Mean of given Arithmetic function

Question

Find the mean of $a, a+d, a+2d, a+3d,\dots,a+nd$

I have no idea what to do in this question but i have tried the following: $$mean\ \bar{x}= \frac{(a)+(a+d)+(a+2d)+(a+3d)+\cdots+(a+nd)}{n+1} $$

This is obvious step we do while dealing with mean.Now i don't know what do but i think we can take $a$ common and can do something ?

Answer 1

there are n+1 times a and take d common from left expression and you get 0+1+2+...+n-1+n whose sum is given by sum of arithmetic progression n(n+1)/2 therefore you can write numerator a(n+1) + d(n)(n+1)/2 so your mean equals a+dn/2

Answer 2

Now distribute to make the numerator $xa+yd$. How many copies of $a$ do you have? What is the total of the coefficients of $d$?