Sequences and Series Test

Result

Sequences and Series Test - 30

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

If $${ b }_{ i }=1-{ a }_{ i },na=\sum _{ i=1 }^{ n }{ { a }_{ i } } ,nb=\sum _{ i=1 }^{ n }{ { b }_{ i } } \quad $$, then $$\sum _{ i=1 }^{ n }{ { { a }_{ i }b }_{ i } } +\sum _{ i=1 }^{ n }{ { \left( { a }_{ i }-a \right) }^{ 2 } } =$$

A
$$ab$$

B
$$-nab$$

C
$$nab$$

D
$$(n+1)ab$$

Solution

$$a_1+b_1=1 \quad \quad na=a_1+a_2+a_3+...+a_n \quad (1) \\ a_2+b_2=1 \quad \quad nb=b_1+b_2+b_3+....+b_n \quad (2) \\ a_3+b_3=1 \\ ----- \\ \underline{a_n+b_n=1} \\ (a_1+a_2+a_3+...+a_n)+(b_1+b_2+b_3+....+b_n)=n \quad (3)$$
From $$(1),(2),(3)$$
$$na+nb=n \\ a+b=1 \quad \quad (4)$$
$${ \sum }_{ i=1 }^{ n }{ a }_{ i }{ b }_{ i }+{ \sum }_{ i=1 }^{ n }{ ({ a }_{ i }-a) }^{ 2 }\\ \Rightarrow { \sum }_{ i=1 }^{ n }{ a }_{ i }(1-{ a }_{ i })+{ \sum }_{ i=1 }^{ n }{ { a }_{ i } }^{ 2 }+{ \sum }_{ i=1 }^{ n }{ a }^{ 2 }-2{ \sum }_{ i=1 }^{ n }{ a }_{ i }a\\ \Rightarrow { \sum }_{ i=1 }^{ n }{ a }_{ i }-{ \sum }_{ i=1 }^{ n }{ { a }_{ i } }^{ 2 }+{ \sum }_{ i=1 }^{ n }{ { a }_{ i } }^{ 2 }+{ a }^{ 2 }{ \sum }_{ i=1 }^{ n }(1)-2a{ \sum }_{ i=1 }^{ n }{ a }_{ i }\\ =(1-2a){ \sum }_{ i=1 }^{ n }{ a }_{ i }+{ a }^{ 2 }{ \sum }_{ i=1 }^{ n }(1)\\ =(1-2a)(na)+{ a }^{ 2 }(n)\\ =na-2{ a }^{ 2 }n+{ a }^{ 2 }n\\ =na-{ a }^{ 2 }n\\ =na(1-a)\\ =nab\quad \quad from(4)$$
Question 2
1 / -0

The sum of the first $$n$$ terms of the series $${ 1 }^{ 2 }+2\cdot { 2 }^{ 2 }+{ 3 }^{ 3 }+2\cdot { 4 }^{ 2 }+{ 5 }^{ 2 }+2\cdot { 6 }^{ 2 }+\cdots $$ is $$\dfrac { n{ \left( n+1 \right) }^{ 2 } }{ 2 } $$ when $$n$$ is even, when $$n$$ is odd the sum is

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

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

C
$$\dfrac { n{ \left( n+1 \right) }^{ 2 } }{ 4 } $$

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

Solution

The sum of $$n$$ terms of given series $$=\dfrac { n{ \left( n+1 \right) }^{ 2 } }{ 2 } $$
Let $$n$$ is odd, i.e., $$n=2m+1$$
Then, $${ S }_{ 2m+1 }\equiv { S }_{ 2m }+\left( 2m+1 \right) $$th term
$$=\dfrac { \left( n-1 \right) { n }^{ 2 } }{ 2 } +n$$th term
$$=\dfrac { \left( n-1 \right) { n }^{ 2 } }{ 2 } +{ n }^{ 2 }$$
$$=\dfrac { \left( n+1 \right) { n }^{ 2 } }{ 2 } $$
Question 3
1 / -0

Directions For Questions

In the questions a number series is given with one term missing shown by question mark(?). This term is one of the four alternatives given under it. Find the right alternative.

...view full instructions

$$4,9,25,?,121,169$$

A
$$36$$

B
$$49$$

C
$$64$$

D
$$81$$

Solution

Given series is $$4,9,25,?,121, 168$$
It can be written as $${2}^{2},{3}^{2},{5}^{2},{7}^{2},{11}^{2},{13}^{2}$$
So, the missing term would be $$7^2$$ i.e. $$49$$.
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

If $$\displaystyle \sum_{r = 1}^{n}t_{n} = \dfrac {n(n +1)(n + 2)(n + 3)}{8}$$, then $$\displaystyle \sum_{r = 1}^{n} \dfrac {1}{t_{1}}$$ equals

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

B
$$\left (\dfrac {1}{(n + 1)(n + 2)} - \dfrac {1}{2}\right )$$

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

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

Solution
Question 6
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 7
1 / -0

