Sequences and Series Test

Result

Sequences and Series Test - 24

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

There is no symmetry in symmetry in legs of Fig

A

B

C

D

E

Solution

There is no symmetry in legs of Fig.(e)
Question 2
1 / -0

The value of $$1^2+2^2+3^2+\cdots+ n^2$$ is

A
$$\displaystyle\frac{n(n+1)}{2}$$

B
$$\displaystyle\frac{n(n+1)(2n+1)}{6}$$

C
$$\displaystyle\frac{n(n+1)(2n-1)}{6}$$

D
$$\displaystyle\left[\frac{n(n+1)}{2}\right]^2$$

Solution

Let $${ S }_{ n }={ 1 }^{ 2 }+{ 2 }^{ 2 }+\dots +{ n }^{ 2 }$$

Consider the identity $${ k }^{ 3 }-{ \left( k-1 \right) }^{ 3 }=3{ k }^{ 2 }-3k+1$$

Putting $$k=1,2,....,n$$ successively, we obtain

$${ 1 }^{ 3 }-{ \left( 0 \right) }^{ 3 }=3{ \left( 1 \right) }^{ 2 }-3\left( 1 \right) +1$$

$${ 2 }^{ 3 }-{ \left( 1 \right) }^{ 3 }=3{ \left( 2 \right) }^{ 2 }-3\left( 2 \right) +1$$

$${ 3 }^{ 3 }-{ \left( 2 \right) }^{ 3 }=3{ \left( 3 \right) }^{ 2 }-3\left( 3 \right) +1$$
$$\vdots $$
$${ n }^{ 3 }-{ \left( n-1 \right) }^{ 3 }=3{ \left( n \right) }^{ 2 }-3\left( n \right) +1$$

Adding both sides we get,

$${ n }^{ 3 }-{ \left( 0 \right) }^{ 3 }=3{ \left( 1^{ 2 }+2^{ 2 }+\dots n^{ 2 } \right) }-3\left( 1+2+\dots +n \right) +n$$

$${ n }^{ 3 }=3{ \sum _{ k=1 }^{ n }{ k^{ 2 } } }-3\sum _{ k=1 }^{ n }{ k+n } $$

Since

$$\sum _{ k=1 }^{ n }{ k } =1+2+\dots +n=\dfrac { n\left( n+1 \right) }{ 2 } $$

Therefore,

$${ S }_{ n }={ \sum _{ k=1 }^{ n }{ k^{ 2 } } }=\dfrac { 1 }{ 3 } \left[ { n }^{ 3 }+\dfrac { 3n\left( n+1 \right) }{ 2 } -n \right] $$

$${ S }_{ n }={ \sum _{ k=1 }^{ n }{ k^{ 2 } } }=\dfrac { 1 }{ 6 } \left( { 2n }^{ 3 }+{ 3n }^{ 2 }+n \right) $$

$${ S }_{ n }=\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } $$

Hence option B is correct.
Question 3
1 / -0

The product $$ \displaystyle \left ( 1-\frac{1}{n} \right )\left ( 1-\frac{1}{n+1} \right )\left ( 1-\frac{1}{n+2} \right )..\left ( 1-\frac{1}{2n} \right ) $$ Equals

A
$$ \displaystyle \frac{n-1}{2n} $$

B
$$ \displaystyle \frac{1}{2n} $$

C
$$ \displaystyle \frac{2n}{n-1} $$

D
$$ \displaystyle \frac{1}{n} $$

Solution

Given product
$$\displaystyle =\left ( 1-\dfrac{1}{n} \right )\left ( 1-\dfrac{1}{n+1} \right )\left ( 1-\dfrac{1}{n+2} \right )\cdot \cdot \cdot \left ( 1-\dfrac{1}{2n} \right )$$
$$\displaystyle =\dfrac{\left ( n-1 \right )}{n}\times \dfrac{n}{\left ( n+1 \right )}\times \dfrac{n+1}{\left ( n+2 \right )}\cdot \cdot \cdot \left ( \dfrac{2n-1}{2n} \right )=\dfrac{\left ( n-1 \right )}{2n}$$
Question 4
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?)
$$0,\,3,\,9,\,18,\,30,\,?$$

A
$$48$$

B
$$42$$

C
$$36$$

D
$$45$$

Solution

The difference between the consecutive numbers increases by $$3$$.
$$3-0 = 3$$
$$9 - 3 = 6$$
$$18-9 = 9$$
$$30-18 = 12$$.
Pattern is adding multiples of 3
Next term is $$30+15=45$$
Question 5
1 / -0

In a certain language '$$ \displaystyle +\div ? $$' means 'where are you' , '$$ \displaystyle @-\div $$' means ' we are here' , and '$$ \displaystyle +@\times $$' means ' you come here' What is the code for ' Where' ?

A
+

B
$$ \displaystyle \div $$

C
?

