Question

find the sum of the AP：1+2+3+4..........+99+100​

Answer 1

Answer:

5050

Step-by-step explanation:

Given,

a=1

a(n)=100

d=1

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

100=1+(n-1)1

99=n-1

[n=100]

S(n)=(n/2)[a+a(n)]

S(100)=100/2 × 1+100

S(100)=(100×101)/2

S(100)=50×101

S(100)=5050