This can be easily proven. Assume x is initial number, y is the difference and there are n numbers after the initial number. For 1,3,5,7,9,11 x=1(initial) y=2(difference) and n=5 (numbers excluding initial) Above sequence can be written as x, x + 1*y, x + 2*y, x + 3*y, x + 4*y, x + 5*y ........... x + n*y Adding this, we get x + n*x + y(1+2+3+4+5....n) ==> x + n*x + y*n*(n+1)/2 ==> x(n+1) + yn(n+1)/2 There are n+1 numbers in the sequence ( first number and then n numbers after first number) Average = x(n+1)/(n+1) + ny(n+1)/2(n+1) ==> x + ny/2 ==> 2x/2 + ny/2 ==> (2x + ny)/2 ==> (x + x+ny)/2 Which is nothing but the short cut described in video.
This can be easily proven.
Assume x is initial number, y is the difference and there are n numbers after the initial number.
For 1,3,5,7,9,11
x=1(initial) y=2(difference) and n=5 (numbers excluding initial)
Above sequence can be written as
x, x + 1*y, x + 2*y, x + 3*y, x + 4*y, x + 5*y ........... x + n*y
Adding this, we get x + n*x + y(1+2+3+4+5....n)
==> x + n*x + y*n*(n+1)/2
==> x(n+1) + yn(n+1)/2
There are n+1 numbers in the sequence ( first number and then n numbers after first number)
Average = x(n+1)/(n+1) + ny(n+1)/2(n+1)
==> x + ny/2
==> 2x/2 + ny/2
==> (2x + ny)/2
==> (x + x+ny)/2
Which is nothing but the short cut described in video.
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