Sequences and Series Test

Result

Sequences and Series Test - 41

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

The sum of the series $$1^{3}+3^{3}+5^{3}+....$$ upto $$20$$ terms is

A
$$319600$$

B
$$321760$$

C
$$306000$$

D
$$347500$$

Solution
Question 2
1 / -0

The $$nth$$ term of the series $$4, 14, 30, 52, 80, 114, ...$$ is

A
$$n^{2} + n + 2$$

B
$$3n^{2} + n$$

C
$$3n^{2} - 5n + 2$$

D
$$(n + 1)^{2}$$

Solution
Question 3
1 / -0

The sum of $$(n+1)$$ terms of $$\frac { 1 }{ 1 } +\frac { 1 }{ 1+2 } +\frac { 1 }{ 1+2+3 } +.......is$$

A
$$\frac { n }{ n+1 } $$

B
$$\frac { 2n }{ n+1 } $$

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

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

Solution

$$ \dfrac{1}{1} + \dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+n(n+1)}$$
$$ \dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{n+1}$$
we find
$$ 1^{st}$$ term = $$ \dfrac{2}{1}-\dfrac{2}{2} = \dfrac{1}{1}$$
$$ 2^{nd}$$ term = $$ \dfrac{2}{2}-\dfrac{2}{3} = \dfrac{1}{3}$$
$$ 3^{rd}$$ term = $$ \dfrac{2}{3}-\dfrac{2}{4} = \dfrac{1}{6}$$
and so on
$$ n^{m}$$ term $$= \dfrac{2}{n}-\dfrac{2}{n+1}$$
$$ (n+1)^{m}$$ term $$ = \dfrac{2}{n+1}-\dfrac{2}{n+2}$$
Now finding the sum of 4 terms;
$$ \dfrac{2}{1}-\dfrac{2}{2}+\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}$$
we get, $$ \dfrac{2}{1} -\dfrac{2}{5}$$
So finding the sum of all $$ n+1 $$ terms
we get, $$ \dfrac{2}{1}-\dfrac{2}{n+2}= 2-\dfrac{2}{n+1}$$
$$ \Rightarrow \dfrac{2(n+1)-2}{n+1}$$
$$ \Rightarrow \dfrac{2n+2-2}{n+1}$$
$$ \Rightarrow \dfrac{2n}{n+1}$$
Hence, option (b) is correct.
Question 4
1 / -0

The three digits in 10! are:

A
$$800$$

B
$$700$$

C
$$500$$

D
$$600$$

Solution

As we know that $$10!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10$$
$$10!=10^2(3\times 4\times 6\times 7\times 8\times 9)$$
$$10!=10^2(3^4\times 2^6\times 7)$$
$$10!=10^2(81\times 64\times 7)$$
The last three digit in $$10!$$ is $$800$$
The correct option is A.
Question 5
1 / -0

The sum of the series $$\dfrac {5}{1 \cdot 2 \cdot 3}+\dfrac {7}{3 \cdot 4 \cdot 5}+\dfrac {9}{5 \cdot 6 \cdot 7}+....$$ is

A
$$\log \dfrac {8}{e}$$

B
$$\log \dfrac {e}{8}$$

C
$$\log2$$

D
$$\log \dfrac {1}{2}$$

Solution
Question 6
1 / -0

If the sum of the series $$1+\dfrac{3}{x}+\dfrac{9}{x^{2}}+\dfrac{27}{x^{3}}+....$$ to $$\infty$$ is a finite number then

A
$$x>3$$

B
$$x<-3$$

C
$$x<-3\ or\ x>3$$

D
$$None\ of\ these$$

Solution

Series: $$1+\dfrac{3}{{x}^{1}}+\dfrac{9}{{x}^{2}}+\dfrac{27}{{x}^{3}}+……\infty $$

