Arithmetic Progression

Asked by  | 19th Sep, 2008, 02:33: PM

Expert Answer:

Sn = n/2 ( 2a+(n-1)d)

Sn-1 = (n-1)/2 ( 2a+(n-2)d)

Sn-2 = (n-2)/2 ( 2a+(n-3)d)

so Sn + Sn-2= n/2 (4a+(2n-4)d ) - ( 2a+(n-3)d) = (n-1+1) (2a+(n-2)d ) - ( 2a+(n-3)d )

= (n-1)( 2a+(n-2)d) + (2a+(n-2)d ) - ( 2a+(n-3)d )

=2Sn-1+d

so k=2

Answered by  | 19th Sep, 2008, 10:20: PM

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