Question

let an AP a,a+d,a+2d....l has n term then Sn n/2 [2a+(n-1)d] can also be written as​

Answer 1

Answer:

let an AP listen step by step on Google

Step-by-step explanation:

OK

Answer 2

Step-by-step explanation:Step-by-step explanation:

Let Sn = Sum of 'n' terms, then,

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

Let l = last term, then,Sn = a + (a+d) + (a+2d) + (a+3d) + ...+ (l-3d) + (l -2d) + (l - d) + l

reversing the order of this gives:Sn = l + (l-d) + (l-2d) + (l-3d) + ...+ (a+3d) + (a +2d) + (a + d) + a

Adding these last two equations gives,

2*Sn = a+l + (a+l) + (a+l) + (a+l) + ...+ (a+l) + (a+l) + (a+l) + a+l2*Sn = (a+l) to n terms

2*Sn = (a+l)*nSn = n/2*(a+l)but l = a+(n-1)d,

so substituting this gives:

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