Question

Prove that if the number of terms of an A. P. is odd then the middle term is the A. M. between the first and last terms.

Answer 1

let, there are n terms in the AP where n is a odd number

then , the first term will be

a₁ = a + (1-1) d = a       -------eqn(1)

and , last term will be

aₙ = a + ( n - 1) d     -------eqn(2)

also,middle term will be

$a_{(n+1)/2} = a + [ (\frac{n+1}{2} - 1 ] d\\\\a_{(n+1)/2} = a + (\frac{n+1-2}{2} )d\\\\a_{(n+1)/2} = a + \frac{(n-1)d}{2}$  

WE have to prove that,

$\frac{a_{1}+a_{n} }{2} = a_{(n+1)/2}$

putting eqn(1) and eqn (2) in LHS

$LHS = \frac{a + a + ( n-1)d}{2} \\\\ = \frac{2a + (n-1)d}{2} = a + \frac{(n-1)d}{2}$

we already have ,

$RHS = a+ \frac{(n-1)d}{2}$

so, LHS = RHS

hence, PROVED.