Tag: binomial theorem, sequence and series

Questions Related to binomial theorem, sequence and series

If a series consists only a finite number of terms it is called a ................

  1. infinite series

  2. finite series

  3. real number

  4. geometric series


Correct Option: B
Explanation:

If a series consists only a finite number of terms it is called a Finite series.

Hence, the answer is finite series.

If the sum of first $75$ terms of an AP is $2625$, then the $38^{th}$ term of an AP is

  1. $39$

  2. $37$

  3. $35$

  4. $38$


Correct Option: C
Explanation:

$S _{75} = 2625$
$\Rightarrow \dfrac {75}{2}(2a + (75 - 1)d) = 2625$
$\Rightarrow 2(a + 37d) = 35\times 2$
$\Rightarrow a + 37d = 35$
$T _{38} = a + (38 - 1)d = a + 37 d$
$= 35$

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. $None\ of\ these$


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 

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

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

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

  4. none


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

  1. x-y

  2. $

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

    $

  3. $

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

    $

  4. $

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

    $


Correct Option: A

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. $50$


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

  1. $1 : 2 : 3$

  2. $1 : 3 : 5$

  3. $2 : 3 : 5$

  4. $1 : 2 : 4$


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

  1. $1020$

  2. $1200$

  3. $716$

  4. $2720$


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

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

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

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

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


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?

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


Correct Option: A