Tag: series

Questions Related to series

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

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

  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. $None\ of\ these$

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Correct Option: B

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 

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

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

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

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

  4. none

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Correct Option: A

$ \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

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

  1. x-y

  2. $

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

    $

  3. $

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

    $

  4. $

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

    $

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Correct Option: A

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

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

  1. $53$

  2. $52$

  3. $51$

  4. $50$

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Correct Option: B
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.

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

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

  1. $1 : 2 : 3$

  2. $1 : 3 : 5$

  3. $2 : 3 : 5$

  4. $1 : 2 : 4$

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Correct Option: A
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, $a : b : c = 1 : 2 : 3$.

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

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

  1. $1020$

  2. $1200$

  3. $716$

  4. $2720$

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Correct Option: A
Explanation:

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

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

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

  1. $S _{n} = T _{n}$

  2. $S _{n} = T _{n}^{4}$

  3. $S _{n} = T _{n}^{2}$

  4. $S _{n} = T _{n}^{3}$

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Correct Option: C
Explanation:

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

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?

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

  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. $(a _9)^2-(a _8)^2={^{n+2}C _{10}}({^{n+1}C _{10}}-{^{n+1}C _9})$

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Correct Option: A

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

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

  1. $2n(n+3)$

  2. $n(n+3)$

  3. $n(n-3)$

  4. $n^2(n+3)$

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Correct Option: B
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 $n(n+3)$.

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

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

  1. $4, 10, 18, 28..$

  2. $4, 12, 18, 28..$

  3. $2, 10, 18, 28..$

  4. $4, 10, 18, 38..$

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Correct Option: A
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 $n(n+3)$.