Question

If sn is the sum n term of an ap then show that d=sn-2sn-1-sn-2

Answer 1

I think question will be

PT d= Sn -2Sn-1 + Sn-2.
Then
d= (Sn - Sn-1)-(Sn-1 - Sn-2)
d= Tn-Tn-1= d.

Answer 2

Answer:

Given : a is first term and d be the common difference.

• Sn = (n/2)[ 2a + ( n -1) d]

Now ;

↠ Sn - 2Sn-1 + Sn + 2

↠ (n/2)[ 2a + ( n -1) d] - 2(n - 1)/2)[ 2a + ( n - 1 -1) d] + (n +2) / 2)[ 2a + ( n + 2 -1) d]

↠ (1/2)[ 2an + n( n -1) d] + [ 4a(n - 1) + 2(n - 1)( n - 2) d] +[ 2a(n + 2)+ ( n + 1) (n + 2) d]

↠ (1/2)[ 2a[ n - 2n + 2 + n + 2] + d [ n2 - n - 2n2 + 6n - 4 + n2 + 3n + 2] ]

↠ (1/2)[ 2a(4) + d(8n - 2) ]

↠ 4a + (4n - 1)d