Let $$r^{th} $$ term of a series is given by, $$T_r = \dfrac {r}{1-3r^2 + r^4} .$$
Then $$ \underset {n \rightarrow \infty}{\lim} \sum_{r=1}^n T_r $$ is

A
$$\cfrac { 3 }{ 2 } $$

B
$$\cfrac { 1 }{ 2 } $$

C
$$\cfrac { -1 }{ 2 } $$

D
$$\cfrac { -3 }{ 2 } $$

Solution

$${ T }_{ r }=\cfrac { r }{ { ({ r }^{ 2 }-1) }^{ 2 }-{ r }^{ 2 } } =\cfrac { r }{ ({ r }^{ 2 }-1+r)({ r }^{ 2 }-1-r) } $$

$$ { T }_{ r }=\cfrac { 1 }{ 2 } \cfrac { ({ r }^{ 2 }+r-1)-({ r }^{ 2 }-r-1) }{ ({ r }^{ 2 }-1+r)({ r }^{ 2 }-1-r) } $$

$$ { T }_{ r }=\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ ({ r }^{ 2 }-1-r) } -\cfrac { 1 }{ ({ r }^{ 2 }-1+r) } \right] $$

$$ { T }_{ 1 }=\cfrac { 1 }{ 2 } \left[ -1-\cfrac { 1 }{ 1 } \right] $$

$$ { T }_{ 2 }=\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ 1 } -\cfrac { 1 }{ 5 } \right] $$

$$ { T }_{ 3 }=\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ 5 } -\cfrac { 1 }{ 11 } \right] $$

$$ { T }_{ n }=\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ ({ n }^{ 2 }-1-n) } -\cfrac { 1 }{ ({ n }^{ 2 }-1+n) } \right] $$

Adding all we get
$${ S }_{ n }=\cfrac { 1 }{ 2 } \left[ -1-\cfrac { 1 }{ ({ n }^{ 2 }-1-n) } \right] $$

$$ \underset { n\rightarrow \infty }{ \lim } { \sum }_{ r=1 }^{ n }{ T }_{ n }=\underset { n\rightarrow \infty }{ \lim } \cfrac { 1 }{ 2 } \left[ -1-\cfrac { 1 }{ ({ n }^{ 2 }-1-n) } \right] =\cfrac { -1 }{ 2 } (1-0)=\cfrac { -1 }{ 2 } $$
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

The value of $$\sum _{ n=1 }^{ \infty }{ { \left( -1 \right) }^{ n+1 }\left( \cfrac { n }{ { 5 }^{ n } } \right) } $$ equals

A
$$\cfrac { 5 }{ 12 } $$

B
$$\cfrac { 5 }{ 24 } $$

C
$$\cfrac { 5 }{ 36 } $$

D
$$\cfrac { 5 }{ 16 } $$

Solution
Question 10
1 / -0

Let S be the infinite sum given by $$S=\displaystyle \sum_{n=0}^{\infty}\frac{a_n}{10^{2n}}$$, where $$(a_n)_{n\geq 0}$$ is a sequence defined by $$a_0=a_1=1$$ and $$a_j=20a_{j-1}$$ for $$j\geq 2$$. If $$S$$ is expressed in the form $$\displaystyle\frac{a}{b}$$, where $$a, b$$ are coprime positive integers, than $$a$$ equals.

A
$$60$$

B
$$75$$

C
$$80$$

D
$$81$$

Solution

Given, $$ S = \sum_{n=0}^\infty \dfrac{a_n}{10^{2n}}$$
Expanding the summation, we get
$$ S = \dfrac{a_0}{1}+\dfrac{a_1}{10^2}+\dfrac{a_2}{10^4}+\dfrac{a_3}{10^6}+\dfrac{a_4}{10^8}+\dots$$
On substituting the given values, we have
$$S = 1+\dfrac{1}{10^2}+\dfrac{20}{10^4}+\dfrac{20^2}{10^6}+\dfrac{20^3}{10^8}+\dots$$
$$S = 1+\dfrac{1}{10^2}+\dfrac{2}{10^3}+\dfrac{2^2}{10^4}+\dfrac{2^3}{10^5}+\dots$$
$$S = 1+\dfrac{1}{10^2}\left\lbrace 1 + \dfrac{2}{10}+\dfrac{2^2}{10^2}+\dfrac{2^3}{10^3}+\dots\right\rbrace$$
$$S = 1+\dfrac{1}{10^2}\left\lbrace 1 + \dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+\dots\right\rbrace$$
Since the terms in the bracket form an infinite geometric progression, we can sum them as
$$S = 1+\dfrac{1}{10^2}\left\lbrace \dfrac{1}{1-\dfrac{1}{5}} \right \rbrace$$
$$S = 1+\dfrac{1}{10^2}\left\lbrace \dfrac{5}{4} \right \rbrace$$
$$S = 1+\dfrac{5}{400}$$
$$S = \dfrac{405}{400}$$
As we have coprime numerator and denominator requirement,
$$\dfrac{a}{b} = \dfrac{81}{80}$$
Hence, the required answer is $$81$$.