Question

The mean of 12, 23, 34, 45, 56, 67, 78, 89, 90 is ?

Answer 1

Hint: Find the sum of the numbers starting from 12 and ending at 89 by using the formula of sum of n terms of an A.P given as : ${{S}_{n}}=\dfrac{n}{2}\left[ {{T}_{1}}+{{T}_{n}} \right]$. Here, ${{S}_{n}}$ is the sum of n terms, n is the total number of terms from 12 to 89, ${{T}_{1}}$ is the first term and ${{T}_{n}}$ is the ${{n}^{th}}$ term. Now, to find the mean, take the ratio $\left( \dfrac{{{S}_{n}}+90}{n+1} \right)$ and get the answer.

Complete step-by-step solution:
Here, we have been provided with the series: 12, 23, 34, 45, 56, 67, 78, 89, 90, and we have to find their mean. So, to find the mean we need to calculate the sum of the given series.
On observing the above series, we can see that the terms starting from 12 and ending at 89 are in A.P, having 12 as its first term and 89 as the last term. So, let us denote the first term with ${{T}_{1}}$ and the last term as ${{T}_{n}}$. Here, n is the number of terms from 12 to 89.
$\Rightarrow {{T}_{1}}=12,{{T}_{n}}=89$
On counting, we get that there are 8 terms from 12 to 89. Therefore, applying the formula for sum of n terms of an A.P, we get,
${{S}_{n}}=\dfrac{n}{2}\left[ {{T}_{1}}+{{T}_{n}} \right]$
Substituting the values of n, ${{T}_{1}}$ and ${{T}_{n}}$, we get,
$\begin{align}
  & \Rightarrow {{S}_{8}}=\dfrac{8}{2}\left( 12+89 \right) \\
 & \Rightarrow {{S}_{8}}=4\times 101 \\
 & \Rightarrow {{S}_{8}}=404 \\
\end{align}$
Now, we have an extra member 90, after these 8 terms, so we must consider it in the sum for the calculation of mean. Therefore, applying the formula to find mean, we get,
$\text{Mean=}\dfrac{\text{Sum of all the terms}}{\text{Total number of terms}}$
Here, sum of all the terms = ${{S}_{8}}+90=404+90=494$
And total number of terms = $8+1=9$
$\begin{align}
  & \Rightarrow \text{Mean}=\dfrac{494}{9} \\
 & \Rightarrow \text{Mean}=54.89 \\
\end{align}$
Hence, the mean of the series is 54.89.

Note: One may note that we have found the number of terms (n) from 12 to 89 by simple counting because we see that there are not many terms which will trouble us while counting. One can also apply the formula for the ${{n}^{th}}$ term to find n. The formula is given as ${{T}_{n}}=a+\left( n-1 \right)d$, where a is the first term and d is a common difference.