Sequences and Series Test

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Sequences and Series Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\displaystyle \sum_{r = 16}^{30}{ (r + 2)(r - 3) } $$ is equal to :

A
$$7782$$

B
$$7787$$

C
$$7790$$

D
$$7780$$

Solution

Let $$S= \displaystyle \sum_{r = 16}^{30}{ (r + 2)(r - 3) } =\displaystyle \sum_{r = 1}^{30}{ (r + 2)(r - 3) } -\sum_{r = 1}^{15}{ (r + 2)(r - 3) }$$

     $$    = \displaystyle \sum_{r = 1}^{30}{ (r^2-r -6) } -\sum_{r = 1}^{15}{ (r^2 -r - 6) }$$

[By using $$ \sum_{i=1}^{n} k^2 =1^2 + 2^2 + 3^2 +… + n^2=\dfrac{n(n+1)(2n+1)}{6}$$]

$$    = \displaystyle \left( \frac{r(r+1)(2r+1)}{6}-\frac{r(r+1)}{2} -6r\right)_{r=30} -\left( \frac{r(r+1)(2r+1)}{6}-\frac{r(r+1)}{2} -6r\right)_{r=15}$$

$$ = \displaystyle \left( \frac{30(30+1)(60+1)}{6}-\frac{30(30+1)}{2} -180\right) -\left( \frac{15(15+1)(30+1)}{6}-\frac{15(15+1)}{2} -90\right)$$

$$=7780$$
Question 2
1 / -0

The sum of the following series $$1+6+\dfrac{9(1^2+2^2+3^2)}{7}+\dfrac{12(1^2+2^2+3^2+4^2)}{9}+\dfrac{15(1^2+2^2+....+5^2)}{11}+.....$$ up to $$15$$ terms is:

A
$$7820$$

B
$$7830$$

C
$$7520$$

D
$$7510$$

Solution

$$T_n=\dfrac{(3+(n-1)\times 3)(1^2+2^2+...+n^2)}{(2n+1)}$$

$$T_n=\dfrac{3n.\dfrac{n(n+1)(2n+1)}{6}}{2n+1}=\dfrac{n^2(n+1)}{2}$$

$$S_{15}=\dfrac{1}{2}\displaystyle \sum_{n=1}^{15}(n^3+n^2)=\dfrac{1}{2}\left[\left(\dfrac{15(15+1)}{2}\right)^2+\dfrac{15\times 16\times 31}{6}\right]$$

$$=7820$$
Question 3
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 $$1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2.6^{2} + .....$$ If $$B - 2A = 100\lambda$$, then $$\lambda$$ is equal to

A
$$464$$

B
$$496$$

C
$$232$$

D
$$248$$

Solution

$$B=1^2+2.2^2+3^2+2.4^2+....+2.40^2$$
$$A=1^2+2.2^2+3^2+2.4^2+.... +2.20^2$$

$$B = 1^2+3^3+5^2+...+39^2+2 [2^2+4^2+...+40^2]$$
$$A = 1^2+3^2+5^2+... + 19^2+2[2^2+4^2+...+20^2]$$

Sum of square of first n odd natural number
$$=\dfrac{n(2n+1)(2n-1)}{3}$$

sum of square of first n even natural number
$$=\dfrac{2n(n+1)(2n+1)}{3}$$

$$B=\left[\dfrac{n(2n+1)(2n-1)}{3}\right]_{n=20} + 2\left[\dfrac{2n(n+1)(2n+1)}{3}\right]_{n=20}$$

$$A=\left[\dfrac{n(2n+1)(2n-1)}{3}\right]_{n=10}+2\left[\dfrac{2n(n+1)2n+1)}{3}\right]_{n=10}$$

$$B = \dfrac{20(41)(39)}{3} + 2\left[\dfrac{(400(21)(41)}{3}\right]$$

$$B = \dfrac{100860}{3}$$

$$A=\dfrac{10(21)(19)}{3}+2\left[\dfrac{20(11)(21)}{3}\right]$$

$$A=\dfrac{13230}{3}$$

$$B-2A = \left[\dfrac{100860-26460}{3}\right] = \dfrac{74400}{3}$$

$$B-2A = 24800= 100\lambda$$

$$\lambda = 248$$
Question 4
1 / -0

If the sum of the first 15 terms of the series $$\lgroup \frac{3}{4} \rgroup^3 + \lgroup 1\frac{1}{2} \rgroup^3 + \lgroup 2\frac{1}{4} \rgroup^3 + 3^3 + \lgroup 3\frac{3}{4} \rgroup^3 + ...........$$ is equal to 225 k. then k is equal to:

