Question

Derive the given formula of sum of A.P:- Sn=n/2[2a+(n-1)d].

Answer 1

S=a1+a2+a3+a4+...+an
S=a1+a2+a3+a4+...+an

S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d]S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d]  →   Equation (1)
 

S=an+an−1+an−2+an−3+...+a1S=an+an−1+an−2+an−3+...+a1

S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d]S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d]  →   Equation (2)
 

Add Equations (1) and (2)
2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)

2S=n(a1+an)2S=n(a1+an)

S=n2(a1+an)S=n2(a1+an)

 

Substitute an = a1 + (n - 1)d to the above equation, we have
S=n2{a1+[a1+(n−1)d]}S=n2{a1+[a1+(n−1)d]}

S=n2[2a1+(n−1)d]S=n2[2a1+(n−1)d]

 

Answer 2

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+l

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

2*Sn = (a+l)*n

Sn = n/2*(a+l)

but l = a+(n-1)d, so substituting this gives:

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