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
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let an AP listen step by step on Google
Step-by-step explanation:
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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)
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