A
9

B
27

C
108

D
54

Solution

$$\lgroup \dfrac{3}{4} \rgroup^3 + \lgroup \dfrac{6}{4} \rgroup^3 + \lgroup \dfrac{9}{4} \rgroup^3 + \lgroup \dfrac{12}{4} \rgroup^3 + .............. 15$$ term

$$S=\displaystyle\sum_{r=1}^{15}\left [\dfrac{3r}{4}\right ]^3=\dfrac{27}{64}\sum r^3$$

$$=\displaystyle\dfrac{27}{64} \sum_{r = 1}^{15} r^3$$

$$ = \dfrac{27}{64}. \Big[ \dfrac{15(15+1)}{2} \Big ]^2$$

$$= 27\times 225$$ (given in question)

$$K = 27$$
Question 5
1 / -0

Let $$S _ { k } = \dfrac { 1 + 2 + 3 + \ldots + k } { k } .$$ If $${ S }_{ { 1 } }^{ { 2 } }+{ S }_{ { 2 } }^{ { 2 } }++{ S }_{ { 10 } }^{ { 2 } }=\dfrac { 5 }{ 12 } { A }$$ then $$A$$ is equal to :

A
$$303$$

B
$$283$$

C
$$156$$

D
$$301$$

Solution

$$\mathrm { S } _ { \mathrm { K } } = \dfrac { \mathrm { K } + 1 } { 2 }$$

$$\Sigma S _ { k } ^ { 2 } = \dfrac { 5 } { 12 } A$$

$$\sum _ { \mathrm { k } = 1 } ^ { 10 } \left( \dfrac { \mathrm { K } + 1 } { 2 } \right) ^ { 2 } = \dfrac { 2 ^ { 2 } + 3 ^ { 2 } + \dots + 11 ^ { 2 } } { 4 } = \dfrac { 5 } { 12 } \mathrm { A }$$

$$\dfrac { 11 \times 12 \times 23 } { 6 } - 1 = \dfrac { 5 } { 3 } \mathrm { A }$$

$$505 = \dfrac { 5 } { 3 } \mathrm { A } , \quad \mathrm { A } = 303$$
Question 6
1 / -0

The sum $$\mathrm{V}_{1}+\mathrm{V}_{2}+\ldots+\mathrm{V}_{\mathrm{n}}$$ is

A
$$\displaystyle \frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}-\mathrm{n}+1)$$

B
$$\displaystyle \frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}+\mathrm{n}+2)$$

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

D
$$\displaystyle \frac{1}{3}(2\mathrm{n}^{3}-2\mathrm{n}+3)$$

Solution

$$\displaystyle \mathrm{V}_{\mathrm{r}}=\frac{\mathrm{r}}{2}[2\mathrm{r}+(\mathrm{r}-1)(2\mathrm{r}-1)]=\frac{1}{2}(2\mathrm{r}^{3}-\mathrm{r}^{2}+\mathrm{r}) $$
$$\therefore \displaystyle \sum \mathrm{V}_{\mathrm{r}}=\frac{1}{2}\left(2 \sum r^3-\sum r^2+\sum r\right)=\frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}+\mathrm{n}+2)$$ .
Question 7
1 / -0

Let $$\mathrm{a}_{1},\ \mathrm{a}_{2},\ \mathrm{a}_{3},\ \ldots,\ \mathrm{a}_{11}$$ be real numbers satisfying $$\mathrm{a}_{1}=15,27-2\mathrm{a}_{2}>0$$ and $$\mathrm{a}_{\mathrm{k}}=2\mathrm{a}_{\mathrm{k}-1}-\mathrm{a}_{\mathrm{k}-2}$$
for $$\mathrm{k}=3,4,\ \ldots, 11.$$ If $$\displaystyle \frac{a_{1}^{2}+a_{2}^{2}+\ldots+a_{11}^{2}}{11}=90$$, then the value of $$\displaystyle \frac{a_{1}+a_{2}+\ldots+a_{11}}{11}$$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

We have: $${ a }_{ k }+{ a }_{ k+2 }=2{ a }_{ k-1 }$$
Thus, the terms are in AP.
Hence, sum of squares of the terms in AP is:
$${ a }^{ 2 }+{ (a+d) }^{ 2 }+...+{ (a+10d) }^{ 2 }=11{ a }^{ 2 }+110ad+385{ d }^{ 2 }=990\\ =>a^{ 2 }+10ad+35{ d }^{ 2 }=90\\ =>35{ d }^{ 2 }+150d+225-90=0\\ =>35{ d }^{ 2 }+150d+135=0\\ =>7{ d }^{ 2 }+30d+27=0\\ =>(7d+9)(d+3)=0=>d=-3,-\dfrac { 9 }{ 7 } $$
Since $${ a }_{ 2 }<13.5$$, $$d=-3$$
Thus, the required answer i.e. the average of $$11$$ terms of an AP = $${ a }_{ 6 }=15+(6-1).(-3)=0$$
Hence, (a) is correct.
Question 8
1 / -0

$$x\left (x + \dfrac {1}{x}\right )^{2} + \left (x^{2} + \dfrac {1}{x^{2}}\right )^{2} + \left (x^{3} + \dfrac {1}{x^{3}}\right )^{2}$$ .... upto $$n$$ terms is

