Sequences and Series Test

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Sequences and Series Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Geometric mean of $$7,\ { 7 }^{ 2 },\ { 7 }^{ 3 }....{ 7 }^{ n }$$ is

A
$$\displaystyle { 7 }^{ \dfrac { \left( n+1 \right) }{ 2 } }$$

B
$${ 7 }^{ { n }/{ 2 } }$$

C
$${ 7 }^{ \dfrac { \left( n-1 \right) }{ 2 } }$$

D
$${ 7 }^{ \left( \dfrac { n+1 }{ 3 } \right) }$$

Solution

$$Gm$$ of two no $$a$$ & $$b=\sqrt {ab}$$
For $$aGP$$
$$GM=\sqrt {a_1. an}$$
$$\Rightarrow \ GM $$ of $$7, 7^2, ..... 7^n$$
$$=\sqrt {7.7^n}$$
$$=7\dfrac {(n+1)}{2}$$
Question 2
1 / -0

Suppose $$a_2,a_3,a_4,a_5,a_6,a_7$$ are integers such that
$$\dfrac {5}{7}=\dfrac {a_2}{2!}+\dfrac {a_3}{3!}+\dfrac {a_4}{4}+\dfrac {a_5}{5!}+\dfrac {a_6}{6!}+\dfrac{a_7}{7!}$$
where $$0 \le a < j$$ for $$ j=2,4,5,6,7.$$ The sum $$a_2+a_3+a_4+a_5+a_6+a_7$$ is

A
8

B
9

C
10

D
11

Solution

$$\cfrac { 5 }{ 7 } =\cfrac { { a }_{ 2 } }{ 2! } +\cfrac { { a }_{ 3 } }{ 3! } +\cfrac { { a }_{ 4 } }{ 4! } +\cfrac { { a }_{ 5 } }{ 5! } +\cfrac { { a }_{ 6 } }{ 6! } +\cfrac { { a }_{ 7 } }{ 7! } \\ \Rightarrow \cfrac { 5 }{ 7 } =\cfrac { 1 }{ 2 } \left( { a }_{ 2 }+\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\left( \cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \right) \\ \cfrac { 10 }{ 7 } ={ a }_{ 2 }+\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\left( \cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \\ 1+\cfrac { 3 }{ 7 } ={ a }_{ 2 }+\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\left( \cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) $$

So, as expression is less than $$1$$

$${ a }_{ 2 }=1\\ \cfrac { 3 }{ 7 } =\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\cfrac { 1 }{ 5 } \left( { a }_{ 5 }+.....\cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \\ 1+\cfrac { 2 }{ 7 } ={ a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\cfrac { 1 }{ 5 } \left( { a }_{ 5 }+\cfrac { 1 }{ 6 } \left( { a }_{ 6 }+\cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \\ { a }_{ 3 }=1$$

Similarly

$${ a }_{ 4 }=1\quad \quad { a }_{ 7 }=2\\ { a }_{ 5 }=0\\ { a }_{ 6 }=4$$

So, Sum $${ a }_{ 7 }-1+1+1+4+2\\ =9$$
Question 3
1 / -0

The value of the sum $$1.2.3+2.3.4+3.4.5+.......$$ upto n terms=

A
$$\frac{1}{6}n^{2}(2n^{2}+1)$$

B
$$\frac{1}{6}n^{2}(n^{2}-1)(2n-1)(2n+3)$$

C
$$\frac{1}{8}(n^{2}+1)(n^{2}+5)$$

D
$$\frac{1}{4}(n)(n+1)(n+2)(n+3)$$

Solution

Sum of $$1.2.3+2.3.4+3.4.5+.......$$ upto $$n$$ terms
$$T_{n}={n}\left(n+1\right)\left(n+2\right)$$
$$S_{n}=$$ Sum of $$T_{n}$$
$$=\sum _{ n=1 }^{ n }{ n\left( n+1 \right) \left( n+2 \right) }$$
$$=\sum{n}^{3}+{3}{n}^{2}+2n$$

$$\sum{n}^{3}+3\sum{n}^{2}+2\sum{n}$$

$$=\dfrac { { \left( n \right) }^{ 2 }{ \left( n+1 \right) }^{ 2 } }{ 4 } +\dfrac { 3\left( n \right) \left( n+1 \right) \left( 2n+1 \right) }{ 6 } +\dfrac { 2\left( n \right) \left( n+1 \right) }{ 2 } $$

$$=\dfrac { 1 }{ 4 } n\left( n+1 \right) \left( n+2 \right) \left( n+3 \right)$$
Hence, it is the correct answer.
Question 4
1 / -0

If $$x\in R$$, find the minimum value of the expression $$3^x+3^{1-x}$$.