This is a $$G.P$$

With $$Q A=1$$

$$r=\dfrac{3}{x}$$

$${ s }_{ n }=\dfrac { A }{ 1-r } =\dfrac { 1 }{ 1-\dfrac { 3 }{ x } } \quad \left\{ \because r<1 \right\} $$

$${s}_{n}=\dfrac{1\left(x\right)}{x-3}$$...... 1

Also $${s}_{n}=\dfrac{A}{r-1}=\dfrac{1}{3.1}=\dfrac{x}{3-x}$$......2

$$\left\{ \because r\quad >\quad 1 \right\} $$

From $$1$$ and $$2$$

We infer:

$${s}_{n}=\dfrac{1\left(-x\right)}{x-3}$$ is true for

$$\left\{ 3\quad <\quad x \right\}$$

And

$${s}_{n}=\dfrac{x}{3-x}$$ is true then

$$\left\{ x\quad <\quad 3 \right\}$$

Option $$x\left\{ x<-3\& x>3 \right\} $$
Question 7
1 / -0

The value of $$\sum_{n=1}^{\infty}\dfrac{1}{(3n-2)(3n+1)}$$ is equal to $$\dfrac{p}{q}$$, where $$p$$ and $$q$$ are relatively prime natural numbers. Then the value of $$(p^2+q^2)$$ is equal to

A
$$4$$

B
$$9$$

C
$$10$$

D
$$13$$

Solution

Given, $$\displaystyle S=\sum_{n=1}^{\infty }\frac{1}{(3n-2)(3n+1)}= \frac{1}{3}\sum_{n=1}^{\infty }[\frac{1}{3n-2}-\frac{1}{3n+1}]$$
$$S=\dfrac{1}{3}[(1-\dfrac{1}{4})+(\dfrac{1}{4}-\dfrac{1}{7})+(\dfrac{1}{7}-\dfrac{1}{10})+....]$$
$$S= \dfrac{1}{3}(1-0)= \dfrac{1}{3}$$
Comparing with $$S= \dfrac{p}{q}\Rightarrow p=1\, q=3$$
$$p^{2}+q^{2}=1^{2}+(3)^{2}=10$$
$$\therefore $$ Option C is correct
Question 8
1 / -0

If $$p:q:r=1:2:3\;then$$ $$\sqrt {5{p^2} + {q^2} + {r^2}} $$ is equal to:

A
4r

B
5p

C
2q

D
5

Solution

Consider the problem

Given, $$p:q:r=1:2:3$$

Therefore, $$\frac{p}{1} = \frac{q}{2} = \frac{r}{4} = k$$ (let)

So, $$p = k,q = 2k,r = 4k$$

Therefore,

$$\begin{array}{l} \sqrt { 5{ p^{ 2 } }+{ q^{ 2 } }+{ r^{ 2 } } } \\ =\sqrt { 5{ k^{ 2 } }+4{ k^{ 2 } }+16{ k^{ 2 } } } \\ =\sqrt { 25{ k^{ 2 } } } \\ 5k=5p \end{array}$$

Hence, Option $$B$$ is the correct answer which is $$5p$$
Question 9
1 / -0

The sum $$1 + 3 + 7 + 15 + 31 + \ldots $$ to $$n$$ term is

A
$${2^n} - 2 - n$$

B
$${2^{n - 1}} - 1 - n$$

C
$${2^{n + 1}} - 2 - n$$

D
None of these

Solution

Given series is 1+3+5+7 so on upto n
nth term of this series is $${2}^{n}-1$$
so,this series can be expressed as ({2}^{1}-1)+({2}^{2}-1)+({2}^{3}-1)so on upto ({2}^{n}-1)
$$\rightarrow({1}^{1}+{2}^{2}+{2}^{3}+---{2}^{n})-n$$
$$\rightarrow\dfrac{2({2}^{n}-1)}{2-1}-n$$
$$\rightarrow{2}^{n+1}-2-n$$
Question 10
1 / -0

Replace the question mark (?) with the correct option.

A
$$2$$

B
$$7$$

C
$$4$$

D
$$9$$

Solution