Arithmetic Progressions Test -12

formula is a(n) = S(n) - S(n-1) 
In general, S(n) = a(1) + a(2) + ........ a(n-1) + a(n) 
In general S(n-1) = a(1) + a(2) + ........a (n-1) 
If you subtract one from the other, you are left with a(n) because all of the A's from a(0) to a(n-1) cancel out. 
a(1) - a(1) = 0
a(2) - a(2) = 0
a(n-1) - a(n-1) = 0
a(n) =0
= a(n)

‘n’ positive integers are 1, 2, 3, 4, ………………., n
Here a = 1,d = 2 − 1 = 1 and n = n

S_n=n/2(2a + (n - 1)d)

S_n=n/2(2a + nd - d)

= n/2(1 + n)
= n(n+1)/2

let a = 1st term , d = common difference  ,

S_n = sum of 1st n terms of an AP

then  S_n = (a) + (a+d) + (a+2d) + ................................{a + (n - 3) d} + { a + (n-2)d} + {a + (n-1)d} .......... (i)

Now Rewrite S_n as follows

S_n = {a +( n-1)d} + {a +( n-2)d }+ {a + (n - 3)d} ..........................(a + 3d) + (a + 2d) + (a + d) + a ............. (ii)

adding the terms  i  and  ii  vertically

adding 1st term of both  we get (a) + {a + (n-1)d}  = 2a + (n-1)d

adding 2nd term of both (a + d) + {a + (n - 2)d} = 2a + (n-1)d

adding 3rd terms of both (a + 2d) + {a + (n-3)d} = 2a + (n-1)d  

since there are n terms in each of the equations i and ii , adding both the equations we get 
2S_n = n{2a + (n-1)d}

S_n = n/2{2a + (n-1)d}.