Question

The mean of the observations 425, 430, 435, ..., 495 is ___________.

Answer 1

The observations 425, 430, 435, ..., 495 are in AP.

Here, a = 425 and d = 5.

Let the number of observation be n.

$$\begin{gathered} \therefore 495=425+\left(n-1\right)\times 5\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 5\left(n-1\right)=495-425=70 \\ \Rightarrow n-1=14 \\ \Rightarrow n=15 \end{gathered}$$

Sum of the given observations $=\frac{15}{2}\left(425+495\right)=\frac{15}{2}\times 920=15\times 460\left[S_n=\frac{n}{2}\left(a+l\right)\right]$

∴ Mean of the given observations

$$\begin{gathered} =\frac{\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{observations}}{\mathrm{Number}\mathrm{of}\mathrm{observations}} \\ =\frac{15\times 460}{15} \\ =460 \end{gathered}$$

The mean of the observations 425, 430, 435, ..., 495 is _____460______.