Tag: binomial theorem, sequence and series

Quizzes (6) Questions (46)

Questions Related to binomial theorem, sequence and 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.