Question

Show that in an A.P. the sum of the terms equidistant from the beginning and end is always same and equal to the sum of first and last terms.

Answer 1

`T_r=a+(r-1)d`
`r-1,x=n-(r-1)`
`x=n-r+1`
`T_(n-r+1)=a+[n-r+1-1]d=a+(n-r)d`
`T_(rt)T_(n-r+1)=T_1+T_n`
`a+(r-1)d+a+(n-r)d`
`2a+d[r-1+n-r]=2a+[n-1]d`
`T_(rt)+T_(n-r+1)=T_1+T_n=2a+(n-1)d`.