Question

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

Answer 1

$\text{Consider the A.P}$

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

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

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

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

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

$\text{Now}$

$S_n-S_{n-1}$

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

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

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

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

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

$=a+(n-1)d$

$=t_n$

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

$\text{which is the required relation}$

Answer 2

Answer:

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

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

t_n=a+(n-1)dtn=a+(n−1)d

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

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

\text{Now}Now

S_n-S_{n-1}Sn−Sn−1

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

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

=\displaystyle\frac{2an+n(n-1)d-2an+2a-n(n-1)d+2(n-1)d}{2}=22an+n(n−1)d−2an+2a−n(n−1)d+2(n−1)d

=\displaystyle\frac{2a+2(n-1)d}{2}=22a+2(n−1)d

=\displaystyle\frac{2[a+(n-1)d]}{2}=22[a+(n−1)d]

=a+(n-1)d=a+(n−1)d

=t_n=tn

\therefore\boxed{\bf\;S_n-S_{n-1}=t_n}∴Sn−Sn−1=tn

\text{which is the required relation}which is the required relation