Tag: maths

Questions Related to maths

The sum of $2n$ terms of a series of which every even term is $'a'$ times the terms before it, and every odd term $'c'$ times the terms before it, the first term being unity, is

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\dfrac { \left( 1-a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac-1 }$

  2. $\dfrac { \left( 1+a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac+1 }$

  3. $\dfrac { \left( 1+a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac-1 }$

  4. $None\ of\ these$

Show answer
Correct Option: C
Explanation:

$\begin{array}{l} { T _{ 1 } }=1 \ { T _{ 2 } }=a \ { T _{ 3 } }=Ca \ { T _{ 4 } }=C{ a^{ 2 } } \ { T _{ 2n } }=a\frac { { 2n } }{ 2 } \cdot C\frac { { 2n } }{ 2 } -1={ a^{ n } }{ C^{ n-1 } } \ { 5 _{ 2n } }=1+a+ca.....{ a^{ n } }{ C^{ n-1 } } \ =1+\left[ { a+c{ a^{ 2 } }+{ c^{ 2 } }{ a^{ 3 } }...{ a^{ n } }{ c^{ n-1 } } } \right]  \ +\left[ { ca+{ c^{ 2 } }{ a^{ 2 } }+{ c^{ 3 } }{ a^{ 3 } }.....{ c^{ n-1 } }{ a^{ n-1 } } } \right]  \ =1+\frac { { a\left( { { a^{ n } }{ c^{ n-1 } } } \right)  } }{ { ac-1 } } +\frac { { ac\left( { { a^{ n-1 } }{ c^{ n-1 } }-1 } \right)  } }{ { ac-1 } }  \ =\frac { { \left( { { a^{ n } }{ c^{ n } }-1 } \right) \left( { a+1 } \right)  } }{ { ac-1 } }  \end{array}$

The sum of $10$ terms of the series $0.7 + .77 + .777 + \ldots \ldots \ldots$ is

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\dfrac { 7 } { 9 } \left( 89 + \dfrac { 1 } { 10 ^ { 10 } } \right)$

  2. $\dfrac { 7 } { 81 } \left( 89 + \dfrac { 1 } { 10 ^ { 10 } } \right)$

  3. $\dfrac { 7 } { 81 } \left( 89 + \dfrac { 1 } { 10 ^ { 9 } } \right)$

  4. $\dfrac { 7 } { 9 } \left( 89 + \dfrac { 1 } { 10 ^ { 9 } } \right)$

Show answer
Correct Option: B
Explanation:
$0.7+0.77+0.777+......$
$=7\left( 0.1+0.11+0.111+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 0.9+0.99+0.999+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 1-0.1+1-0.1+1-0.001+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 10-\left( 0.1+0.01+0.001+...... \right)  \right) $
$=\dfrac { 7 }{ 9 } \left( 10-\dfrac { 0.1\left( 1-{ 10 }^{ -10 } \right)  }{ 1-0.1 }  \right) =\frac { 7 }{ 9 } \left( 10-\dfrac { 0.1\left( { 10 }^{ 10 }-1 \right)  }{ 0.9\times { 10 }^{ 10 } }  \right) =\dfrac { 7 }{ 9 } \left( 10-\dfrac { 1 }{ 9 } +\dfrac { 1 }{ 9\times { 10 }^{ 10 } }  \right) $
$=\dfrac { 7 }{ 9 } \left( \dfrac { 89 }{ 9 } +\dfrac { 1 }{ 9\times { 10 }^{ 10 } }  \right) $
$=\dfrac { 7 }{ 81 } \left( 89+\dfrac { 1 }{ { 10 }^{ 10 } }  \right) $      [B]

The sum of series $\displaystyle \frac{3}{4} + \frac{15}{16} + \frac{63}{64}+ ..... $ up to $n$ terms is

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\displaystyle n - \frac{4^n}{3} - \frac{1}{3}$

  2. $\displaystyle n + \frac{4^{-n}}{3} - \frac{1}{3}$

  3. $\displaystyle n + \frac{4^n}{3} - \frac{1}{3}$

  4. $\displaystyle n - \frac{4^{-n}}{3} - \frac{1}{3}$

Show answer
Correct Option: B
Explanation:

For $n=1$, we have
$\displaystyle n - \dfrac{4^n }{3} - \dfrac{1}{3} = 1  - \dfrac{4}{3} - \dfrac{1}{3} = - \dfrac{2}{3}$
$\displaystyle n + \dfrac{4^n}{3} - \dfrac{1}{3} = 1 + \dfrac{4}{3} - \dfrac{1}{3} = 2$
$n - \displaystyle \dfrac{4^{-n}}{3} + \dfrac{1}{3} = 1 - \dfrac{4^{-1}}{3} + \dfrac{1}{3}= \dfrac{5}{4}$
Also, for $n = 2$, we have
$ \displaystyle n + \dfrac{4^{-n}}{3} - \dfrac{1}{3} = 2 + \dfrac{1}{48} - \dfrac{1}{3} = \dfrac{27}{16}$ and $\displaystyle \dfrac{3}{4} + \dfrac{15}{16} = \dfrac{27}{16}$
Hence, option (b) is correct.
ALTER We have,
$\displaystyle \dfrac{3}{4} + \dfrac{15}{16} + \dfrac{63}{64}+ ..... $ to n terms
$= \displaystyle \dfrac{2^2 - 1}{2^2} + \dfrac{2^4 - 1}{2^4} + \dfrac{2^6 - 1}{2^6}+ .... $ to n terms.
$= \displaystyle \left ( 1 - \dfrac{1}{2^2} \right ) + \left ( 1 - \dfrac{1}{2^4} \right ) + \left( 1 - \dfrac{1}{2^6} \right ) + ..... $ to n terms
$= n - \left \{ \dfrac{1}{2^2} + \dfrac{1}{2^4} + \dfrac{1}{2^6} + .... \text{to n terms} \right \}$
$= n \displaystyle - \dfrac{1}{2^2} \left \{ \dfrac{1 - \left (\dfrac{1}{2^2} \right )^n }{1 - \dfrac{1}{2^2}} \right \}$
$= \displaystyle n - \dfrac{1}{3} (1 - 4^{-n})$
$= n + \displaystyle \dfrac{4^{-n}}{3} - \dfrac{1}{3}$

If the sum of $n$ terms of a GP (with common ratio $r$) beginning with the $\displaystyle p^{th}$ term is $k$ times the sum of an equal number of the same series beginning with the $\displaystyle q^{th}$ term, then the value of $k$ is

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\displaystyle r^{p/q}$

  2. $\displaystyle r^{q/p}$

  3. $\displaystyle r^{p-q}$

  4. $\displaystyle r^{p+q}$

Show answer
Correct Option: C
Explanation:

$p^{th}$ term of the series  $=ar^{p-1}$     ($a$ is first term)

$q^{th }$ term of th series $= ar^{q-1}$
Sum of $n$ term beginning with $p^{th} $ term 
$=\dfrac{ar^{p-1}(r^n - 1)}{r-1}$
Sum of $n$ term beginning with $q^{th} $ term 
$=\dfrac{ar^{q-1}(r^n - 1)}{r-1}$
Sum of $n$ term beginning with $p^{th} $ term $= k$ (sum of $n$ term beginning with $q^{th} $ term )
Thus $\dfrac{ar^{p-1}(r^n - 1)}{r-1}$$=k\dfrac{ar^{q-1}(r^n - 1)}{r-1}$
$\Rightarrow k = \dfrac{r^{p-1}}{r^{q-1}}$
$\Rightarrow k= r^{p-q}$

The sum of $1 + \dfrac {2}{5} + \dfrac {3}{5^{2}} + \dfrac {4}{5^{3}} + ....$ up to $n$ terms is

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\dfrac {25}{16} - \dfrac {4n + 5}{16\times 5^{n - 1}}$

