Sequences and Series Test

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Sequences and Series Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

Three bells commenced to toll at the same time and tolled at intervals of $$20, 30, 40$$ seconds respectively. If they toll together at $$6$$ am, then which of the following is the time at which they can toll together

A
$$6:55$$am

B
$$6:56$$am

C
$$6:57$$am

D
$$6:59$$am

Solution
Question 2
1 / -0

The first term of an AP is $$148$$ and the common difference is $$-2$$. If the AM of first $$n$$ terms of the AP is $$125$$, then the value of $$n$$ is

A
$$18$$

B
$$24$$

C
$$30$$

D
$$36$$

E
$$48$$

Solution

Given, $$a=148, d=-2$$
AM of $$n$$ terms $$=125$$
$$\Rightarrow \dfrac { { a }_{ 1 }+{ a }_{ 2 }+\cdots +{ a }_{ n } }{ n } =125$$
$$\Rightarrow \dfrac { \dfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] }{ n } =125$$
$$\Rightarrow 2a+\left( n-1 \right) d=250$$
$$\Rightarrow 2\times 148+\left( n-1 \right) \left( -2 \right) =250$$
$$\Rightarrow 296-2n+2=250$$
$$\Rightarrow 298-2n=250$$
$$\Rightarrow n=24$$
Question 3
1 / -0

Find the sum of the series
$$\displaystyle \frac{1}{2\cdot 3}+\frac {1}{4\cdot 5}+\frac {1}{6\cdot 7}+ ...$$

A
$$log\, \frac {e}{2}$$

B
$$log\, \frac {e}{4}$$

C
$$log\, \frac {2}{3}$$

D
$$log\, \frac {2}{4}$$

Solution

$$\displaystyle \frac {1}{2\cdot 3}+ \frac {1}{4\cdot 5}+\frac {1}{6\cdot 7}$$

$$\displaystyle \frac {1}{2}- \frac {1}{3}+ \frac {1}{4}- \frac {1}{5}+\frac {1}{6}-\frac {1}{7} + ...$$

$$\displaystyle =1-1 + \frac {1}{2}-\frac {1}{3}+ \frac {1}{4}- \frac {1}{5}+...$$

$$\displaystyle = 1-\left ( 1- \frac {1}{2}+\frac {1}{3}-\frac {1}{4}+ \frac {1}{5}\right )$$

$$= 1- log \, 2 = log\, e - log\, 2$$

$$=log \frac {e}{2}$$
Question 4
1 / -0

If the natural numbers are divided into groups of {1}, {2, 3}, {4, 5, 6}, {7, 8, 9, 10} ....Then the /sum of 50th group is

A
65225

B
56225

C
62525

D
53625

Solution

Pattern:
1
2 3
4 5 6
7 8 9 10
..........
Series for starting number:
1, 2, 4, 7, ...
Let $$T_{n}=an^{2}+bn+c$$
a+b+c=1 ......(1)
4a+2b+c=2 ......(2)
9a+3b+c=4 .......(3)
From eq (1), (2), (3), we get:
$$a=\dfrac{1}{2},b=-\dfrac{1}{2} ,c=1$$
$$\therefore T_{n}=\dfrac{n^{2}}{2}-\dfrac{n}{2}+1$$
$$\therefore T_{50}=\dfrac{50^{2}}{2}-\dfrac{50}{2}+1=1226$$
Now, elements in 50th group:
1226, 1227, 1228,.....(50 terms)
$$\therefore S_{n}=\dfrac{50}{2}[2\times 1226+(50-1)\times 1]=62525$$
Question 5
1 / -0

If $$\displaystyle \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ..... $$ upto $$\displaystyle \infty = \frac{\pi^2}{6},$$ then $$\displaystyle \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + .... =$$

A
$$\dfrac{\pi^2}{12}$$

B
$$\dfrac{\pi^2}{24}$$

C
$$\dfrac{\pi^2}{8}$$

D
$$\dfrac{\pi^2}{4}$$

Solution

Let $$S=\sum ^{ \infty }{ \dfrac { 1 }{ { n }^{ 2 } } } =\dfrac { { \pi }^{ 2 } }{ 6 } \Rightarrow \dfrac { S }{ { 2 }^{ 2 } } =\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ 4^{ 2 } } +\dfrac { 1 }{ 6^{ 2 } } +...\infty $$

$$\therefore { S }^{ ' }=S-\dfrac { S }{ 4 } =\dfrac { 3S }{ 4 } =\dfrac { 1 }{ { 1 }^{ 2 } } +\left(\dfrac { 1 }{ 2^{ 2 } } -\dfrac { 1 }{ 2^{ 2 } } \right)+\dfrac { 1 }{ 3^{ 2 } } +\left(\dfrac { 1 }{ 4^{ 2 } } -\dfrac { 1 }{ 4^{ 2 } } \right)+...\infty $$

ie, $$\therefore { S }^{ ' }=\dfrac { 1 }{ { 1 }^{ 2 } } +\dfrac { 1 }{ 3^{ 2 } } +\dfrac { 1 }{ { 5 }^{ 2 } } +...\infty =\dfrac { 3S }{ 4 } =\dfrac { 3{ \pi }^{ 2 } }{ 4\times 6 } =\dfrac { { \pi }^{ 2 } }{ 8 } $$

