Question

Find the sum of all integers lying between 0 and 50

Answer 1

a = 1

d = 1

L = 49

=> a + (n-1) d = 49

=> 1 + (n-1) 1 = 49

=> n-1 = 48

=> n = 49

Now,

Sn = n/2 [ a +L]

S49 = 49 /2 [ 1+ 49]

= 49 /2 × 50

= 49 × 25

= 1225

Answer 2

Now the required A. P. is
0,1,2,...50
a=0,n=50,d=1
$s(n) = \frac{n}{2}(2a + (n - 1)d) \\ \: \: \: = \frac{50}{2}(2(0) + (50 - 1)1) \\ = 25(0 + 49) = 25(49) \\ = 1225$
Therefore the sum of integers between 0 to 50 is 1225.