Question

Relationship between Tn, Sn and Sn-1 where n is greater than 1 and is a natural number

Answer 1

Therefore$S_n-S_{n-1}=T_n$

Step-by-step explanation:

$S_n$ is the sum of n terms which are in A.P.

$S_{n-1}$ is the sum of (n-1) terms which are in A.P.

$T_n$ is the nth term of the series

$T_{n-1}$ is the (n-1)th term.

A.P is a sequence of number.

a =the first term of a A.P

and d=common difference of a A.P

The sum of n terms which are in a.p is

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

The sum of (n-1) terms which are in a.p is

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

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

And   $T_n=a+(n-1)d$

Adding equation (1) and (2)

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

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

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

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

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

                $=a+(n-1)d$

                $=T_n$

Therefore$S_n-S_{n-1}=T_n$