So, Option C is the correct one.
Question 6
1 / -0

The value of a for which side of nth square equals the diagonals of $$(n + 1)^{th}$$ square is

A
1/3

B
1/4

C
1/2

D
$$1\sqrt{2}$$

Solution

Given $$(Diagonal)_{n+1}=(Side)_n \\ \sqrt{2}a_1r^n=a_1r^{n-1} \\ \sqrt{2}r=1 \\ r=\cfrac{1}{\sqrt{2}}$$
Question 7
1 / -0

$$\displaystyle \sum_{k = 1}^{\infty} \dfrac {6^{k}}{(3^{2k + 1} + 2^{2k + 1}) - (3^{k}2^{k + 1} + 2^{k}3^{k + 1})}$$ is equal to

A
$$3$$

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

C
$$\dfrac {4}{3}$$

D
$$\dfrac {6}{5}$$

Solution

$${ \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ ({ 3 }^{ 2k+1 }+{ 2 }^{ 2k+1 })-({ 3 }^{ k }{ 2 }^{ k+1 }+{ 2 }^{ k }{ 3 }^{ k+1 }) } \\ ={ \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ 3\cdot { 3 }^{ 2k }-3\cdot { 3 }^{ k }{ 2 }^{ k }+{ 2 }^{ 2 }{ 2 }^{ 2k }-2\cdot { 3 }^{ k }{ 2 }^{ k } } \\ ={ \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ 3\cdot { 3 }^{ k }({ 3 }^{ k }-{ 2 }^{ k })-{ 2 }\cdot { 2 }^{ k }({ 3 }^{ k }-{ 2 }^{ k }) } \\ ={ \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ \cfrac { { 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } =\cfrac { { 3 }^{ k }A }{ ({ 3 }^{ k }-{ 2 }^{ k }) } +\cfrac { { 3 }^{ k }B }{ ({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ \cfrac { { 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } =\cfrac { (3A+B){ 3 }^{ 2k }B-(2A+B){ 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ 3A+B=0\Rightarrow 3A=-B\quad \quad (1)\\ 2A+B=-1\\ 2A-3A=-1\quad from(1)\\ -A=-1\\ A=1\Rightarrow B=-3A=-3\\ { \sum }_{ k=1 }^{ \infty }\cfrac { { 6 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k })({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } =\sum \cfrac { { 3 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k }) } -\sum \cfrac { { 3 }^{ k+1 } }{ ({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ { T }_{ 1 }=\cfrac { 3 }{ 3-2 } -\cfrac { { 3 }^{ 2 } }{ { 3 }^{ 2 }-{ 2 }^{ 2 } } \\ { T }_{ 2 }=\cfrac { { 3 }^{ 2 } }{ { 3 }^{ 2 }-{ 2 }^{ 2 } } -\cfrac { { 3 }^{ 3 } }{ { 3 }^{ 3 }-{ 2 }^{ 3 } } \\ { T }_{ n }=\cfrac { { 3 }^{ k } }{ ({ 3 }^{ k }-{ 2 }^{ k }) } -\cfrac { { 3 }^{ k+1 } }{ ({ 3 }^{ k+1 }-{ 2 }^{ k+1 }) } \\ { S }_{ n }=3-\cfrac { { 3 }^{ n+1 } }{ { 3 }^{ n+1 }-{ 2 }^{ n+1 } } \\ \underset { n\rightarrow \infty }{ \lim } { S }_{ n }=3$$
Question 8
1 / -0

Let $$S = \displaystyle \sum_{n = 1}^{99} = \dfrac {5^{100}}{5^{100} + 25^{n}}$$ then find the value of $$[S]$$, where $$[.] = G.I.F.$$

A
$$99$$

B
$$100$$

C
$$25$$

D
$$49$$

Solution
Question 9
1 / -0

If $$\alpha =1 / 4$$ and $$P_n$$ denotes the perimeter of the nth square then$$\sum_{n=1}^{\infty } P_n$$ equals

