Tag: series

Quizzes (0) Questions (44)

Questions Related to series

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

If  in traingle ABC $\cos 2B=\dfrac {\cos (A+C)}{\cos (A-C)}$, then 

  1. $\tan A, \tan B, \tan C$ are in $A.P$
  2. $\tan A, \tan B, \tan C$ are in $G.P$
  3. $\tan A, \tan B, \tan C$ are in $H.P$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given cos(2B) = cos(A+C)/cos(A-C). Using componendo and dividendo, (1-cos(2B))/(1+cos(2B)) = (cos(A-C)-cos(A+C))/(cos(A-C)+cos(A+C)). This simplifies to tan^2(B) = tan(A)tan(C), meaning tan(A), tan(B), tan(C) are in G.P.

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

A gentlemen invites a party of m + n $(m \neq n)$ friends to a dinner and places m at one table $T _1$ and n at another table $T _2$, the table being round. If not all people shall have the same neighbour n any two arrangement, then the number of ways in which he can arrange the guests, is 

  1. $\dfrac{(m+n)!}{4mn}$
  2. $\dfrac{1}{2} \dfrac{(m+n)!}{mn}$
  3. $2\dfrac{(m+n)!}{mn}$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The number of ways to arrange m people at a round table is (m-1)!. For two tables, it is (m-1)!(n-1)!. Dividing by the symmetry of the tables and considering the distinctness of the arrangements, the formula results in (m+n)!/(4mn).

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

$ \left{ a _ { n } \right} $ and $ \left{ b _ { n } \right} $ are two sequences given by $ a _ { n } = ( x ) ^ { 1 / 2 ^ { \circ } } + ( y ) ^ { 1 / 2 ^ { \circ } } $ and $ b _ { n } = ( x ) ^ { 1 / 2 ^ { 2 } } - ( y ) ^ { 1 / 2 ^ { \circ } } $ for all $ \mathrm { n } \in \mathrm { N } . $ The value of $ \mathrm { a } _ { 1 } \mathrm { a } _ { 2 } \mathrm { a } _ { 3 } \dots \ldots \ldots \mathrm { a } _ { \mathrm { n } } $ is equal to

  1. x-y
  2. $

    \frac { x + y } { b _ { n } }

    $
  3. $

    \frac { x - y } { b _ { n } }

    $
  4. $

    \frac { x y } { b _ { n } }

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

If $\displaystyle f(n+1)=\frac {2f(n)+1}{2}, n=1,2, .....$ and $f(1)=2$, then $f(101)= ..........$

  1. $53$
  2. $52$
  3. $51$
  4. $50$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:
 $\Rightarrow f(n+1)=\dfrac {2f(n)+1}{2}$
and $\Rightarrow  f(1)=2$
For  $n=1$,$ f(2)=\dfrac {2f(1)+1}{2}=\dfrac {5}{2}$


For  $n=2$, $f(3)=\dfrac {2f(2)+1}{2}=3$

For $n=3$,$ f(4)=\dfrac {2f(3)+1}{2}=\dfrac {6+1}{2}=\dfrac {7}{2}$

So, $\Rightarrow  f(1), f(2), f(3), f(4), ....=2, \dfrac {5}{2}, 3, \dfrac {7}{2},.....$

$\therefore  f(n)=\dfrac {3+n}{2}$

$\Rightarrow  f(101)=\dfrac {3+101}{2}=52$

$\Rightarrow  f(101)=52$
Hence, option 'B' is correct.

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

If $a, b, c$ are in AP, $b - a, c - b$ and $a$ are in GP, then $a : b : c$ is

  1. $1 : 2 : 3$
  2. $1 : 3 : 5$
  3. $2 : 3 : 5$
  4. $1 : 2 : 4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $a,b,c$ are in AP and $b-a, c-b, a$ are in GP.
Therefore, $2b = a + c$ and $(c - b)^{2} = (b - a)a$
$\Rightarrow  (b - a)^{2} = (b - a)a$
$\Rightarrow b = 2a$
$\Rightarrow c = 3a$
Thus, $a : b : c = 1 : 2 : 3$.

