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]
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]

Answer 2

Thank you for asking this question:

We know that Sn = n(n+1)/2

now it is given that

S2n = 3Sn

=> 2n(2n+1)/2 = 3n(n+1)/2

=> 2(2n+1) = 3(n+1)

=> 4n+2 = 3n+3

=> n = 1

then S3n/Sn = [3n(3n+1)/2]/[n(n+1)/2] = 3*4/1*2 = 6

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