Tag: arithmetic mean of ap

Questions Related to arithmetic mean of ap

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

maths average arithmetic mean of ap introduction to averages means

  1. $n$

  2. $2n$

  3. $n/2$

  4. $n - 1$

Show answer
Correct Option: A
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
$= n^2/n=n$

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

maths average arithmetic mean of ap introduction to averages means

  1. $n$

  2. $2n$

  3. $\dfrac n2$

  4. $n - 1$

Show answer
Correct Option: A
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 = $\dfrac{n^2}n = n$

What is the average of the first $300$ terms of the given sequence?
$1, -2, 3, -4, 5, -6, ....., n.(-1)^{n + 1}$

maths average arithmetic mean of ap introduction to averages means

  1. $-1$

  2. $0.5$

  3. $0$

  4. $-0.5$

Show answer
Correct Option: D
Explanation:
Avg $=\cfrac{[1+3+5+7+...+(2n-1)]-[2+4+6+...+2n]}{n}$
Here $n=300$
No. of even terms $=150$
No. of odd terms $=150$
Now,
$1+3+5+7+...+150\;terms \\ S _{n}=\cfrac{n}{2}[2a+(n-1)d] \\ S _{n1}=\cfrac{150}{2} [2\times 1 +149\times 2]=22500$
and $2+4+6+8+....+150\;terms \\ S _{n}=\cfrac{n}{2}[2a+(n-1)d] \\ S _{n2}=\cfrac{150}{2}[2\times 2+149\times 2]=22650$
Average $=\cfrac{S _{n1}-S _{n2}}{300}=\cfrac{22500-22650}{300} \\ =-0.5$

The A.M. of 'n' observations is M. If the sum of $(n - 4)$ observation is 'a', what is the mean of remaining $4$ observations?

maths average arithmetic mean of ap introduction to averages means

  1. $nM + a$

  2. $\dfrac {nM - a}{2}$

  3. $\dfrac {nM + a}{2}$

  4. $\dfrac {nM - a}{4}$

Show answer
Correct Option: D
Explanation:
Given $\cfrac { { a } _{ 1 }+{ a } _{ 2 }+{ a } _{ 3 }+{ a } _{ 4 }+....+{ a } _{ n } }{ n } =M\quad \quad \quad (1)\\ \left( { a } _{ 1 }+{ a } _{ 2 }+{ a } _{ 3 }+{ a } _{ 4 }+....+{ a } _{ n-4 } \right) =a\quad \quad \quad (2)$

From $(1)$ and $(2)$
$a+{ a } _{ n-3 }+{ a } _{ n-2 }+{ a } _{ n-1 }+{ a } _{ n }=nM\\ { a } _{ n-3 }+{ a } _{ n-2 }+{ a } _{ n-1 }+{ a } _{ n }=(nM-a)\\ Mean=\cfrac { { a } _{ n-3 }+{ a } _{ n-2 }+{ a } _{ n-1 }+{ a } _{ n } }{ 4 } =\cfrac { nM-a }{ 4 } \\ Mean=\cfrac { (nM-a) }{ 4 } $

The mean  of five numbers in AP is $89$. The product of first and last terms is $7885$. The AM of first, third and fifth term is

maths average arithmetic mean of ap introduction to averages means

  1. $83$

  2. $86$

  3. $89$

  4. $90$

Show answer
Correct Option: C
Explanation:
Let the numbers in A.P. be  (a-2d), (a-d), a, (a+d), (a+2d)
$\therefore \dfrac{(a-2d)+(a-d)+a+(a+d)+(a+2d)}{5}=89$
$\Rightarrow \dfrac{5a}{5}=89$
$\therefore a=89$
$\therefore$ The AM of 1st, 3rd & 5th term is:
$=\dfrac{(a-2d)+a+(a+2d)}{3}$
$=a=89$                           

The sum of four numbers in AP is $176$. The product of 1st and last is $1855$.  The mean of middle two is 

maths average arithmetic mean of ap introduction to averages means

  1. $42$

  2. $41$

  3. $44$

  4. $53$

Show answer
Correct Option: C
Explanation:
Let the numbers in A.P. be  (a-3d), (a-d), (a+d), (a+3d)
$\therefore (a-3d)+(a-d)+(a+d)+(a+3d)=176$
$\Rightarrow 4a=176$
$\therefore a=44$
$\therefore$ The mean of middle two is:
$=\dfrac{(a-d)+(a+d)}{2}$
$=a=44$                           

The arithmetic mean of 1, 8, 27, 64, ....... up to n terms is given by

maths average arithmetic mean of ap introduction to averages means

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

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

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

  4. $\dfrac{n^2(n+1)^2}{4}$

Show answer
Correct Option: C
Explanation:

Solution:

Given:
$1,8,27,64,...........$ upto $n$ terms
or,$1,2^3,3^3,4^3,..............$ upto $n$ terms
$\therefore AM=\cfrac{1+2^3+3^3+4^3............+n^3}{n}=\cfrac{\left[\cfrac{n(n+1)}2\right]^2}{n}$
$=\cfrac{n^2(n+1)^2}{4n}=\cfrac{n(n+1)^2}4$
Hence, C is the correct option.

Say true or false.
A.M. of any $n$ positive numbers $a _1, a _2, a _3, ...., a _n$ is $A$ then $A=\dfrac {a _1+a _2+....+a _n}{n}$

maths average arithmetic mean of ap introduction to averages means

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Arithmetic mean is the average of all the numbers.
Hence for n positive numbers:
$a _1$,$a _2$,$a _3$,....$a _n$, the arithmetic mean is $\dfrac{a _1+a _2+a _3+...+a _n}{n}$.

Say true or false.

The A.M. between $(a-b)^2$ and $(a+b)^2$  is $a^2+b^2 $.

maths average arithmetic mean of ap introduction to averages means

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

If the two terms are $x$ and $y$, then their AM is $\dfrac{x+y}{2}$
AM between $(a-b)^2$ and $(a+b)^2$
$=$ $\dfrac{(a-b)^2 + (a+b)^2}{2}$
$=$ $\dfrac{(a^2 +b^2 - 2ab + a^2+b^2 + 2ab)}{2}$
$=$ $a^2 + b^2$

If the arithmetic mean of n numbers of a series is $\overline{x}$ and the sum of the first (n-1) numbers is k, then the nth number is

maths average arithmetic mean of ap introduction to averages means

  1. n+k

  2. $n\overline{x}+k$

  3. $n\overline{x}-k$

  4. n-k

Show answer
Correct Option: C
Explanation:

$Sum$ $of$ $all$ $the$ $n$ $numbers$ $= Mean \times no. of numbers = n \times \overline x$
$Sum$ $of$ $n-1$ $numbers$ $= k$
$Thus$ $nth$ $number$ $= n \times \overline x - k$