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

Let $x _{1}, x _{2}, .....x _{n}$ be in an AP of $x _{1} + x _{4} + x _{9} + x _{11} + x _{20} + x _{22} + x _{27} + x _{30} = 272$, then $x _{1} + x _{2} + x _{3} + ..... + x _{30}$ is equal to

  1. $1020$
  2. $1200$
  3. $716$
  4. $2720$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If an AP consist of $30$ terms, Then $x _{1} + x _{30} = x _{4} + x _{27} = x _{9} + x _{22} = x _{11} + x _{20}$
$\because x _{1} +x _{4} + x _{9} + x _{11} + x _{20} + x _{27} + x _{30} = 272$
$\Rightarrow (x _{1} + x _{30}) + (x _{4} + x _{27}) + (x _{9} + x _{22}) + (x _{11} + x _{26}) = 272$
$\Rightarrow 4(x _{1} + x _{30}) = 272$
$\Rightarrow x _{1} + x _{30} = \dfrac {272}{4} = 68$
$S _{30} = \dfrac {30}{2} (x _{1} + x _{30}) = 15\times 68 = 1020$

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

$S _{n} = 1^{3} + 2^{3} + ..... + n^{3}$ and $T _{n} = 1 + 2 + ..... + n$, then

  1. $S _{n} = T _{n}$
  2. $S _{n} = T _{n}^{4}$
  3. $S _{n} = T _{n}^{2}$
  4. $S _{n} = T _{n}^{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$S _{n} = 1^{3} + 2^{3} + ...... + n^{3} = \sum n^{3}$
$T _{n} = 1 + 2 + ..... + n = \sum n$
$S _{n} = \sum n^{3} = \left [\dfrac {n(n + 1)}{2}\right ]^{2}$
$\Rightarrow S _{n} = \left {\sum n\right }^{2} = T _{n}^{2}$

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

If for $n\in I, n > 10; 1+(1+x)+(1+x)^2+.....+(1+x)^n=\displaystyle\sum^n _{k=0}a _k\cdot x^k, x\neq 0$ then?

  1. $\displaystyle\sum^n _{k=0}a _k=2^{n+1}$
  2. $a _{n-2}=\dfrac{n(n+1)}{2}$
  3. $a _p > a _{p-1}$ for $p < \dfrac{n}{2}, p \in N$
  4. $(a _9)^2-(a _8)^2={^{n+2}C _{10}}({^{n+1}C _{10}}-{^{n+1}C _9})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum is a geometric series: ( (1+x)^(n+1) - 1 ) / ( (1+x) - 1 ) = ( (1+x)^(n+1) - 1 ) / x. The sum of coefficients a_k is the value of the polynomial at x=1, which is ((1+1)^(n+1) - 1) / 1 = 2^(n+1) - 1. Option A is the standard result for this series.

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

Identify the function for the following sequence $4, 10, 18, 28...$

  1. $2n(n+3)$
  2. $n(n+3)$
  3. $n(n-3)$
  4. $n^2(n+3)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When $n = 1$, $1(1+3)=4$ i.e. $1(1+3)$
When $n = 2$, $2(2+3)=10$ i.e. $2(2+3)$
When $n = 3$, $3(3+3)=18$ i.e. $3(3+3)$
When $n = 4$, $4(4+3)=28$ i.e. $4(4+3)$
So, $4, 10, 27, 28..$ is the function for the sequence is $n(n+3)$.

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

Identify the sequence for the following function $n(n+3)$.

  1. $4, 10, 18, 28..$
  2. $4, 12, 18, 28..$
  3. $2, 10, 18, 28..$
  4. $4, 10, 18, 38..$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given function is $n(n+3)$
When $n = 1$, $1(1+3)=4$
When $n = 2$, $2(2+3)=10$
When $n = 3$, $3(3+3)=27$
When $n = 4$, $4(4+3)=28$....
So, $4, 10, 27, 28..$ is the function for the sequence is $n(n+3)$.