Question

If sn denotes the sum of n terms of an A.P whose common difference is d and first term is a find sn-2sn-1+sn+2 ??

Answer 1

Sn = (n/2)[ 2a + ( n -1) d]
then 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]

Answer 2

Answer:

Step-by-step explanation:

Tn = Sn - Sn-1

Tn-1 =S n-1 - Sn-2

Sn - 2Sn-1 + Sn-2= Sn - Sn-1 - Sn-1 + Sn-2

= (Sn- Sn-1 ) -(Sn-1 - Sn-2)

= Tn - Tn-1 = d