Sequences and Series Test

Result

Sequences and Series Test - 46

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

$$14, 12, 21, 59, 231, 1149, ?$$

A
$$6987$$

B
$$6787$$

C
$$6887$$

D
$$6687$$

E
$$6587$$

Solution

$$14,12,21,54,231,1149,..............$$
$$14\times 1-2=14-2=12$$
$$12\times 2-3=24-3=21$$
$$21\times 3-4=63-4=59$$
$$59\times 4-5=236-5=231$$
$$231\times 5-6=1155-6=1149$$
$$1149\times 6-7=6894-7=6887$$
ANSWER is C
Question 2
1 / -0

Find the next term of the series $$1728, 2744, 4096, 5832, 8000, 10648, ?$$

A
$$2167$$

B
$$13824$$

C
$$15625$$

D
$$9261$$

E
$$17576$$

Solution

$$1728,2744,4096,5832,8000,10648.....$$
$$12^3\rightarrow 1728$$
$$14^3\rightarrow 2744$$
$$16^3\rightarrow 4096$$
$$18^3\rightarrow 5832$$
$$20^3\rightarrow 8000$$
$$22^3\rightarrow 10648$$
$$24^3\rightarrow 13824$$
ANSWER IS B
Question 3
1 / -0

Find the missing number from the given alternatives.

A
$$39,116$$

B
$$52,156$$

C
$$30,117$$

D
$$31,116$$

Solution

$$ 13\quad\quad 39\\26\quad\quad 78\\?\quad\quad ?$$
The pattern satisfies
$$x\quad\quad 3x$$
Therefore correct option is
$$52\quad\quad 156$$
Question 4
1 / -0

If $$1,\ a_{1},\ a_{2},\ a_{3},......$$ and $$a_{8}$$ are nine, ninth roots of unity (taken in counter- clockwise sequence ) then $$|(2-a_{1})(2-a_{3})(2-a_{7})|=$$

A
$$\sqrt {225}$$

B
$$\sqrt {511}$$

C
$$\sqrt {1023}$$

D
$$15$$

Solution

undefined
Question 5
1 / -0

$$8, 31, 122, 485, 1936, 7739, ?$$

A
$$30950$$

B
$$46430$$

C
$$34650$$

D
$$42850$$

E
$$38540$$

Solution
Question 6
1 / -0

The sum of infinite series $$\begin{vmatrix} 1 & 2 \\ 6 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 2 } & 2 \\ 2 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 4 } & 2 \\ \frac { 2 }{ 3 } & 4 \end{vmatrix}+.........$$

A
$$-10$$

B
$$0$$

C
$$10$$

D
$$\infty$$

Solution
Question 7
1 / -0

If $$\dfrac{\pi}{4}-1+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{1}{13}+=0$$ then value of $$\dfrac{1}{1\times3}+\dfrac{1}{5\times7}+\dfrac{1}{9\times11}+\dfrac{1}{13\times15}+..$$ is

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

B
$$\dfrac{\pi}{6}$$

C
$$\dfrac{\pi}{4}$$

D
$$\dfrac{\pi}{34}$$
Question 8
1 / -0

Let $$t_{r}=\frac{r}{1+r^{2}+r^{4}}$$ then, $$\lim_{n\rightarrow \infty }\sum_{r=1}^{n}t_{r}$$ equals

A
$$\frac{1}{4}$$

B
1

C
$$\frac{1}{2}$$

D
None of these

Solution

$$t_r=\dfrac{r}{1+r^2+r^4}$$
$$=\dfrac{r}{r^4+2r^2+1-r^2}$$
$$=\dfrac{r}{(r^2+1)^2-r^2}=\dfrac{r}{(r^2+1-r)(r^2+1+r)}$$
$$=\dfrac{1}{2}\dfrac{2r}{(r^2+1-r)(r^2+r+1)}$$
$$=\dfrac{1}{2}\dfrac{(r^2+r+1)-(r^2-r+1)}{(r^2-r+1)(r^2+r+1)}$$
$$=\dfrac{1}{2}\left[\dfrac{1}{r^2-r+1}-\dfrac{1}{r^2+r+1}\right]$$
$$=\dfrac{1}{2}\left[\dfrac{1}{r^2-r+1}-\dfrac{1}{(r+1)^2-(r+1)+1}\right]$$
$$\displaystyle\sum^n_{r=1}t_r=\dfrac{1}{2}\left[\dfrac{1}{1-1+1}-\dfrac{1}{(n+1)^2-(n+1)+1}\right]$$
$$=\dfrac{1}{2}\left[1-\dfrac{1}{n^2+n+1}\right]=\dfrac{1}{2}\left[\dfrac{n^2+n+1-1}{n^2+n+1}\right]$$
$$\displaystyle\sum t_n=\dfrac{1}{2}\left[\dfrac{1+\dfrac{1}{n}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}\right]$$
$$\underset{n\rightarrow \infty}{lt}\displaystyle\sum t_r=\dfrac{1}{2}\left[\dfrac{1+0}{1+0+0}\right]$$
$$=\dfrac{1}{2}$$.
Question 9
1 / -0

$$499, 622, 868, 1237, 1729, 2344, ?$$

A
$$3205$$

B
$$3082$$

C
$$2959$$

D
$$3462$$

E
$$2876$$

Solution

$$499, 622, 868, 1237, 1729, 2344,$$ ______
$$499+1\times 123=499+123=622$$
$$622+2\times 123=622+146=868$$
$$868+3\times 123=868+369=1237$$
$$1237+4\times 123=1237+492=1729$$
$$1729+5\times 123=1729+615=2344$$
$$2344+6\times 123=2344+738=3082$$.
Question 10
1 / -0

the value of $$\sum\limits_{n = 1}^\infty {\frac{{(3n + 1)}}{{n(4n + 4){4^n}}}} $$ is equal to

A
$$\frac{1}{5}$$

B
$$\frac{1}{4}$$

C
$$\frac{1}{25}$$

D
$$\frac{1}{16}$$