A
$$3 \sqrt 2$$

B
$$2 \sqrt 3$$

C
$$3 \sqrt 3$$

D
none of these

Solution

We know,
$$AM > GM$$
Therefore,
$$\dfrac{3^x+3^{1-x}}{2} \geq \sqrt{3^x \times 3^{1-x}}$$ for all $$x\in R$$.

$$\dfrac{3^x+3^{1-x}}{2} \geq \sqrt{3}$$ for all $$x\in R$$.

$$3^x+3^{1-x} \geq 2\sqrt{3}$$ for all $$x\in R$$

Hence, the required value is $$2\sqrt{3}$$.
Question 5
1 / -0

The sum of the first n terms of the series $$1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 +$$ ...... is $$\dfrac{n(n + 1)^2}{2}$$ when n is even.When n is odd the sum is -

A
$$\displaystyle\frac{3n(n + 1)}{2}$$

B
$$ \cfrac { 8{ n }^{ 3 }+6{ n }^{ 2 }+n }{ 3 } $$

C
$$\displaystyle\frac{n(n + 1)^2}{4}$$

D
$$\displaystyle\left [ \frac{n(n + 1)}{4} \right ]^2$$

Solution

$${ 1 }^{ 2 }+2.{ 2 }^{ 2 }+{ 3 }^{ 2 }+2.{ 4 }^{ 2 }+{ 5 }^{ 2 }+2.{ 6 }^{ 2 }+.......\\ ({ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.....{ n }^{ 2 })+2({ 2 }^{ 2 }+{ 4 }^{ 2 }+{ 6 }^{ 2 }+......)\\ =({ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.....{ n }^{ 2 })+2.2({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+......)\\ =({ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.....{ n }^{ 2 })+{ 2 }^{ 2 }({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+......)\\ =\cfrac { n(4{ n }^{ 2 }-1) }{ 3 } +4[\cfrac { n(n+1)(2n+1) }{ 6 } ]\\ =\cfrac { n(4{ n }^{ 2 }-1) }{ 3 } +\cfrac { 2({ n }^{ 2 }+n)(2n+1) }{ 3 } \\ =\cfrac { 4{ n }^{ 3 }-n+(4n+2)({ n }^{ 2 }+n) }{ 3 } \\ =\cfrac { 4{ n }^{ 3 }-n+4{ n }^{ 3 }+4{ n }^{ 2 }+2{ n }^{ 2 }+2n }{ 3 } \\ =\cfrac { 8{ n }^{ 3 }+6{ n }^{ 2 }+n }{ 3 } $$
Question 6
1 / -0

If $$28\div 11 = 8, 39\div 21 = 7, 45\div 27 = 4$$, then $$95\div 25 = ?$$

A
$$11$$

B
$$9$$

C
$$16$$

D
$$5$$

Solution

$$28\div 11 \Rightarrow [(2 + 8) \div (1 + 1)] + 3$$
$$= (10\div 2) + 3 = 5 + 3 = 8$$
$$39\div 21 \Rightarrow [(3 + 9) \div (2 + 1)] + 3$$
$$= (12\div 3) + 3 = 4 + 3 = 7$$
$$45\div 27 \Rightarrow [(4 + 5) \div (2 + 7)] + 3$$
$$= (9\div 9) + 3 = 1 + 3 = 4$$
$$\therefore 95\div 25\Rightarrow [(9 + 5) \div (2 + 5)] + 3$$
$$= (14\div 7) + 3 = 2 + 3 = 5$$.
Question 7
1 / -0

The odd natural numbers have been divided in groups as $$(1, 3); (5, 7, 9, 11); (13, 15, 17, 19, 21, 23); .....$$ Then the sum of numbers in the $$10^{th}$$ group is

A
$$4000$$

B
$$4003$$

C
$$4007$$

D
$$4008$$

Solution

Numbers in first group $$= 2$$
Numbers in second group $$= 4$$
Numbers in $$9^{th}$$ group $$= 18$$
Numbers in $$10^{th}$$ group $$= 20$$
Total numbers in $$1^{st}$$ to $$9^{th}$$ group
$$= 2 + 4 + .....18$$
$$= 2 (1 + ..... 9) = 2\times \dfrac {9\times 10}{2} = 90$$
Total numbers in $$10^{th}$$ group =110
Sum of first $$N$$ odd numbers $$=n^2$$
sum of numbers in $$10^{th}$$ group = sum till $$10^{th}$$ group - Sum till $$9^{th}$$ group
$$= 110^{2} - 90^{2}$$
$$= (110 + 90)(110 - 90)$$
$$= 200\times 20 = 4000$$.
Question 8
1 / -0

In a row of girls, Mridula is $$18^{th}$$ from the right and Sanjana is $$18^{th}$$ from the left. If both of them interchange their position, Sanjana becomes $$25^{th}$$ from the left, how many girls are there in the row?