A
$$\dfrac{8}{3}\left ( 4+\sqrt{10} \right )$$

B
$$\dfrac{8}{3}$$

C
$$\dfrac{16}{3}$$

D
None of these

Solution

Parameter of $$n^{th}$$ square $$=4a_n=4a_1r^{n-1}$$
$$\sum_{n=1}^{\infty}$$$$=P_1+P_2+P_3+....+P_n \\ =4a_1(1+r+r^2+r^3+....infty)$$
Sum of infinite GP $$=\cfrac{a}{1-r}$$
$$\sum_{n=1}^{\infty}=4a_1\cfrac{1}{1-r}=\cfrac{4a_1}{1-\sqrt{2\alpha ^2-2/alpha+1}}$$
Given $$a_1=1, \quad \alpha=\cfrac{1}{4}$$
$$\sum_{n=1}^{\infty}$$$$=\cfrac { 4 }{ 1-\sqrt { \cfrac { 2 }{ 16 } -\cfrac { 2 }{ 4 } +1 } } =\cfrac { 4 }{ 1-\cfrac { \sqrt { 2-8+16 } }{ 4 } } =\cfrac { 16 }{ 4-\sqrt { 10 } } \\ =\cfrac { 16 }{ 4-\sqrt { 10 } } \times \cfrac { 4+\sqrt { 10 } }{ 4+\sqrt { 10 } } =\cfrac { 16(4+\sqrt { 10 } ) }{ 6 } =\cfrac { 8 }{ 3 } (4+\sqrt { 10 } )$$
Question 10
1 / -0

If $$ a_1 \in R - \left \{ 0 \right \}, i = 1, 2, 3, 4$$ and $$x \in R$$ and $$\left ( \sum_{i=1}^{3}a_i^2\right ) x^2-2x \left ( \sum_{i=1}^{3}a_i a_{i+1}\right )+\sum_{i=2}^{4}a_i^2\leq 0$$, then $$a_1, a_2, a_3, a_4$$ are in

A
A.P

B
G.P

C
H.P

D
A.G.P

Solution

$$({ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+{ { a }_{ 3 } }^{ 2 }){ x }^{ 2 }-2x({ a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+{ a }_{ 3 }{ a }_{ 4 })+({ { a }_{ 2 } }^{ 2 }+{ { a }_{ 3 } }^{ 2 }+{ { a }_{ 4 } }^{ 2 })\le 0\\ ({ x }^{ 2 }{ { a }_{ 1 } }^{ 2 }-2x{ a }_{ 1 }{ a }_{ 2 }+{ { a }_{ 2 } }^{ 2 })+({ { a }_{ 2 } }^{ 2 }{ x }^{ 2 }-2{ a }_{ 2 }{ a }_{ 3 }+{ { a }_{ 3 } }^{ 2 })+({ { a }_{ 3 } }^{ 2 }{ x }-2x{ a }_{ 3 }{ a }_{ 4 }+{ { a }_{ 4 } }^{ 2 })\le 0\\ { ({ a }_{ 1 }x-{ a }_{ 2 }) }^{ 2 }+{ ({ a }_{ 2 }x-{ a }_{ 3 }) }^{ 2 }+{ ({ a }_{ 3 }x-{ a }_{ 4 }) }^{ 2 }\le 0$$
Since each one is a square term
These sum of term cannot be less than $$0$$
Only possibility they can be zero
For
$$a_1x-a_2=0 \Rightarrow x=\cfrac{a_2}{a_1} \\ a_2x-a_3=0 \Rightarrow x=\cfrac{a_3}{a_2} \\ a_3x-a_4=0 \Rightarrow x=\cfrac{a_4}{a_3} \\ x=\cfrac{a_2}{a_1} = \cfrac{a_3}{a_2} = \cfrac{a_4}{a_3}=k$$
Hence, $$a_1,a_2,a_3,a_4$$ are in GP

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Sequences and Series Test - 29

Result

Sequences and Series Test - 29

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Rank

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SHARING IS CARING

Question 1

$$6:55$$am

$$6:56$$am

$$6:57$$am

$$6:59$$am

Solution

Question 2

$$18$$

$$24$$

$$30$$

$$36$$

$$48$$

Solution

Question 3

$$log\, \frac {e}{2}$$

$$log\, \frac {e}{4}$$

$$log\, \frac {2}{3}$$

$$log\, \frac {2}{4}$$

Solution

Question 4

65225

56225

62525

53625

Solution

Question 5

$$\dfrac{\pi^2}{12}$$

$$\dfrac{\pi^2}{24}$$

$$\dfrac{\pi^2}{8}$$

$$\dfrac{\pi^2}{4}$$

Solution

Question 6

1/3

1/4

1/2

$$1\sqrt{2}$$

Solution

Question 7

$$3$$

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

$$\dfrac {4}{3}$$

$$\dfrac {6}{5}$$

Solution

Question 8

$$99$$

$$100$$

$$25$$

$$49$$

Solution

Question 9

$$\dfrac{8}{3}\left ( 4+\sqrt{10} \right )$$

$$\dfrac{8}{3}$$

$$\dfrac{16}{3}$$

None of these

Solution

Question 10

A.P

G.P

H.P

A.G.P

Solution

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