Question

Find the sum of intergers which are divisible by 5 from 1 to 100.​

Answer 1

$<b>$

Ap:5,10,15....100

a=5

d=5

let the no. of terms be n

tn=a+(n-1)d=100

5+(n-1)5=100

n-1=95/5

n=19+1

n=$\huge\red{\boxed{\mathfrak20}}$

sum: n/2(2a+(n-1)d)

=10(10+95)

=$\huge\red{\boxed{\mathfrak{1050}}}$

Answer 2

Step-by-step explanation:

AP=5,10,15,20,25...100  ,Nth term =100 ,A=5 ,d=5

Nth term=a+(n-1)d

So,n=20

Sum of Nth term=20/2[5+100]

10×105=SUM

1050=SUM