A
$$40$$

B
$$41$$

C
$$42$$

D
$$35$$

Solution

Total number of girls $$=$$ (Position of Sanjana from left $$+$$ Position of Sanjana from right) $$- 1 = (25 + 18) - 1 = 42$$.
Question 9
1 / -0

A series whose $$n^{th}$$ term is $$\lgroup \dfrac{n}{x}\rgroup + y$$, the sum of r terms will be

A
$$\{ \dfrac{r (r + 1 )}{2x} \} + ry$$

B
$$\{ \dfrac{r (r - 1 )}{2x} \} $$

C
$$\{ \dfrac{r (r - 1 )}{2x} \} - ry$$

D
$$\{ \dfrac{r (r + 1 )}{2y} \} - rx$$

Solution

$$a_n = \dfrac{n}{x}+y$$

Sum of r terms

$$= \sum_{n=1}^{r}\left ( \dfrac{n}{x}+y \right )$$

$$=\sum_{n=1}^{r} \dfrac{n}{x}+\sum_{n=1}^{r} y $$

$$=\left ( \dfrac{1}{x}+\dfrac{2}{x}+.........+\dfrac{r}{x} \right )+\left ( y+y+.........+y \right )$$

$$= \dfrac{1}{x}(1+2+3+...........+r)+ry$$

$$= \dfrac{r(r+1)}{2x} +ry$$
Question 10
1 / -0

Choose the correct answer from the alternatives given.
A loss of $$20%$$ is incurred when $$6$$ articles are sold for a rupee. To gain $$20% $$ how many articles should be sold for a rupee?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

: Given that, SP = $$1$$. Loss$$ % = 20%$$ We know that, $$SP = 0.8$$ $$\times$$ $$CP$$
$$CP \, = \, \dfrac{5}{4}$$
To gain$$ 20%$$,SP=1.2 $$\displaystyle \times \, CP \, = \, \dfrac{120}{100} \, \times \, \dfrac{5}{4} \, = \, \dfrac{3}{2} \, \Rightarrow \, = \, \dfrac{3}{2}$$
For$$ Rs. 3/2$$ number of articles sold
For $$Rs. 1$$ number of articles to be sold $$= 6 $$ $$\times$$ $$ \dfrac{2}{3}$$ = $$ 4 $$ articles.

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Sequences and Series Test - 32

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Sequences and Series Test - 32

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SHARING IS CARING

Question 1

$$\displaystyle { 7 }^{ \dfrac { \left( n+1 \right) }{ 2 } }$$

$${ 7 }^{ { n }/{ 2 } }$$

$${ 7 }^{ \dfrac { \left( n-1 \right) }{ 2 } }$$

$${ 7 }^{ \left( \dfrac { n+1 }{ 3 } \right) }$$

Solution

Question 2

8

9

10

11

Solution

Question 3

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

$$\frac{1}{6}n^{2}(n^{2}-1)(2n-1)(2n+3)$$

$$\frac{1}{8}(n^{2}+1)(n^{2}+5)$$

$$\frac{1}{4}(n)(n+1)(n+2)(n+3)$$

Solution

Question 4

$$3 \sqrt 2$$

$$2 \sqrt 3$$

$$3 \sqrt 3$$

none of these

Solution

Question 5

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

$$ \cfrac { 8{ n }^{ 3 }+6{ n }^{ 2 }+n }{ 3 } $$

$$\displaystyle\frac{n(n + 1)^2}{4}$$

$$\displaystyle\left [ \frac{n(n + 1)}{4} \right ]^2$$

Solution

Question 6

$$11$$

$$9$$

$$16$$

$$5$$

Solution

Question 7

$$4000$$

$$4003$$

$$4007$$

$$4008$$

Solution

Question 8

$$40$$

$$41$$

$$42$$

$$35$$

Solution

Question 9

$$\{ \dfrac{r (r + 1 )}{2x} \} + ry$$

$$\{ \dfrac{r (r - 1 )}{2x} \} $$

$$\{ \dfrac{r (r - 1 )}{2x} \} - ry$$

$$\{ \dfrac{r (r + 1 )}{2y} \} - rx$$

Solution

Question 10

$$1$$

$$2$$

$$3$$

$$4$$

Solution

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