Question

Identify a relationship among Tn , Sn and Sn-1 where n>1

Answer 1

$\text{Consider the A.P}$

$\text{a, a+d, a+2d,...........}$

$\text{The n th term of the A.P is }$

$t_n=a+(n-1)d$

$\text{The sum of n terms of the A.P is }$

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

$\text{Now}$

$S_n-S_{n-1}$

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

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

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

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

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

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

$=a+(n-1)d$

$=t_n$

$\therefore\boxed{\bf\;S_n-S_{n-1}=t_n}$

$\text{which is the required relation}$