A
$$\dfrac {x^{2n} - 1}{x^{2} - 1}\times \dfrac {x^{2n + 2} + n}{x^{2n}} + 2n$$

B
$$\dfrac {x^{2n} + 1}{x^{2} + 1}\times \dfrac {x^{2n + 2} - n}{x^{2n}} - 2n$$

C
$$\dfrac {x^{2n} - 1}{x^{2} - 1}\times \dfrac {x^{2n - 2} - n}{x^{2n}} - 2n$$

D
None of these

Solution

The series is
$$(x^{2} + x^{4} + x^{6} + ....) + \left (\dfrac {1}{x^{2}} + \dfrac {1}{x^{4}} + \dfrac {1}{x^{6}} + ....\right ) + (2 + 2 + ....)$$
$$= \dfrac {x^{2}(x^{2n} - 1)}{x^{2} - 1} + \dfrac {\dfrac {1}{x^{2}}\left (1 - \dfrac {1}{x^{2n}}\right )}{1 - \dfrac {1}{x^{2}}} + 2n$$
$$= \dfrac {x^{2}(x^{2n} - 1)}{x^{2} - 1} + \dfrac {x^{2n} - 1}{(x^{2} - 1)x^{2n}} + 2n$$
$$= \dfrac {x^{2n} - 1}{x^{2} - 1}\times \dfrac {x^{2n + 2} + 1}{x^{2}} + 2n$$
Question 9
1 / -0

Which letter will be the fifth from the right if the first and second, the third and fourth and so on are interchanged among each other in the word "COMPANIONATE"?

A
A

B
I

C
N

D
O

Solution

When we interchange the first and the second, third and the fourth, and so on letters in the word COMPANIONATE, the new word becomes OCPMNAOIANET.
Therefore, 5th letter from right is I
Answer is Option B
Question 10
1 / -0

Mean of the first $$n$$ terms of the A.P. $$a, (a + d), (a + 2d), ........$$ is

A
$$\displaystyle a + \frac{nd}{2}$$

B
$$\displaystyle a + \frac{(n - 1)d}{2}$$

C
$$a + (n - 1) d$$

D
$$a + nd$$

Solution

Required mean $$= \displaystyle \frac{a + (a + d) + (a + 2d) + ....... + \{ a + (n - 1) d \}}{n}$$
$$\displaystyle = \frac{\displaystyle \frac{n}{2} [a + a + (n - 1) d]}{n} = a + \frac{(n - 1)d}{2}$$

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Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 10

Result

Sequences and Series Test - 10

Score

-

Rank

-

SHARING IS CARING

Question 1

$$7782$$

$$7787$$

$$7790$$

$$7780$$

Solution

Question 2

$$7820$$

$$7830$$

$$7520$$

$$7510$$

Solution

Question 3

$$464$$

$$496$$

$$232$$

$$248$$

Solution

Question 4

9

27

108

54

Solution

Question 5

$$303$$

$$283$$

$$156$$

$$301$$

Solution

Question 6

$$\displaystyle \frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}-\mathrm{n}+1)$$

$$\displaystyle \frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}+\mathrm{n}+2)$$

$$\displaystyle \frac{1}{2}\mathrm{n}(2\mathrm{n}^{2}-\mathrm{n}+1)$$

$$\displaystyle \frac{1}{3}(2\mathrm{n}^{3}-2\mathrm{n}+3)$$

Solution

Question 7

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 8

$$\dfrac {x^{2n} - 1}{x^{2} - 1}\times \dfrac {x^{2n + 2} + n}{x^{2n}} + 2n$$

$$\dfrac {x^{2n} + 1}{x^{2} + 1}\times \dfrac {x^{2n + 2} - n}{x^{2n}} - 2n$$

$$\dfrac {x^{2n} - 1}{x^{2} - 1}\times \dfrac {x^{2n - 2} - n}{x^{2n}} - 2n$$

None of these

Solution

Question 9

A

I

N

O

Solution

Question 10

$$\displaystyle a + \frac{nd}{2}$$

$$\displaystyle a + \frac{(n - 1)d}{2}$$

$$a + (n - 1) d$$

$$a + nd$$

Solution

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