find the sum of all natural numbers from 1 to 150
Answers
Answered by
30
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
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
Answered by
17
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
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
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