Question

What is the average of integers from $25$ to $41$ ?

Answer 1

Hint: The given problem requires us to find the sum of a series of some integers. The first term and the last term of the series is given to us in the question. We should know the formula for finding the average of given numbers as ${\text{Average = }}\dfrac{{{\text{Sum of numbers}}}}{{{\text{Total number}}}}$. We know that the consecutive integers form an arithmetic progression. So, to find the sum of all the numbers, we need to know the formula for the sum of $n$ terms of arithmetic progression. We can find out the common difference of an arithmetic progression by knowing the difference of any two consecutive terms of the series.

Complete step by step answer:
So, we have to find the average of all the integers from $25$ to $41$. The difference of any two consecutive integers is constant. So, the given sequence of integers from $25$ to $41$ is an arithmetic progression. Now, we have to find the sum of this arithmetic progression.
Here, first term $ = a = 25$.
Last term $ = l = 41$
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio \[ = d = 26 - 25 = 1\]
So, $d = 1$

We also need to know the number of terms in the arithmetic progression in order to find the value of the sum of arithmetic progression. We know that the formula for the nth terms of an AP is ${a_n} = a + \left( {n - 1} \right)d$.So, considering $41$ as the nth term of the AP, we can find the number of terms in the AP and then evaluate the sum of the arithmetic progression.
So, ${a_n} = a + \left( {n - 1} \right)d = 41$
Substituting the values of a and d in the above equation, we get,
$ \Rightarrow 25 + \left( {n - 1} \right)1 = 41$
Now, solving the above equation for the value of n, we get,
$ \Rightarrow \left( {n - 1} \right) = 41 - 25$
$ \Rightarrow n = 17$
So, there are $17$ integers from $25$ to $41$.

Now, we can find the sum of the given arithmetic progression using the formula,
$S = \dfrac{n}{2}\left[ {a + l} \right]$
Hence, the sum of integers from $25$ to $41$ $ = S = \dfrac{{17}}{2}\left[ {25 + 41} \right]$
Opening the brackets and cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow S = \dfrac{{17}}{2} \times 66$
Simplifying the expression, we get,
$ \Rightarrow S = 17 \times 33$
$ \Rightarrow S = 561$
So, the sum of the integers from $25$ to $41$ is $561$.
Now, we can calculate the average of the integers from $25$ to $41$ as $\left( {\dfrac{{{\text{Sum of numbers}}}}{{{\text{Total number}}}}} \right)$.
Now, substituting the values of total number and sum of numbers, we get,
$ \Rightarrow {\text{Average}}\,{\text{ = }}\dfrac{{561}}{{17}}$
$ \therefore {\text{Average}}\,{\text{ = 33}}$

Hence, the average of integers from $25$ to $41$ is 33.

Note: We must know the formula for mean and average of some given numbers to solve the problem. Arithmetic progression is a series where any two consecutive terms have the same difference between them. The common difference of an arithmetic series can be calculated by subtraction of any two consecutive terms of the series. The sum of n terms of an arithmetic progression can be calculated if we know the first term, the number of terms and difference of the arithmetic series as: $S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.