Question

find the sum of all natural numbers from 1 to 150

Answer 1

first term (a)=1
last term (t n)=150
difference(d)=1
sum(sn)=?
now
tn = a+ (n-1)*d
150=1+(n-1)*1
149=n-1
n=150
so
sn= n/2[2*a+(n-1)*d]
= 150/2[2*1+(150-1)*1]
= 75*(2+149)
= 75*151
= 11325

Answer 2

Hi !

Natural no:s = 1,2,3,4 ..... ,150

First term = a = 1
Common difference = d = 1
Last term = l = 150

No: of terms = n

l = a + (n - 1)d
150 = 1 + (n - 1)1
149 = n - 1
n = 150

No: of terms = n = 150

Sum of natural no:s = Sn

Sn  = n/2 [ a + l ]
      = 150/2 [ 1 + 150 ]
      = 75 x 151
      = 11325

Therefore ,
sum of all natural numbers from 1 to 150 = 11325