  2. $\dfrac {3}{4} - \dfrac {2n + 5}{16\times 5^{n + 1}}$

  3. $\dfrac {3}{7} - \dfrac {3n + 5}{16\times 5^{n - 1}}$

  4. $\dfrac {1}{2} - \dfrac {5n + 1}{3\times 5^{n + 2}}$

Show answer
Correct Option: A
Explanation:
Let $S=1+\cfrac { 2 }{ 5 } +\cfrac { 3 }{ { 5 }^{ 2 } } +\cfrac { 4 }{ { 5 }^{ 3 } } +...\quad \quad (1)$
Multiplying $S$ with $\cfrac{1}{5}$ we get
$\cfrac { 1 }{ 5 } S=\cfrac { 1 }{ 5 } +\cfrac { 2 }{ { 5 }^{ 2 } } +\cfrac { 3 }{ { 5 }^{ 3 } } +\cfrac { 4 }{ { 5 }^{ 4 } } +...\quad \quad (2)$
Subtracting $(2)$ from $(1)$
$\quad \quad \quad S\;\;=1+\cfrac { 2 }{ 5 } +\cfrac { 3 }{ { 5 }^{ 2 } } +\cfrac { 4 }{ { 5 }^{ 3 } } +....{ T } _{ n }\\ \underline { \quad \quad -\cfrac { 1 }{ 5 } S=-\left[ \cfrac { 1 }{ 5 } +\cfrac { 2 }{ { 5 }^{ 2 } } +\cfrac { 3 }{ { 5 }^{ 3 } } +...{ T } _{ n-1 } \right] -{ T } _{ n } } \\ \left( 1-\cfrac { 1 }{ 5 }  \right) S=1+\cfrac { 1 }{ 5 } +\cfrac { 1 }{ { 5 }^{ 2 } } +\cfrac { 1 }{ { 5 }^{ 3 } } +.....-{ T } _{ n-1 }\\ \left( 1-\cfrac { 1 }{ 5 }  \right) S=\cfrac { (1)\left( 1-\cfrac { 1 }{ { 5 }^{ n } }  \right)  }{ \left( 1-\cfrac { 1 }{ 5 }  \right)  } -\cfrac { n }{ { 5 }^{ n-1 } } \\ \cfrac { 4 }{ 5 } S=\cfrac { 5 }{ 4 } -\cfrac { (4n+5) }{ 4\times { 5 }^{ n } } \\ S=\cfrac { 25 }{ 16 } -\cfrac { (4n+5) }{ 16\times { 5 }^{ n-1 } } $

In a $G.P$. the ratio of the sum of the first eleven terms to the sum of last eleven terms is $\displaystyle \frac{1}{8}$ and the ratio of the sum of all terms without the first nine to the sum of all the terms without the last nine is $2$. Then the number of terms of the $G.P$ is

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $15$

  2. $43$

  3. $38$

  4. $56$

Show answer
Correct Option: C
Explanation:

We have: 
$\dfrac { \frac { a({ r }^{ 11 }-1) }{ r-1 }  }{ \frac { a{ r }^{ n-11 }({ r }^{ 11 }-1) }{ r-1 }  } =\dfrac { 1 }{ 8 }$
$\Rightarrow { r }^{ n-11 }=8$ ...(i)
Also:
$\dfrac { \frac { a{ r }^{ 9 }({ r }^{ 11 }-1) }{ (r-1) }  }{ \frac { a({ r }^{ n-9 }-1) }{ (r-1) }  } =2$
$\Rightarrow { r }^{ 9 }=2 $
$\Rightarrow r={ 2 }^{ \frac { 1 }{ 9 }  }$ ...(ii)
Substituting (ii) in (i):
${ 2 }^{ \frac { n-11 }{ 9 }  }={ 2 }^{ 3 }$
$\Rightarrow \dfrac { n-11 }{ 9 } =3$
$\Rightarrow n=38$
Hence, (c) is correct.

In a single throw of die, what is the probability of getting a number greater than 3 ?

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. $\dfrac12$

  2. $\dfrac23$

  3. $\dfrac13$

  4. none of these

Show answer
Correct Option: A

An urn contains 2 red, 3 green and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue.

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. $\dfrac57$

  2. $\dfrac{10}{21}$

  3. $\dfrac27$

  4. $\dfrac{11}{21}$

Show answer
Correct Option: B

If the chance that a vessel arrives safely at a port is $\dfrac 9{10}$ then what is the chance that out of $5$ vessels expected at least $4$ will arrive safely?

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. $\dfrac {14 \times 9^4}{10^5}$

  2. $\dfrac {15 \times 9^5}{10^4}$

  3. $\dfrac {14 \times 9^3}{10^4}$

  4. $\dfrac {14 \times 9^6}{10^5}$

Show answer
Correct Option: A

A box contains nine bulbs out of which $4$ are defective. If four bulbs are chosen at random, find the probability that all the four bulbs are defective.

maths introduction of probability theory types of events some more terms in probability events and its algebra

  1. $\dfrac{62}{63}$

  2. $\dfrac{125}{126}$

  3. $\dfrac{1}{63}$

  4. $\dfrac{1}{126}$

Show answer
Correct Option: D