ =n{{.2}^{n-1}} $$

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

If in a series $$t_n=\dfrac{n+1}{(n+2)!}$$ then $$\displaystyle\sum^{10}_{n=0}t_n$$ is equal to?

A
$$1-\dfrac{1}{10!}$$

B
$$1-\dfrac{1}{11!}$$

C
$$1-\dfrac{1}{12!}$$

D
None of these

Solution

$$ \begin{array}{l} \text { Solution - It is given that } \\ \qquad \begin{aligned} t_{n} &=\frac{n+1}{(n+2) !} \\ t n &=\frac{n+1+1-1}{(n+2) !}=\frac{(n+2)-1}{(n+2) !} \\ & \text { th }=\frac{n+2}{(n+2) !}-\frac{1}{(n+2) !} \end{aligned} \end{array} $$

$$ t n=\frac{1}{(n+1) !}-\frac{1}{(n+2) !} $$ $$ \sum_{n=0}^{10} t_{n}=\sum_{n=0}^{10}\left(\frac{1}{(n+1) !}-\frac{1}{(n+2) !}\right) $$

$$ \begin{aligned} &=\frac{1}{1}\left(-\frac{1}{2 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{3 !}-\frac{1}{4 !} \cdots \cdots-\frac{1}{11 !}+\frac{1}{11 !}\right)-\frac{1}{12 !} \\ &=1-\frac{1}{12 !} \\ \text { Hence, } &(\mathrm{C}) \text { is the correct option. } \end{aligned} $$
Question 7
1 / -0

insert the missing number: 5,8,12,17,23,___,38

A
$$29$$

B
$$30$$

C
$$32$$

D
$$25$$

Solution

$$5,8,12,17,23,......38$$
$$5+3=8;8+4=12,12+5=17,17+6=23$$
$$23+7=30,30+8=30$$
$$\therefore 30$$
Question 8
1 / -0

There is a certain relationship between the pair of figure on either side of ::. Identify the relationship on the left side and find the missing figure.

A

B

C

D

Solution

As per the first relation, in the first figure, there is a square on which there is 3 circle and in the second figure, there is a circle on which there is $$3$$ square.

Following the same in second relation, the correct option is $$B$$.
Question 9
1 / -0

Rahul told Anand, "Yesterday I defeated the only brother of the daughter of my grandmother." Whom did Rahul defeat ?

A
Son

B
Father

C
Brother

D
Cousin

Solution

Daughter of Grandmother $$\rightarrow$$ Aunt Aunt's only brother $$\rightarrow$$ Father
Question 10
1 / -0

Which number replaces the question mark ?

A
10

B
12

C
14

D
16

Solution

In each triangle, the central value equals the average of the 3 values around the outside.

Hence, the required value = $$\dfrac{15+17+4}{3}=12$$

Therefore, option B is correct.