Tag: arithmetic mean of ap

Questions Related to arithmetic mean of ap

Arithmetic mean of $2$ and $8$ is

maths average arithmetic mean of ap introduction to averages means

  1. $5$

  2. $10$

  3. $16$

  4. $3.2$

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Correct Option: A
Explanation:

Arithmetic mean of $a$ and $b$ is $\dfrac{a+b}{2}$

$\therefore$ arithmetic mean of 2 and 8 is $\dfrac{2+8}{2}=\dfrac{10}{2}=5$
Hence, option $A$ is correct.

State the following statement is True or False
Arithmetic mean of first five natural numbers is $3$.

maths average arithmetic mean of ap introduction to averages means

  1. True

  2. False

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Correct Option: A
Explanation:

First five natural numbers are $1 ,2 , 3 ,4, 5$

arithmetic mean = $\dfrac { sum\quad of\quad numbers }{ count\quad of\quad numbers } $
$ = \dfrac{1 + 2 + 3 + 4  + 5 }{ 5} $
$=\dfrac { 15 }{ 5 } =3$
Arithmetic mean of first five natural number is 3.
Hence the given statement is true.

The arithmetic mean of the squares of the first $n$ natural numbers is

maths average arithmetic mean of ap introduction to averages means

  1. $\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } $

  2. $\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 2 } $

  3. $\dfrac { \left( n+1 \right) \left( 2n+1 \right) }{ 6 } $

  4. $\dfrac { \left( n+1 \right) \left( 2n+1 \right) }{ 3 } $

Show answer
Correct Option: C
Explanation:

Sum of squares of the first $n$ natural numbers $=\dfrac { n(n+1)(2n+1) }{ 6 } $


Their arithmetic mean is $\dfrac { Sum }{ n } =\dfrac { n(n+1)(2n+1) }{ 6n } =\dfrac { (n+1)(2n+1) }{ 6 } $
Hence, C is correct.

The middle terms , if four different numbers are in proportion are called ______ .

maths average arithmetic mean of ap introduction to averages means

  1. Antecedents

  2. Means

  3. Extremes

  4. Consequents

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Correct Option: B
Explanation:

$\Rightarrow$  The middle terms  of four different numbers are in proportion are called $Means.$

$\Rightarrow$  A proportion is simply a statement that two ratios are equal. It can be written in two ways:  $\dfrac{a}{b} = \dfrac{c}{d}$ or  $a:b = c:d.$
$\Rightarrow$  Here, outer terms $a$ and $d$ called extremes.
$\Rightarrow$  Middle terms $b$ and $c$ are called means.

The arithmetic mean between $\cfrac { x+a }{ x } $ and $\cfrac { x-a }{ x } $ when $x\ne 0$, is (the symbol $\ne$ means "not equal to"):

maths average arithmetic mean of ap introduction to averages means

  1. $2$, if $a\ne 0$

  2. $1$

  3. $1$, only if $a=0$

  4. $\dfrac {a}{x}$

  5. $x$

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Correct Option: B

The arithmetic mean (average) of a set of $50$ numbers is $38$. If two numbers, namely, $45$ and $55$, are discarded, the mean of the remaining set of numbers is :

maths average arithmetic mean of ap introduction to averages means

  1. $36.5$

  2. $37$

  3. $37.2$

  4. $37.5$

  5. $37.52$

Show answer
Correct Option: D
Explanation:

$Arithmetic\quad sum=\cfrac { sum\quad of\quad numbers }{ number\quad of\quad numbers } $
$\therefore S=50\times 38=1900;\quad \therefore x=\cfrac { 1900-45-55 }{ 48 } =37.5$

If $A _1,A _2$ be two arithmetic means between $\dfrac{1}{3}$ and $\dfrac{1}{24}$, then their value are 

maths average arithmetic mean of ap introduction to averages means

  1. $\dfrac{7}{72},\dfrac{5}{36}$

  2. $\dfrac{17}{72},\dfrac{5}{36}$

  3. $\dfrac{7}{36},\dfrac{5}{72}$

  4. $\dfrac{5}{72},\dfrac{17}{72}$

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Correct Option: B

Sum of $4$ numbers in GP is $60$. And the AM of first and last no. is $18$ find the first term and common difference of the GP

maths average arithmetic mean of ap introduction to averages means

  1. $a=4, r=2$

  2. $a=32, r=\dfrac {1}{2}$

  3. $a=3, r=1$

  4. $a=6, r=3$

Show answer
Correct Option: A,B
Explanation:

Let 4 numbers in G.P. be $a,ar,{ ar }^{ 2 }{ ,ar }^{ 3 }\ a(1+r+{ r }^{ 2 }+{ r }^{ 3 })=60\ \cfrac { a+{ ar }^{ 3 } }{ 2 } =18=>a+{ ar }^{ 3 }=36\ a(1+r)(1+{ r }^{ 2 })=60\ a(1+r)(1+{ r }^{ 2 }-r)=36\ \cfrac { 1+{ r }^{ 2 }-r }{ 1+{ r }^{ 2 } } =\cfrac { 36 }{ 60 } \ 5+5{ r }^{ 2 }-5r=3+3{ r }^{ 2 }\ 2{ r }^{ 2 }-5r+2=0\ r=2,\cfrac { 1 }{ 2 } \ if\quad r=2,a(1+8)=36=>a=4\ if\quad r=\cfrac { 1 }{ 2 } ,a(1+\cfrac { 1 }{ 8 } )=36=>a=32\ a=4,r=2\quad (or)\quad a=32,r=\cfrac { 1 }{ 2 } $

The A.M. of the observations $1.3.5, 3.5.7, 5.7.9,...,(2n-1)(2n+1)(2n+3)$ is $(\forall n\in N)$

maths average arithmetic mean of ap introduction to averages means

  1. $2n^3+6n^2+7n-2$

  2. $n^3+8n^2+7n-2$

  3. $2n^3+5n^2+6n-1$

  4. $2n^3+8n^2+7n-2$

Show answer
Correct Option: A
Explanation:
$1.3.5, 3.5.7, 5.7.9, ........, (2n-1)(2n+1)(2n+3) $
 $x _1$       $x _2$       $x _3$                                   $x _n=n \,  terms$

$AM=\displaystyle\sum _{i=1}^{n}\dfrac{x _{i}}{n}$

$\displaystyle\sum _{i=1}^{n}x _{i}=\sum _{r=1}^{n}(2r-1)(2r+1)(2r+3)$

$=\displaystyle\sum _{r=1}^{n} 8r^{3}-2r+12r^{2}-3$

$=\displaystyle 8\sum _{r=1}^{n}+12\sum _{r=1}^{n}r^{2}-2\sum _{r=1}^{n}-3\sum _{r=1}^{n}(1) ....... (1)$

$1^{2}+2^{2}+.........+x^{2}=\dfrac{n(2n+1)(n+1)}{6}=\displaystyle\sum _{r=1}^{n}r^{2}$

$\displaystyle\sum _{r=1}^{n}r^{3}=1^{3}+2^{3}+......+ n^{3}=\left(\dfrac{n(n+1)}{2}\right)^{2}$

$\displaystyle\sum _{r=1}^{n}r=1+2+3+.......+ n=\dfrac{n(n+1)}{2}$

from $(1)$

$8\left[\dfrac{n(n+1)}{2}\right]^{2}+12\left[\dfrac{n(2n+1)(n+1)}{6}\right]-\dfrac{2n(n+1)}{2}-3n$

$=2(n(n+1))^{2}+2n(2n+1)(n+1)-n(n+1)-3n$

$=n(n+1)[2n^{2}+2n+4n+2-1]-3n$

$\displaystyle\sum =n(n+1)[2x^{2}+6x+1]-3n$

$A.M=\dfrac{n(n+1)(2n^{2}+6n+1)-3n}{n}$

$=(n+1)(2n^{2}6n+1)-3$

$=2n^{3}+6n^{2}+n-2n^{2}+6n+1-3=2n^{3}+8n^{2}+7n-2$

If $n\ AM's$ are inserted between $1$ and $31$ and ratio of ${7}^{th}$ and $(n-1)^{th}$ $A.M.$ is $5:9$ then $n$ equals ?

maths average arithmetic mean of ap introduction to averages means

  1. $12$

  2. $13$

  3. $14$

  4. $None$

Show answer
Correct Option: C