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Proof for Sn=n/2(a+L)

Hi!
I'm aware of the proof for Sn=n/2[2a + (n-1)d] , but I'd like to know how you would go about proving the second formula for sum of n terms using the last term:

Sn= n/2 (a + L)

Thanks

Reply 1

Christophicus

abbasi

Hi!
I'm aware of the proof for Sn=n/2[2a + (n-1)d] , but I'd like to know how you would go about proving the second formula for sum of n terms using the last term:

Sn= n/2 (a + L)

Thanks

same proof, just say that L = a + (n-1)d
Sn = 0.5n(2a + (n-1)d)
Sn = 0.5n(a + a + (n-1)d)
Sn = 0.5n(a + L)

Reply 2

chats

Sn = a+ (a+d) + (a+2d) +.........+(L-d)+L

Sn= L+(L-d)+(L-2d)+..........+(a+d)+a

2Sn= (a+L) + (a+L) + (a+L)

2Sn= n(a+L)

Sn= n/2 (a+L)

Reply 3

Malik

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

edit: damn you guys just to fast!