D
@

Solution

s.no code sentence
1. $$ \displaystyle '+\div ?' $$ Where are you
2. $$ \displaystyle '@-\div ' $$ We are here
3. $$ \displaystyle '+@\times $$ you come here

As we can see that where is only in sentence $$1$$ [Where is the word for which we have to find the code ] Therefore we need to gather the codes for $$are$$ & $$you$$ to find out the code for where sentence 1 & 2 have the word $$are$$ in common and the symbol $$ \displaystyle \div $$ in common Therefore $$ \displaystyle \div $$ is the symbol for $$are$$
Sentence 1 & 3 have the word you in common and the symbol + in common therefore + stands for $$you$$

thus, leaving only ?, which stands for $$where$$
Question 6
1 / -0

Directions For Questions

In a certain code, ' il be pee ' means 'roses are blue', silk hee means ' red flowers' and ' pee mit hee' means ' flowers are vegctables'

...view full instructions

How is 'red written in that code ?

A
hee

B
silk

C
be

D
cannot be determined

Solution

Type Direct coding (Jumbled fashion)
S.no. code sentence
1. il be pee roses are blue
2. silk hee red flowers
3. pee mit hee Flowers are Vegetables
common word in sentences 1 & 3 $$ \displaystyle \rightarrow $$ 'are' and code $$ \displaystyle \rightarrow $$ 'pee'
common word in sentences 2 & 3 $$ \displaystyle \rightarrow $$ flowers and code $$ \displaystyle \rightarrow $$ 'hee'
Therefore
Codes words
pee are
hee flowers
silk red
niit vegetables
il roses/ blue
be blue./ rose
Question 7
1 / -0

7th term of $$\displaystyle\frac{1}{2}$$, $$\displaystyle\frac{1}{4}$$, $$\displaystyle\frac{1}{6}$$ $$\dots\dots$$ is

A
$$\displaystyle\frac{1}{10}$$

B
$$\displaystyle\frac{1}{12}$$

C
$$\displaystyle\frac{1}{14}$$

D
$$\displaystyle\frac{1}{16}$$

Solution

We can see that the series has terms such that $$ \dfrac {1}{2(1)}, \dfrac {1}{2(2)}, \dfrac {1}{2(3)}..... $$

So, the $$7th$$ term will be $$ \dfrac {1}{2(7)} = \dfrac {1}{14} $$
Question 8
1 / -0

Find the sum to 90 terms of the series $$5+55+555+\cdots\cdots$$

A
$$\displaystyle\frac{50}{81}\left[10^{90}-82\right]$$

B
$$\displaystyle\frac{50}{81}\left[10^{90}-83\right]$$

C
$$\displaystyle\frac{50}{81}\left[10^{90}-80\right]$$

D
$$\displaystyle\frac{50}{81}\left[10^{90}-90\right]$$

Solution

Given
$$5+55+555+5555+........+\underset{n fives}{\underbrace{55......5}}$$

$$=\dfrac{5}{9}(9+99+999+9999+.......+\underset{n \, nines}{\underbrace{99......9}}$$

$$=\dfrac{5}{9}(10^1-1+10^2-1+10^3-1+.......+10^n-1)$$

$$=\dfrac{5}{9}(10^1+10^2+10^3+......+10^n-n)$$

$$=\dfrac{5}{9}\left(\dfrac{10^{n+1}-10}{9}-n\right)$$

Asked for $$90$$ terms

$$n=90$$

$$=\dfrac{5}{9}\left(\dfrac{10^{91}-10}{9}-90\right)$$

$$=\dfrac{5}{9}\left(\dfrac{10(10^{90}-1)}{9}-90\right)$$

$$=\dfrac{50}{9}\left(\dfrac{10^{90}-1}{9}-9\right)$$

$$=\dfrac{50}{9}\left(\dfrac{10^{90}-1-81}{9}\right)$$

$$=\dfrac{50}{81}(10^{90}-82)$$.
Question 9
1 / -0

How is 'vegetables are red flowers' written in this code ?

A
Pec silk mit hee

B
silk peehee be

C
il silk mit hee

D
none

Solution

Type Direct coding (Jumbled fashion)
S.no. code sentence
1. il be pee roses are blue
2. silk hee red flowers
3. pee mit hee Flowers are Vegetables
common word in sentences 1 & 3 $$ \displaystyle \rightarrow $$ 'are' and code $$ \displaystyle \rightarrow $$ 'pee'
common word in sentences 2 & 3 $$ \displaystyle \rightarrow $$ flowers and code $$ \displaystyle \rightarrow $$ 'hee'
Therefore
Codes words
pee are
hee flowers
silk red
niit vegetables
il roses/ blue
be blue./ rose
nit silk hee pee
Question 10
1 / -0

In a series, $$T_n=2n+5$$, find $$S_4$$.

A
40

B
30

C
20

D
10

Solution

$$T_n=2n+5$$
$$T_1=2(1)+5=7$$;
Similarly $$T_2=9$$, $$T_3=11$$ and $$T_4=13$$
$$\therefore S_4=T_1+T_2+T_3+T_4=40$$.