Sequences and Series Test

Result

Sequences and Series Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

The series $$\dfrac{8}{5}+\dfrac{16}{65}+\dfrac{24}{325}+.........$$ upto infinity has the sum equal to

A
$$\dfrac{16}{5}$$

B
$$2$$

C
$$4$$

D
$$5$$

Solution
Question 2
1 / -0

How many Pythagorean triangles are there with integer sides and one leg is Ramanujan Number 1729 .

A
$$10$$

B
$$11$$

C
$$12$$

D
$$13$$

E
None
Question 3
1 / -0

Find the missing number, if a certain rule is followed
$$1$$ $$9$$ ?
$$2$$ $$12$$ $$14$$
$$3$$ $$117$$ $$105$$
either row-wise or column-wise.

A
$$7$$

B
$$8$$

C
$$11$$

D
$$9$$

Solution
Question 4
1 / -0

$$\sum\limits_{r = 1}^n {{{\sin }^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {r - 1} }}{{\sqrt {r\left( {r + 1} \right)} }}} \right)} $$ is equal to

A
$${\tan ^{ - 1}}\left( {\sqrt n } \right) - \frac{\pi }{4}$$

B
$${\tan ^{ - 1}}\left( {\sqrt n + 1} \right) - \frac{\pi }{4}$$

C
$${\tan ^{ - 1}}\left( {\sqrt n } \right)$$

D
$${\tan ^{ - 1}}\left( {\sqrt n + 1} \right)$$

Solution
Question 5
1 / -0

Let $$A$$ be the sum of the first $$20$$ terms and $$B$$ be the sum of the first $$40$$ terms of the series $$1$$ + $$2.2^2$$ + $$3^2$$ + $$2.4^2$$ + $$5^2+2.6^2$$ +......... Find the value of $$A$$.

A
$$496$$

B
$$232$$

C
$$248$$

D
$$464$$

Solution

Sum of first $$20$$ terms -
$$\Rightarrow \left( { 1 }^{ 2 }+{ 3 }^{ 2 }+\_ \_ \_ \_ +{ 19 }^{ 2 } \right) +2\left( { 2 }^{ 2 }+{ 4 }^{ 2 }+\_ \_ \_ \_ +{ 20 }^{ 2 } \right) $$
$$\Rightarrow A=\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+3^2\_ \_ \_ \_ +{ 19 }^{ 2 }+20^2 \right) +\left( { 2 }^{ 2 }+{ 4 }^{ 2 }+\_ \_ \_ \_ +{ 20 }^{ 2 } \right) $$
$$\Rightarrow A=\dfrac{20\times 41\times 21}{6}+1^2\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+\_ \_ \_ \_ +{ 10 }^{ 2 } \right) $$
$$\Rightarrow A=2870+4\times \dfrac{10\times 11\times 21}{6}$$
$$\Rightarrow A=2870+1540$$
$$\Rightarrow A=4410.$$
Hence, the answer is $$4410.$$
Question 6
1 / -0

If $$1^{2}+2^{2}+3^{2}++(2003) ^{2}=(2003)(4007)(334)$$ and $$(1)(2003)+ (2)(2002)= (2003)(334)(x)$$ then $$x$$ equals

A
$$2005$$

B
$$2004$$

C
$$2003$$

D
$$2001$$

Solution
Question 7
1 / -0

If $$\left( { 1-y } \right) ^{ 35 }\left( 1+y \right) ^{ 45 }={ A }_{ 0 }+{ A }_{ 1 }Y+{ A }_{ 2 }{ Y }^{ 2 }+{ A }_{ 3 }{ Y }^{ 3 }+.....+{ A }_{ 80 }{ Y }^{ 80 },$$ then

A
$$\frac { { A }_{ 2 } }{ A_{ 1 } } <2$$

B
$${ A }_{ 1 }={ A }_{ 2 }$$

C
$$\frac { { A }_{ 2 } }{ { A }_{ 1 } } <1$$

D
$$1<\frac { { A }_{ 2 } }{ { A }_{ 1 } } <2$$
Question 8
1 / -0

If $$n\in N$$ then $$\displaystyle \sum \limits_{ r=0 }^{ n }{ { \left( -1 \right)^r }\;^{ n }{ C }_{ 2 } } \left[ \dfrac { 1 }{ { 2 }^{ r} }+ \dfrac { { 3 }^{ r } }{ { 2 }^{ 2r } }+ \dfrac { { 7 }^{ r } }{ { 2 }^{ 3r } } +\;.\;.\;.\;\;upto\quad m\quad terms \right] $$ is equal to

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

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

C
$$\dfrac { { 2 }^{ mn }-1 }{ \left( { 2 }^{ n }-1 \right) { 2 }^{ mn } } $$

D
None of these

Solution
Question 9
1 / -0

If $$ '-'$$ denotes $$'\div ',\ '\div '$$ denotes $$'\times ',\ '+'$$ denotes' - 'and $$'\times '$$ denotes $$'+'$$, then find the value of $$116+9\div 52-4\times 5$$.

A
$$16$$

B
$$8$$

C
$$9$$

D
$$4$$

Solution

$$-$$ denoted $$\div$$
$$\div$$ denoted $$\times $$
$$+$$ denoted $$-$$
$$\times $$ denoted $$+$$ $$\boxed{BODMAS}$$
$$116+9\div 52-4\times 5$$
$$116-9\times 52\div 4+5$$
$$116-9\times 13+5$$
$$116-117+5$$
$$116-112=4$$
Question 10
1 / -0

$$1+\dfrac{1}{2}(1+2)+\dfrac{1}{3}(1+2+3)+\dfrac{1}{4}(1+2+3+4)+.....$$ upto $$20$$ terms is

A
$$110$$

B
$$111$$

C
$$115$$

D
$$116$$

Solution

Given
$$S=1+\cfrac{1}{2}(1+2)+\cfrac{1}{3}(1+2+3)+.....+\cfrac{1}{20}(1+2+3+....+20)$$
$$\displaystyle=\sum _{ n=1 }^{ 20 }{ \cfrac { 1 }{ n } } \sum { n } $$
$$\displaystyle \sum _{ n=1 }^{ 20 }{ \cfrac { 1 }{ n } } .\cfrac{n(n+1)}{2}=\cfrac{1}{2}\sum _{ n=1 }^{ 20 }{ (n+1) } $$
$$\cfrac { 1 }{ 2 } { \left[ \cfrac { n(n+1) }{ 2 } +n \right] }_{ 0 }^{ 20 }$$
$$=\cfrac{1}{2}\left[ \cfrac { 20(21) }{ 2 } +20 \right] $$
$$=5(21)+10=115$$