Question

in an AP prove that d=Sn-2Sn-1+Sn-2

Answer 1

For an Arithmetic Progression,

S n = 2n

.[2a+(n−1)d]S n−1 = 2n−1

.[2a+(n−2)d] S n−2 = 2n−2

[2a+(n−3)d]

Now, let's take the RHS :

RHS= S n −2S n−1 +S n−2 =an+ 2n(n−1)

.d−2a(n−1)−(n−1)(n−2)d+a(n−2)+ 2(n−2)(n−3)

.d

=a[n−2(n−1)+n−2]+d[

2

n(n−1)

−(n−1)(n−2)+

2

(n−2)(n−3) ]=0+ 2d [(n2 −n)−2(n 2 −3n+2)+(n −5n+6)]

= 2d

(−4+6)= 2d ×2=d =LHS

LHS=RHS

Thus it is proved.

I hope my answer will help you .