Sequences and Series Test

Result

Sequences and Series Test - 61

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

Find the missing term in the series given below. $$12,\ 13,\ 18,\ 19,\ 24,\ 25 $$?

A
$$27$$

B
$$30$$

C
$$29$$

D
$$26$$

Solution

$$12,13,18,19,24,25$$
difference between $$12,13=1$$
difference between $$13,18=5$$
Similarly
difference between $$18,19=1$$
difference between $$19,24=5$$
similarly
difference between $$24,25=1$$
so to get next number of $$25$$ add $$5$$ i.e., $$25+5=30$$
Question 2
1 / -0

The sum of first $$9$$ terms of the series
$$\dfrac { { 1 }^{ 3 } }{ 1 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 } }{ 1+3 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 } }{ 1+3+5 } +\dots$$

A
$$96$$

B
$$142$$

C
$$192$$

D
$$71$$

Solution
Question 3
1 / -0

Sum of the series $$ S = 1^{2}-2^{2}+3^{2}-4^{2}+......-2002^{2}+2003^{2}$$ is

A
2007006

B
1005004

C
2000506

D
none of these

Solution
Question 4
1 / -0

If $$\dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+........\infty =\dfrac{\pi^{2}}{6}$$ then $$\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+......\infty$$

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

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

C
$$\dfrac{\pi^{2}}{3}$$

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

Solution
Question 5
1 / -0

Sum of series $$\displaystyle \sum^{n}_{r=1}(r^{2}+1)_{r!}$$ is

A
$$(n+1)!$$

B
$$(n+2)!-1$$

C
$$n(n+2)!$$

D
$$(n+1)!-1$$
Question 6
1 / -0

Sum of the series
$$1+2.2+3.2^{2}+...+100.2^{10}=$$

A
$$100.2^{100}+1$$

B
$$99.2^{100}+1$$

C
$$99.2^{100}-1$$

D
$$100.2^{100}-1$$

Solution
Question 7
1 / -0

$$\dfrac {1}{1.4}+\dfrac {1}{4.7}+\dfrac {1}{7.10}+...+\dfrac {1}{(3n-5)(3n-2)}$$

A
$$\dfrac {n-1}{3n+2}$$

B
$$\dfrac {n-1}{3n-2}$$

C
$$\dfrac {n-1}{3n-1}$$

D
$$\dfrac {n-1}{3n-4}$$

Solution
Question 8
1 / -0

The sum of the series $$1+\frac { 1.3 }{ 6 } +\frac { 1.3.5 }{ 6.8 } +...\infty $$ is

A
$$4$$

B
$$0$$

C
$$\infty $$

D
$${ (4) }^{ \frac { 2 }{ 3 } }$$

Solution
Question 9
1 / -0

Sum series $$S=1+\frac{3}{2}+\frac{5}{2^{2}}+\frac{7}{2^{3}}+....\infty $$ is

A
12

B
6

C
3

D
10

Solution
Question 10
1 / -0

The $${ 2006 }^{ th }$$ digit in the sequence $$12345678910111213$$.....is

A
$$4$$

B
$$7$$

C
$$0$$

D
$$5$$

Solution

All the natural numbers starting from $$1$$ are enlisted:
So number of $$1$$ digit numbers $$= 9$$
Number of $$2$$ digit numbers $$= 90$$
To calculate number of digits we need to multiply by $$2$$
Number of $$3$$ digit numbers starting from $$100$$ till $$699 = 600$$(to calculate number of digits we need to multiply by $$3$$)
So, starting from $$1$$ to $$699$$, total number of digits will be
$$9 + 2\times 90 + 3\times 600 = 1989$$.
If we also add $$700-704$$ we get total $$1989 + 15 = 2004$$ digits enlisted.
Now the $$2006$$ digit in the sequence is clearly evident that would be the second digit of $$705$$ which is $$0$$.