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Multiple choice maths average arithmetic mean of ap introduction to averages means

For a certain frequency table which has been partly reproduced here, the arithmetic mean was found to be Rs.28.07

Income (in Rs.) 15 20 25 30 35 40
Number of workers 8 12 ? 16 3 10

If he total number of workers is 75, then the missing frequencies are?

  1. 14, 15

  2. 15, 14

  3. 13, 16

  4. 12, 17
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice maths average arithmetic mean of ap introduction to averages means

The arithmetic mean of first ten natural numbers is

  1. $5.5$
  2. $6$
  3. $7.5$
  4. $10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The arithmetic mean of first ten natural numbers is the average of these natural numbers.


$\therefore$ arithmetic mean

   $=\dfrac{1+2+3+4+5+6+7+8+9+10}{10}$

   $=\dfrac{n(\dfrac{n+1}{2})}{10}\,\,\,\,\,\,\,\rightarrow Here \,\boxed{n=10}$

   $=\dfrac{10\times \dfrac{11}{2}}{10}$ 

   $=\dfrac{11}{2}=\boxed{5.5}\rightarrow $ option -A

Multiple choice maths average arithmetic mean of ap introduction to averages means

If AM between $\displaystyle p^{th}$ and $\displaystyle q^{th}$ terms of an AP be equal to the AM between $\displaystyle r^{th}$ and $\displaystyle s^{th}$ term of the AP, then $p + q$ is equal to

  1. $r + s$
  2. $\displaystyle \frac{r-s}{r+s}$
  3. $\displaystyle \frac{r+s}{r-s}$
  4. $r + s + 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know A.P formula for nth terms with 'a' as the first term and 'd' as the common difference as shown below:


${ t } _{ n }=a+\left( n-1 \right) d$

Also AM is given between two numbers a and b. 
       $A=\dfrac { a+b }{ 2 } $

So arithmetic mean of pth and qth terms of AP is as shown below:

$=\dfrac { a+\left( p-1 \right) d+a+\left( q-1 \right) d }{ 2 } $

Similarly we can have AM of rth term and sth term of AP as shown below:

$=\dfrac { a+\left( r-1 \right) d+a+\left( s-1 \right) d }{ 2 } $

Applying the given conditions we get,

$\dfrac { a+\left( p-1 \right) d+a+\left( q-1 \right) d }{ 2 } =\dfrac { a+\left( r-1 \right) d+a+\left( s-1 \right) d }{ 2 } $

      $\dfrac { a+pd-d+a+qd-d }{ 2 } =\dfrac { a+rd-d+a+sd-d }{ 2 } $

$a+pd-d+a+qd-d=a+rd-d+a+sd-d$

         $2a+d\left( p+q \right) -2d=2a+d\left( r+s \right) -2d$

                           $d\left( p+q \right) =d\left( r+s \right) d$

                                 $p+q=r+s$ 

Hence option A is correct.

Multiple choice maths average arithmetic mean of ap introduction to averages means
If $a+b+c+d+e+f=12$ then the maximum value of $ab+bc+cd+de+ef+fa$ is (a, b, c, d, e, f are non negative real numbers)
  1. $36$
  2. $24$
  3. $30$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
AM $\ge$ GM
$\cfrac { a+b+c+d+e+f }{ 6 } \ge { (abcdef) }^{ 1/6 }\\ { 2 }^{ 2 }\ge { (abcdef) }^{ 1/3 }\quad \quad \quad (1)\\ \cfrac { ab+bc+cd+de+ef+fa }{ 6 } \ge { (abcdef) }^{ 2/6 }\\ ab+bc+cd+de+ef+fa\ge 6{ (abcdef) }^{ 1/3 } \quad (2)$
Dividing $(2)$ by $(1)$
$ab+bc+cd+de+ef+fa \ge 24$
Multiple choice maths average arithmetic mean of ap introduction to averages means

The arithmetic mean of 1, 2, 3, ..., n, is

  1. $\displaystyle \frac{n-1}{2}$
  2. $\displaystyle \frac{n+1}{2}$
  3. $\displaystyle \frac{n}{2}$
  4. $\displaystyle \frac{n}{2}+1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Rightarrow$   We have the sequence, $1,2,3.......n$

$\Rightarrow$   This is an AP, with the initial term $a=1$ and the common difference $d=1$.
$\therefore$   The sum of $n$ terms of an AP is given by,
$\Rightarrow$  $S _n=\dfrac{n}{2}[2a+(n-1)d]$

$\Rightarrow$  $S _n=\dfrac{n}{2}[2\times 1+(n-1)\times 1]$

$\Rightarrow$  $S _n=\dfrac{n}{2}[2+(n-1)]$

$\Rightarrow$  $S _n=\dfrac{n}{2}[n+1]$
$\rightarrow$   Arithmetic mean of $n$ numbers $a _1,a _2,a _3,a _4,... a _n$ is given by the formula
$\Rightarrow$  $Arithmetic\,mean=\dfrac{a _1+a _2+a _3+a _4+...+a _n}{n}$

$\Rightarrow$  $Arithmetic\,mean=\dfrac{S _n}{n}$

$\Rightarrow$  $Arithmetic\, mean=\dfrac{\dfrac{n}{2}[n+1]}{n}$

$\therefore$     $Arithmetic\, mean=\dfrac{n+1}{2}$

Multiple choice maths average arithmetic mean of ap introduction to averages means

Find the arithmetic mean of the series $1, 3, 5,...........
(2n - 1)$

  1. $n$
  2. $2n$
  3. $n/2$
  4. $n - 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the given Arithmetic Progression,
First term $ = a  = 1 $
Common difference $ = 3 - 1 = 2 $

Let $ 2n-1 $ be the $  k $ th term.

Then $ {x} _{n} = a + (n-1)d $
$ => 2n-1 = 1 + (k-1)2  $
$ => 2n -1 = 1 +2k-2 $
$ => k = n $

So,  $ 2n-1 $ is the $ n $ th term.

Now, Sum of the given series upto 'n' terms $ = \frac {n}{2} (2a+(n-1)d) =n^2 $

so mean is = sum of series/total no of terms
$= n^2/n=n$

Multiple choice maths average arithmetic mean of ap introduction to averages means

Find the arithmetic mean of the series: $1,3,5 ........... (2n - 1)$

  1. $n$
  2. $2n$
  3. $\dfrac n2$
  4. $n - 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1,3,5,..., (2n-1)$

Total number of terms: $n$
Arithmetic mean: $\dfrac{1+3+5+\cdots+(2n-1)}{n}$

Using sum of first n terms of an AP:
$1+3+5+\cdots+(2n-1) = n^2$
$\therefore$ AM = $\dfrac{n^2}n = n$