Question

An Arithmetic Series is defined as
$f\left(n\right)=a+\left(n-1\right)d,n\in N,$ prove that A.M. of $f\left(1\right)$ and $f\left(2n-1\right)$ is $f\left(n\right)$.

Answer 1

Solution:

Given $f\left(n\right)=a+(n-1)d,n\in N$

then $f\left(1\right)=a$

$f(2n-1)=a+(2n-2)d=a+2(n-1)d$

Now $f\left(1\right)+f(2n-1)=a+a+(2n-2)d$

$=2[a+(n-1\left)d\right]=2f\left(n\right)$

$\Rightarrow A.M$ of $f\left(1\right)$ and $f(2n-1)$ is $f\left(n\right)$.

Hence Proved